Contoh Soal 1 Statistika

$\begin{aligned}&\textrm{Perhatikanlah tabel berikut}\\ &\textrm{Nilai ulangan kelas XII}\\ &\textrm{untuk menjawab soal no. 1 sampai 5}\\ &\begin{array}{|c|c|c|}\hline \colorbox{magenta}{Nilai}&\colorbox{yellow}{Matematika}&\colorbox{cyan}{Bahasa Inggris}\\\hline 30-39&1&0\\\hline 40-49&4&2\\\hline 50-59&6&7\\\hline 60-69&17&18\\\hline 70-79&10&12\\\hline 80-89&7&6\\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ 1.&\textrm{Total datum pada data tabel}\\ &\textrm{distribusi frekuensi di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&30\\ \textrm{b}.&35\\ \textrm{c}.&40\\ \color{red}\textrm{d}.&45\\ \textrm{e}.&50 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}&\textrm{Total datum pada tabel di atas}\\ &\textrm{sama dengan total frekuensi yaitu}: \\ &=1+4+6+17+10+7=\color{black}45,\\ &\color{red}\textrm{atau}\\ &=0+2+7+18+12+6=\color{black}45 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Banyak kelas interval pada tabel}\\ &\textrm{distribusi frekuensi tersebut adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4\\ \textrm{b}.&5\\ \color{red}\textrm{c}.&6\\ \textrm{d}.&7\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\textrm{Data di atas terbagai dalam 6}\\ &\textrm{kelas intervalnya, yaitu}:\\ &\textrm{kelas pertama : 30-39}\\ &\textrm{kelas kedua : 40-49}\\ &\textrm{kelas ketiga : 50-59}\\ &\textrm{kelas keempat : 60-69}\\ &\textrm{kelas kelima : 70-79}\\ &\textrm{kelas keenam : 80-89}\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Panjang kelas interval pada tabel}\\ &\textrm{di atas adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&9\\ \color{red}\textrm{b}.&10\\ \textrm{c}.&11\\ \textrm{d}.&12\\ \textrm{e}.&13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Panjnag kelas interval pada}\\ &\textrm{tabel distribusi frekuensi di atas} \\ &\textrm{ambil contoh kelas pertama yaitu}:\\ &\textrm{pada}\: \: 30-39\: \: \textrm{ada}\\ &\color{red}=(39-30)+1=10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Titik tengah dari kelas interval ke enam}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&84\\ \color{red}\textrm{b}.&84,5\\ \textrm{c}.&85\\ \textrm{d}.&85,5\\ \textrm{e}.&86 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}&\textrm{Titik tengah interval kelas keenam}\\ &\textrm{pada tabel distribusi frekuensi di atas} \\ &\textrm{yaitu}:\\ &\color{red}=\displaystyle \frac{1}{2}(80+89)=\frac{169}{2}=84,5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Tepi bawah dan tepi atas dari kelas}\\ &\textrm{interval dari tabel distribusi frekuensi}\\ &\textrm{di atas yang tepat adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&30,5\: \: \textrm{dan}\: \: 39,5\\ \color{red}\textrm{b}.&39,5\: \: \textrm{dan}\: \: 49,5\\ \textrm{c}.&50,5\: \: \textrm{dan}\: \: 59,5\\ \textrm{d}.&60,5\: \: \textrm{dan}\: \: 70,5\\ \textrm{e}.&79,05\: \: \textrm{dan}\: \: 89,05\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{purple}\begin{aligned}&\textrm{Tepi bawah dan tepi atas dari kelas}\\ &\textrm{interval pada tabel distribusi frekuensi}\\ &\textrm{di atas yaitu}:\\ &\begin{cases} \textrm{tepi bawah} & =x_{i}-0,5 \\ \textrm{tepi atas} & =x_{i}+0,5 \end{cases}\\ &\textrm{Berikut tabelnya}\\ &\begin{aligned} &\color{blue}\begin{array}{|c|c|c|}\hline \textrm{Nilai}&\textrm{tepi bawah}&\textrm{tepi atas}\\\hline 30-39&30-0,5=29,5&39+0,5=39,5\\\hline \color{red}40-49,5&\color{red}40-0,5=39,5&\color{red}49+0,5=49,5\\\hline 50-59&.....=49,5&....=59,5\\\hline 60-69&....=59,5&....=69,5\\\hline 70-79&....=69,5&....=79,5\\\hline 80-89&....=79,5&\: \, ....=89,5\\\hline \end{array} \end{aligned} \end{aligned} \end{array}$

Contoh Soal 4 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{ll}\\ 16.&\textrm{Diketahui suatu fungsi kuadrat}\\ &f(x)=ax^{2}+bx+c.\: \: \textrm{Jika fungsi}\\ &(-1,0),(1,4),\: \textrm{dan}\: \: (2,9),\: \: \textrm{maka}\\ &\textrm{fungsi yang dimaksud adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle f(x)=x^{2}-2x+3\\ \textrm{b}.&f(x)=x^{2}+2x+3\\ \textrm{c}.&f(x)=x^{2}+2x-3\\ \textrm{d}.&f(x)=x^{2}-2x-3\\ \color{red}\textrm{e}.&f(x)=x^{2}+2x+1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} (-1,0)\Rightarrow f(-1)=a-b+c=0\: ....\color{red}(1)\\ (1,4)\Rightarrow f(1)=a+b+c=4\: ....\color{red}(2)\\ (2,9)\Rightarrow f(2)=4a+2b+c=9\: ....\color{red}(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)\&(2),\: \textrm{didapatkan}\\ &b=2\: \: ...............\color{blue}(4)\\ &\textrm{Saat}\: \: (1)\&(3),\: \textrm{didapatkan}\\ &\color{blue}\begin{array}{llll}\\ 4a+2b+c&=9&\\ \: \: \: \: a-b+c&=0&-\\\hline \quad\qquad \qquad 3a+3b&=9&\\ \: \: \: \quad\qquad \qquad a+b&=3&...(5) \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{cases} a &=1 \\ c & =1 \end{cases}\\ &\color{blue}\textrm{Jadi},\: \: f(x)=ax^{2}+bx+c=x^{2}+2x+1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Diketahui persamaan}\begin{cases} x-y & =2 \\ kx+y & =3 \end{cases}\\ &\textrm{memiliki solusi}\: \: (x,y)\: \: \textrm{di kuadran I}\\ &\textrm{Jika dan hanya jika nilai}\: \: k\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle k=-1\\ \textrm{b}.&k>-1\\ \textrm{c}.&k<\displaystyle \frac{3}{2}\\ \textrm{d}.&0<k<\displaystyle \frac{3}{2}\\ \color{red}\textrm{e}.&-1<k<\displaystyle \frac{3}{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x-y=2\: \: \: \quad....(1)\\ kx+y=3\quad\: ....(2)\end{matrix}\right.\\ &\textrm{Dengan metode matriks didapatkan}\\ &\color{blue}x=\displaystyle \frac{\begin{vmatrix} 2 & -1\\ 3& 1 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{2-(-3)}{1+k}=\frac{5}{k+1}\\ &\textrm{Dengan cara yang sama pula}\\ &\color{blue}y=\displaystyle \frac{\begin{vmatrix} 1 & 2\\ k & 3 \end{vmatrix}}{\begin{vmatrix} 1 & -1\\ k & 1 \end{vmatrix}}=\displaystyle \frac{3-2k}{k+1}\\ &\textrm{Supaya memiliki solusi di kwadran I},\\ &\textrm{maka baik}\: \: x\: \: \textrm{maupun}\: \: y\\ &\textrm{haruslah positif, akibatnya}:\\ &\color{red} k+1>0\Rightarrow k>-1\\ &\textrm{Sebagai akibat yang lain adalah}:\\ &3-2k>0\Rightarrow k<\displaystyle \frac{3}{2}\\ &\color{blue}\textrm{Jadi},\: \: -1<k<\displaystyle \frac{3}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Diketahui sistem persamaan}\\ &y+\displaystyle \frac{2}{x+z}=4\\ &5y+\displaystyle \frac{18}{2x+y+z}=18\\ &\displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\\ &\textrm{Nilai}\: \: y+\sqrt{x^{2}-2xz+y^{2}}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 3\\ \textrm{b}.&5\\ \textrm{c}.&7\\ \textrm{d}.&9\\ \textrm{e}.&11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} y+\displaystyle \frac{2}{x+z}=4\qquad\quad\\ 5y+\displaystyle \frac{18}{2x+y+z}=18\\ \displaystyle \frac{8}{x+z}-\frac{6}{2x+y+z}=3\end{matrix}\right.\\ &\textrm{Jika disederhanakan beberapa bagian}\\ &\begin{cases} y+2A & =4\: ....(1) \\ 5y+18B & =18\: ....(2) \\ 8A-6B & =3\: ....(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2)\&(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llllll}\\ y+2A&=4&\left | \times 5 \right |&5y+10A=20\\ 5y+3(8A-3)&=18&\left | \times 1 \right |&5y+24A=27&-\\\hline &&&\: \: \quad-14A=-7\\ &&&\: \: \: \: \: \: \: \qquad A=\displaystyle \frac{1}{2}...(4)\\ \textrm{maka}\: B=\displaystyle \frac{1}{6}\: \& &y=3&&\\ \textrm{akibatnya}\\ \begin{cases} x &=1 \\ z &=1 \end{cases} \end{array} \\ &\color{blue}\textrm{Jadi},\: \: y+\sqrt{x^{2}-2xz+z^{2}}=3+0=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Diberikan}\: \: a,b,\: \textrm{dan}\: \: c \: \: \textrm{adalah angka-angka}\\ &\textrm{dari bilangan 3 digit yang memenuhi}\\ &49a+7b+c=286.\: \: \textrm{Nilai dari}\: \: a+b+c\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&16\\ \textrm{b}.&17\\ \textrm{c}.&18\\ \textrm{d}.&19\\ \textrm{e}.&20 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{blue}49a+7b+c=286\\ &\textrm{Nilai maksimum}\: \: a\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}49\times 5=245,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}245+7b+c=286\Rightarrow 7b+c=286-245=41\\ &\textrm{Nilai maksimum}\: \: b\: \textrm{adalah}\: \: \color{blue}5\\ &\color{red}7\times 5=35,\: \: \color{black}\textrm{akibatnya}:\\ &\color{blue}35+c=41\Rightarrow c=41-35=6\\ &\color{black}\textrm{Sehingga}\: \: \color{blue}a,b,\: \: \color{black}\textrm{dan}\: \: \color{blue}c\: \: \color{black}\textrm{adalah}\: \: \color{blue}5,5,\: \: \color{black}\textrm{dan}\: \: \color{blue}6\\ &\textrm{Jadi},\: \textrm{nilai}\: \: \color{red}a+b+c=5+5+6=16 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahui sistem persamaan}\\ &(2x+3y)^{.^{\log (x-y+2z)}}=1\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &5x+3y+8z=2\\ &\textrm{Himpunan penyelesaian yang}\\ &\textrm{memenuhi adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ \displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \}\\ \textrm{b}.&\left \{ -\displaystyle \frac{17}{12},\frac{1}{2},\frac{7}{6} \right \}\\ \textrm{c}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{2},-\frac{7}{6} \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{17}{12},\frac{1}{12},\frac{7}{6} \right \}\\ \color{red}\textrm{e}.&\left \{ -\displaystyle \frac{17}{12},-\frac{1}{12},\frac{7}{6} \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\color{blue}\textrm{Untuk persamaan}\: \: (1)\\ &(2x+3y)^{.^{\log (x-y+2z)}}=(2x+3y)^{0}\\ &\Leftrightarrow (x-y+2z)=10^{0}=1\\ &\color{blue}\textrm{Untuk persamaan}\: \: (2)\\ &3^{2x+y+z}\times 27^{3z+2y+x}=81\\ &\Leftrightarrow 3^{2x+y+z+3(3z+2y+x)}=3^{4}\\ &\Leftrightarrow 5x+7y+10z=4\\ &\color{blue}\textrm{Sehingga sistem persamaan akan terlihat}\\ &\left\{\begin{matrix} x-y+2z=1\: \: \qquad....(1)\\ 5x+7y+10z=4\quad\: ....(2)\\ 5x+3y+8z=2\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (2)\&(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ 5x+7y+10z&=4&\\ 5x+3y+8z&=2&-\\\hline \qquad 4y\quad+2z&=2\\ \qquad 2y\quad+z&=1\: ...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ 5x-5y+10z&=5&\\ 5x+7y+10z&=4&-\\\hline \quad -12y\quad&=1\\ \: \: \: \: \qquad y\quad&=-\displaystyle \frac{1}{12}\: ...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (5)\: \: \textrm{disubstistusikan ke}\: \: (4)\\ &\color{blue}\begin{aligned}2y+z&=1\\ 2\left ( -\displaystyle \frac{1}{12} \right )+z&=1\\ z&=1+\displaystyle \frac{1}{6}\\ z&=\displaystyle \frac{7}{6} \end{aligned}\\ &\textrm{Cukup jelas juga}\: \: x=....\\ &\color{blue}\textrm{Jadi},\: \textrm{pilihannya adalah}\: \: e \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Bintari, N., Gunarto, D. 2007. Panduan Menguasai Soal-Soal Olimpiade MAtematika Nasional dan Internasional. Yogyakarta: INDONESIA CERDAS.
2. Kanginan, M. 2016. Matematika untuk SMA-MA/SMK-MAK Kelas X. Bandung: SRIKANDI EMPAT WIDYA UTAMA
3. Kurnianingsih, S. 2008. SPM Matematika SMA dan MA Program IPS Siap Tuntas Menghadapi Ujian. Jakarta: ESIS
4. Susianto, B. 2011. Soal dan Pembahasan Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO
5. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: TIGA SERANGKAI PUSTAKA MANDIRI

Contoh Soal 3 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{ll}\\ 11.&\textrm{Suatu bilangan terdiri atas 3 angka. Jumlah}\\ &\textrm{ketiga angka tersebut adalah 9. Angka kedua}\\ &\textrm{dikurangi angka pertama dan angka ketiga }\\ &\textrm{sama dengan 1. Dua kali angka pertama sama}\\ &\textrm{dengan jumlah angka kedua dan angka ketiga.}\\ &\textrm{Angka puluhan pada bilangan tersebut adalah}\\ &....\\ &\begin{array}{llll}\\ \textrm{a}.&3\\ \textrm{b}.&4\\ \color{red}\textrm{c}.&5\\ \textrm{d}.&6\\ \textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Model matematikanya}\\ &\left\{\begin{matrix} A+B+C=9\: \: \qquad....(1)\\ 2B-A-C=1\qquad\: ....(2)\\ 2A=B+C\: \: \: \: \qquad\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle A+B+C&=9\\ \displaystyle -A+B-C&=1&+\\\hline \qquad2B&=10\\ \: \: \: \: \qquad\qquad B&=5&...(4)\\ \end{array}\\ &\color{blue}\textrm{Jadi},\: \textrm{bilangan kedua adalah}\: =\: B=5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&(\textbf{SIMAK UI 2010})\\ &\textrm{Jika}\: \: x+y+2z=K,\: x+2y+z=K,\\ &2x+y+z=K\: \: \textrm{dengan}\: \: K\neq 0,\: \textrm{maka}\\ &x^{2}+y^{2}+z^{2}\: \: \textrm{bila dinyatakan dalam}\: \: K\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{16}K^{2}\\ \color{red}\textrm{b}.&\displaystyle \frac{3}{16}K^{2}\\ \textrm{c}.&\displaystyle \frac{4}{17}K^{2}\\ \textrm{d}.&\displaystyle \frac{3}{8}K^{2}\\ \textrm{e}.&\displaystyle \frac{2}{3}K^{2} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+2z=K\: \: \qquad....(1)\\ x+2y+z=K\qquad\: ....(2)\\ 2x+y+z=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\color{black}\textrm{maka}\\ &\color{red}\left\{\begin{matrix} z+(x+y+z)=K\: \: \qquad....(1)\\ y+(x+y+z)=K\qquad\: ....(2)\\ x+(x+y+z)=K\: \: \qquad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle x+y+2z&=K\\ \displaystyle x+2y+z&=K&\\ \displaystyle 2x+y+z&=K&+\\\hline 4x+4y+4z&=3K\\ x+y+z&=\displaystyle \frac{3}{4}K&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (4)\: \: \textrm{disubstitusikan ke}\: \: (1),(2),\: \textrm{dan}\: (3)\\ &\textrm{Jelas bahwa akan didapatkan}\\ &x=y=z=\displaystyle \frac{1}{4}K\\ &\color{blue}\textrm{Jadi},\: \: x^{2}+y^{2}+y^{2}=3\left ( \displaystyle \frac{1}{4}K \right )^{2}=\displaystyle \frac{3}{16}K^{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Diketahui}\: \: 0,15252525252...=\displaystyle \frac{p}{2q+r}\\ &\textrm{Jika jumlah}\: \: p\: \: \textrm{dan}\: \: q=\textrm{3 kali}\: \: r,\: \textrm{maka}\\ &\textrm{masing-masing harga}\: \: p,q, \: \textrm{dan}\: \: r=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 152,2819,2584\\ \textrm{b}.&\displaystyle 252,\displaystyle \frac{5638}{7},\frac{8102}{21}\\ \color{red}\textrm{c}.&\displaystyle 151,\frac{2819}{7},\frac{1292}{7}\\ \textrm{d}.&\displaystyle 151,\displaystyle \frac{2819}{7},\frac{2584}{7}\\ \textrm{e}.&\displaystyle 152,\frac{2819}{14},\frac{1292}{7} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari soal diketahui}\\ &\color{red}\begin{cases} 0,1\overline{5252}& =\displaystyle \frac{p}{2q+r}\: .....(1)\\ \quad p+q & =3r\: ............(2) \end{cases}\\ &\color{black}\textrm{dan}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \qquad \color{red}x&=0,15252525252...\\ \displaystyle 1000\color{red}x&=152,5252525252...\\ \displaystyle \quad10\color{red}x&=\: \: \: \: \: 1,5252525252...&-\\\hline \: \: 990\color{red}x&=151\\ \qquad \color{red}x&=\displaystyle \frac{151}{990},\: \: \color{black}\textrm{maka}\\ \displaystyle \frac{p}{2q+r}&=\displaystyle \frac{151}{990}\\ &\begin{cases} p &=151 \: \: .......(3)\\ 2p+r &=990 \: \: .......(4) \end{cases} \end{array}\\ &\textrm{Dari}\: \: (3)\: \textrm{diperoleh}:q=3r-p=3r-151\: ....\color{blue}(5)\\ &\textrm{Dari}\: \: (5)\: \: \textrm{disubstitusikan ke}\: \: (4)\\ &\begin{aligned}2q+r&=990\\ 2(3r-151)+r&=990\\ 6r-302+r&=990\\ 7r&=990+302=1292\\ r&=\displaystyle \frac{1292}{7}\: .....\color{blue}(6) \end{aligned}\\ &\textrm{Dari}\: \: (3)\&(6)\: \: \textrm{disubstitusikan ke}\: \: (2)\\ &\color{purple}\begin{aligned}p+q&=3r\\ 151+q&=3\left ( \displaystyle \frac{1292}{7} \right )\\ q&=\displaystyle \frac{3876}{7}-151\\ &=\displaystyle \frac{3876-1057}{7}\\ &=\displaystyle \frac{2819}{7}\: .....\color{blue}(7) \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: p,q,r\: \: \textrm{adalah}\: :\: \displaystyle 151,\frac{2819}{7},\frac{1292}{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah sistem persamaan berikut}\\ &\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\textrm{agar sistem persamaan ini tidak}\\ &\textrm{memiliki penyelesaian, maka nilai}\: \: k=....\\ &\begin{array}{llll}\\ \textrm{a}.&-4\\ \textrm{b}.&2\\ \textrm{c}.&3\\ \color{red}\textrm{d}.&4\\ \textrm{e}.&6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Agar sistem persamaan}\\ &\color{red}\begin{cases} 3x+2y-5z & =3 \\ 2x-6y+kz & =9 \\ 5x-4y-z & =5 \end{cases}\\ &\color{black}\textrm{tidak berpenyelesaian, maka}\\ &\color{black}\textrm{ingat penyelesaian metode matrik}\\ &\color{black}\textrm{buatlah penyebutnya}=0,\: \: \textrm{yaitu}:\\ &\color{blue}\begin{vmatrix} 3 & 2 & -5\\ 2 & -6 & k\\ 5 & -4 & -1 \end{vmatrix}=0\\ &\textrm{Selanjutnya}\\ &3\begin{vmatrix} -6 & k\\ -4 & -1 \end{vmatrix}-2\begin{vmatrix} 2 & k\\ 5 & -1 \end{vmatrix}-5\begin{vmatrix} 2 & -6\\ 5 & -4 \end{vmatrix}=0\\ &3(6+4k)-2(-2-5k)-5(-8+30)=0\\ &18+12k+4+10k+40-150=0\\ &22x=88\\ &\quad \color{blue}x=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Diketahui}\\ &\begin{pmatrix} \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5}\\ \displaystyle \frac{1}{5} & \displaystyle \frac{1}{5} & -\displaystyle \frac{4}{5}\\ -\displaystyle \frac{2}{5} & \displaystyle \frac{1}{10} & \displaystyle \frac{1}{10} \end{pmatrix}\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} 1\\ 2\\ 0 \end{pmatrix}\\ &\textrm{Nilai}\: \: x,y,\: \: \textrm{dan}\: \: z\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{5},\frac{4}{5},-\frac{1}{10}\\ \textrm{b}.&-1,5,1\\ \color{red}\textrm{c}.&1,5,-1\\ \textrm{d}.&-1,1,5\\ \textrm{e}.&5,1,-1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\color{red}\left\{\begin{matrix} \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\quad \quad....(1)\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z=2\qquad\: ....(2)\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z=0\: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y-\frac{4}{5}z&=2&-\\\hline \quad\qquad \qquad \displaystyle \frac{5}{5}z&=-1&\\ \: \: \: \quad\qquad \qquad \displaystyle z&=-1&...(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{lllllll}\\ \displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z&=1&\left | \times 1 \right |&\displaystyle \frac{1}{5}x+\frac{1}{5}y+\frac{1}{5}z=1\\ -\displaystyle \frac{2}{5}x+\frac{1}{10}y+\frac{1}{10}z&=0&\left | \times 2 \right |&-\displaystyle \frac{4}{5}+\frac{1}{5}y+\frac{1}{5}z=0&-\\\hline &&&\: \: \: \: \displaystyle \frac{5}{5}x\qquad\qquad\: =1&\\ &&&\: \: \: \quad x=-1\: ........(5) \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{akan didapatkan}\\ &y=5\\ &\color{blue}\textrm{Jadi},\: \: (x,y,z)=(1,5,-1) \end{aligned} \end{array}$

Contoh Soal 2 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{ll}\\ 6.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad\\ 2x-y+2z=9\quad\: \\ x+3y-z=7\: \: \: \: \quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{3}\\ \textrm{b}.&\displaystyle \frac{3}{4}\\ \color{red}\textrm{c}.&\displaystyle \frac{13}{12}\\ \textrm{d}.&\displaystyle \frac{5}{4}\\ \textrm{e}.&\displaystyle \frac{7}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y-z=1\: \: \qquad....(1)\\ 2x-y+2z=9\quad\: ....(2)\\ x+3y-z=7\: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ x+y-z&=1&\\ 2x-y+2z&=9&+\\\hline 3x\: \: \qquad+z&=10&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llllll}\\ x+y-z&=1&\left | \times 3 \right |&3x+3y-3z&=3&\\ x+3y-z&=7&\left | \times 1 \right |&\quad x+3y-z&=7&-\\\hline &&&2x\quad \: \: \quad-2z&=-4&\\ &&&\: \: x\quad \: \: \: \: \quad-z&=2&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{array}{lll}\\ 3x+z&=10&\\ x-z&=-2&+\\\hline 4x&=8&\\ \qquad\quad x&=2&.....(6)\\ \color{red}\textrm{didapat pula}&z&=4......(7) \end{array} \\ &\textrm{Dari persamaan}\: \: (1)\&(3)\: \: \textrm{didapatkan juga}\\ &\color{blue}\begin{array}{lll}\\ x+y-z&=1&\\ x+3y-z&=-7&-\\\hline \quad -2y&=-6&\\ \qquad\qquad y&=3&....(8) \end{array}\\ &\color{blue}\textrm{Jadi},\: \: \displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad\\ x+y-4z=10\quad \\ -2x+y+z=0 \quad \end{matrix}\right.\\ &\textrm{Nilai dari}\: \: \displaystyle \frac{xz}{y}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle -\frac{6}{13}\\ \color{red}\textrm{b}.&\displaystyle -\frac{5}{13}\\ \textrm{c}.&\displaystyle -\frac{1}{13}\\ \textrm{d}.&\displaystyle \frac{1}{13}\\ \textrm{e}.&\displaystyle \frac{7}{13} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=5\: \: \: \: \quad.....(1)\\ x+y-4z=10\quad .....(2)\\ -2x+y+z=0 \quad .....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ x+y+z&=5&\\ x+y-4z&=10&-\\\hline \: \: \qquad \: \: \: \: \: 5z&=-5&\\ \: \: \qquad\quad \: \: \: z&=-1&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llllll}\\ x+y+z&=5&\\ -2x+y+z&=0&-\\\hline 3x\quad \: \quad&=5&\\ \: \: \quad \: \: \: \: \quad x&=\displaystyle \frac{5}{3}&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}x+y+z&=5\\ \displaystyle \frac{5}{3}+y-1&=5\\ y&=5+1-\displaystyle \frac{5}{3}=\frac{13}{3} \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: \displaystyle \frac{xz}{y}=\displaystyle \frac{\left ( \displaystyle \frac{5}{3} \right ).(-1)}{\displaystyle \frac{13}{3}}=-\displaystyle \frac{5}{13} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Himpunan penyelesaian dari}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8 \\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10 \\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4 \end{cases}\\ &\textrm{adalah}\: \: \left \{ (x,y,z) \right \},\: \textrm{maka}\: \: x+3z=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&1\\ \color{red}\textrm{d}.&3\\ \textrm{e}.&5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\begin{cases} \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z} & =8\: ....(1)\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z} & =10\: .....(2)\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z} & =4\: ...........(3) \end{cases}\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}&=8\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=10&-\\\hline -\displaystyle \frac{1}{x}\: \: \: \: \: -\frac{1}{z}&=-2\\ \displaystyle \frac{1}{x}\: \: \: \: \: \: \: \: +\frac{1}{z}&=2&...(4)\\ \end{array} \\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{lllllll}\\ \displaystyle \frac{2}{x}+\frac{2}{y}+\frac{4}{z}&=8&\left | \times 2 \right |&\displaystyle \frac{4}{x}+\frac{4}{y}+\frac{8}{z}&=16\\ \displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&\left | \times 1 \right |&\displaystyle \frac{2}{x}+\frac{4}{y}+\frac{2}{z}&=4&-\\\hline &&&\displaystyle \frac{2}{x}\: \: \: \: \: +\frac{6}{z}&=12\\ &&\Leftrightarrow &\displaystyle \frac{1}{x}\: \: \: \: \: +\frac{3}{z}&=6&...(5)\\ \end{array}\\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{array}{llll}\\ \displaystyle \frac{1}{x}+\frac{3}{z}&=6\\ \displaystyle \frac{1}{x}+\frac{1}{z}&=2\: \: \: -\\\hline \qquad\displaystyle \frac{2}{z}&=4&\\ \qquad z&=\displaystyle \frac{1}{2}\: \: ......(6)\\ \qquad x&=2-\displaystyle \frac{1}{2}=\displaystyle \frac{3}{2} \end{array}\\ &\color{blue}\textrm{Jadi},\: \: x+3z=\displaystyle \frac{3}{2}+3.\frac{1}{2}=3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Diketahui tiga buah bilangan berturut-turut}\\ &a,\: b,\: \textrm{dan}\: c.\: \textrm{Rata-rata dari ke tiga bilangan}\\ &\textrm{itu adalah 12. Bilangan kedua sama dengan}\\ &\textrm{jumlah bilangan yang lain dikurangi 12}.\\ &\textrm{Jika bilangan ke tiga sama dengan jumlah}\\ &\textrm{bilangan yang lain, maka nilai}\: \: 2a+b-c=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle 42\\ \textrm{b}.&-\displaystyle 36\\ \textrm{c}.&-\displaystyle 18\\ \textrm{d}.&-\displaystyle 12\\ \color{red}\textrm{e}.&-\displaystyle 6 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\textrm{Model matematika dari persamaan di atas}\\ &\left\{\begin{matrix} a+b+c=36\: \: \qquad....(1)\\ -a+b-x=12\quad\: ....(2)\\ a+b-c=0\: \: \: \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llll}\\ a+b+c&=36&\\ -a+b-c&=12&+\\\hline \: \: \: \: \: \: \: \: \: \: \: 2b&=48&\\ \: \: \qquad\quad \: \: \: b&=24&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llllll}\\ a+b+c&=36&\\ a+b-c&=0&-\\\hline \quad\qquad2c &=36&\\ \: \: \quad \: \: \: \: \quad c&=18&....(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{aligned}a+b+c&=36\\ a+24+18&=36\\ a&=36-42\\ &=-6 \end{aligned} \\ &\color{blue}\textrm{Jadi},\: \: 2a+b-c=2(-6)+24-18=-6 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jumlah uang terdiri atas koin pecahan}\: \: Rp500,00\\ &Rp200,00\: \: dan\: \: Rp100,00\: \: \textrm{dengan nilai total}\\ &Rp100.000,00.\: \textrm{Jika nilai uang pecahan 500-an}\\ &\textrm{setengah dari nilai uang pecahan 200-an, tetapi}\\ &\textrm{tiga kali uang pecahan 100-an, maka banyak koin}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&460\\ \textrm{b}.&440\\ \textrm{c}.&420\\ \textrm{d}.&380\\ \textrm{e}.&350 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Model matematika dari kasus di atas}\\ &\left\{\begin{matrix} A(500)+B(200)+C(100)=100.000\: ....(1)\\ A(500)=\displaystyle \frac{1}{2}B(200)\qquad\qquad\qquad\qquad\: ....(2)\\ A(500)=3C(100)\qquad\qquad\qquad\: \: \, \: \: \: \: \quad ....(3)\end{matrix}\right.\\ &\textrm{Dari persamaan}\: \: (2)\: \textrm{didapatkan}\\ &2A(500)=B(200)\\ &\textrm{Dari persamaan}\: \: (3)\: \textrm{akan didapatkan}\\ &\displaystyle \frac{1}{3}A(500)=C(100)\\ &\textrm{Dari persamaan}\: \: (1)\: \: \textrm{maka},\\ &A(500)+B(200)+C(100)=100.000\\ &A(500)+2A(500)+\displaystyle \frac{1}{3}A(500)=100.000\\ &\displaystyle \frac{10}{3}A(500)=100.000\Leftrightarrow A(500)=30.000\\ &\textrm{maka akan didapatkan}\\ &B(200)=2(30.000)=60.000\\ &C(100)=\displaystyle \frac{1}{3}(30.000)=10.000\\ &\color{red}\begin{cases} A(500) &=30.000\Rightarrow \color{black}A=\displaystyle \frac{30.000}{500}=60 \\ B(200) &=60.000\Rightarrow \color{black}B=\displaystyle \frac{60.000}{200}=300 \\ C(100) &=10.000\Rightarrow \color{black}C=\displaystyle \frac{10.000}{100}=100 \end{cases}\\ &\color{blue}\textrm{Jadi},\: \: A+B+C=60+300+100=460 \end{aligned} \end{array}$

Contoh Soal 1 Sistem Persamaan Linear Tiga Variabel

$\begin{array}{ll}\\ 1.&\textrm{Suatu unit pekerjaan dapat diselesaikan oleh A}\\ &\textrm{B, dan C bersama-sama dalam 2 jam saja.}\\ &\textrm{Jika pekerjaan itu dapat diselesaikan oleh A dan}\\ &\textrm{B bersama-sama dalam 2 jam 24 menit, dan oleh}\\ &\textrm{B dan C bersama-sama dalam waktu 3 jam,}\\ &\textrm{maka sistem persamaan berikut yang memenuhi}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\begin{cases} A+B+C&=2 \\ A+B & =\displaystyle \frac{12}{5} \\ B+C &=3 \end{cases}\\ \textrm{b}.&\begin{cases} A+B+C&=\displaystyle \frac{1}{2} \\ A+B & =\displaystyle \frac{5}{12} \\ B+C &=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{c}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{12}{5} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=3 \end{cases}\\ \color{red}\textrm{d}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{2} \\ \displaystyle \frac{1}{A}+\frac{1}{B}& =\displaystyle \frac{5}{12} \\ \displaystyle \frac{1}{B}+\frac{1}{C}&=\displaystyle \frac{1}{3} \end{cases}\\ \textrm{e}.&\begin{cases} \displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=2 \\ \displaystyle \frac{1}{A}+\frac{1}{B}-\frac{1}{C}& =\displaystyle \frac{12}{5} \\ \displaystyle -\frac{1}{A}+\frac{1}{B}+\frac{1}{C}&=3 \end{cases} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Per}&\textrm{hatikan bahwa}:\color{red}\textrm{Waktu penyelesaian}\\ \color{red}\textrm{sua}&\color{red}\textrm{tu pekerjaan adalah termasuk}\\ \color{red}\textrm{per}&\color{red}\textrm{bandingan berbalik nilai},\: \color{blue}\textrm{maka}\\ \bullet \: \: \: &A,B,\: \textrm{dan}\: C \: \textrm{dalam 2 jam, artinya}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{2},\: \color{blue}\textrm{demikian juga}\\ \bullet \: \: \: &A\: \textrm{dan}\: B\: \textrm{bersama-sama selesai dalam}\\ &\textrm{2 jam 24 menit atau}\: \displaystyle \frac{12}{5}\: \textrm{jam}:\\ &\color{black}\displaystyle \frac{1}{A}+\frac{1}{B}=\frac{5}{12}\\ \bullet \: \: \: &B\: \textrm{dan}\: C\: \textrm{selesai dalam 3 jam}:\\ &\color{black}\displaystyle \frac{1}{B}+\frac{1}{C}=\frac{1}{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Himpunan penyelesaian dari}\\ &\left\{\begin{matrix} x+y+4z=15\quad\\ x-y+z=2\qquad\\ x+2y-3z=-4 \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ (-1,1,3) \right \}\\ \color{red}\textrm{b}.&\left \{ (1,2,3) \right \}\\ \textrm{c}.&\left \{ (-2,1,1) \right \}\\ \textrm{d}.&\left \{ (3,2,-1) \right \}\\ \textrm{e}.&\left \{ (1,-2,3) \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\textrm{Semunya dikerjakan dengan metode}\\ &\color{blue}\textrm{matriks}\: (\color{black}\textbf{Cara Cramer})\\ &\begin{aligned} \color{blue}x&=\displaystyle \frac{\begin{vmatrix} 15 & 1 & 4\\ 2& -1 & 1\\ -4& 2 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{15\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}+4\begin{vmatrix} 2 & -1\\ -4 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{15(3-2)-1(-6+4)+4(4-4)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{15(1)-1(-2)+4(0)}{1(1)-1(-4)+4(3)}=\frac{17}{17}=1 \\ \color{blue}y&=\displaystyle \frac{\begin{vmatrix} 1 & 15 & 4\\ 1& 2 & 1\\ 1& -4 & -3 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} 2 & 1\\ -4 & -3 \end{vmatrix}-15\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(-6+4)-15(-3-1)+4(-4-2)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(-2)-15(-4)+4(-6)}{1(1)-1(-4)+4(3)}=\frac{34}{17}=2\\ \color{blue}z&=\displaystyle \frac{\begin{vmatrix} 1 & 1 & 15\\ 1& -1 & 2\\ 1& 2 & -4 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 4\\ 1 & -1 & 1\\ 1 & 2 & -3 \end{vmatrix}}\\ &=\displaystyle \frac{1\begin{vmatrix} -1 & 2\\ 2 & -4 \end{vmatrix}-1\begin{vmatrix} 1 & 2\\ 1 & -4 \end{vmatrix}+15\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}{1\begin{vmatrix} -1 & 1\\ 2 & -3 \end{vmatrix}-1\begin{vmatrix} 1 & 1\\ 1 & -3 \end{vmatrix}+4\begin{vmatrix} 1 & -1\\ 1 & 2 \end{vmatrix}}\\ &=\displaystyle \frac{1(4-4)-1(-4-2)+15(2+1)}{1(3-2)-1(-3-1)+4(2+1)}\\ &=\displaystyle \frac{1(0)-1(-6)+15(3)}{1(1)-1(-4)+4(3)}=\frac{51}{17}=3 \end{aligned} \end{array}$

$.\quad\quad \color{blue}\textrm{Cara di atas}$  full matriks-Cramer

$\begin{array}{ll}\\ 3.&\textrm{Hasil dari}\: \: xyz\: \: \textrm{yang memenuhi}\\ &\left\{\begin{matrix} x+y+z=2\quad\\ x-y+z=-2\: \\ x-y-z=2\quad \end{matrix}\right.\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-8\\ \textrm{b}.&-4\\ \textrm{c}.&2\\ \textrm{d}.&4\\ \textrm{e}.&8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=2\quad.....(1)\\ x-y+z=-2\: .....(2)\\ x-y-z=2\quad .....(3) \end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llc}\\ x+y+z&=2&\\ x-y+z&=-2&-\\\hline \: \, \quad2y&=4&\\ \qquad\quad y&=2&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{lcc}\\ x+y+z&=2&\\ x-y-z&=2&+\\\hline 2x&=4&\\ \qquad\quad x&=2&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (2)+(2)+z&=2\\ z&=-2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: xyz=(2).(2).(-2)=-8 \end{aligned} \end{array}$

$.\quad\: \:  \color{black}\textrm{Cara di atas}$  full eliminasi-substitusi

$\begin{array}{ll}\\ 4.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+y+z=-6\quad\\ x-2y+z=3\quad\: \\ -2x+y+z=9\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-30\\ \textrm{b}.&-15\\ \textrm{c}.&5\\ \color{red}\textrm{d}.&30\\ \textrm{e}.&35 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+y+z=-6\quad ....(1)\\ x-2y+z=3\quad\: ....(2)\\ -2x+y+z=9\quad ....(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llc}\\ x+y+z&=-6&\\ x-2y+z&=3&-\\\hline \: \: \: \quad 3y&=-9&\\ \qquad\quad y&=-3&....(4) \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{llc}\\ x+y+z&=-6&\\ -2x+y+z&=9&-\\\hline 3x&=-15&\\ \qquad\quad x&=-5&....(5) \end{array} \\ &\textrm{Persamaan}\: \: (4)\&(5)\: \: \textrm{ke}\: \: (1)\\ &\color{red}\begin{aligned}x+y+z&=2\\ (-5)+(-3)+z&=-6\\ z&=2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: xyz=(-5).(-3).(2)=30 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui sistem persamaan berikut}\\ &\left\{\begin{matrix} x+2y+z=4\: \: \qquad\\ 3x+y+2z=-5\quad\: \\ x-2y+2z=-6\quad \end{matrix}\right.\\ &\textrm{Nilai}\: \: xyz=....\\ &\begin{array}{llll}\\ \textrm{a}.&-96\\ \color{red}\textrm{b}.&-24\\ \textrm{c}.&24\\ \textrm{d}.&32\\ \textrm{e}.&96 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\textrm{Diketahui sistem persamaan}\\ &\left\{\begin{matrix} x+2y+z=4\: \qquad.......(1)\\ 3x+y+2z=-5\quad\: ......(2)\\ x-2y+2z=-6\quad .......(3)\end{matrix}\right.\\ &\textrm{Saat}\: \: (1)+(2),\: \textrm{maka}\\ &\color{blue}\begin{array}{llclll}\\ x+2y+z&=4&\left | \times 1 \right |&\: \: x+2y+z&=4\\ 3x+y+2z&=-5&\left | \times 2 \right |&6x+2y+4z&=-10&-\\\hline &&&-5x\: \: \quad-3z&=14&...(4)\\ \end{array}\\ &\textrm{Saat}\: \: (1)+(3),\: \textrm{maka}\\ &\color{blue}\begin{array}{lll}\\ x+2y+z&=4&\\ x-2y+2z&=-6&+\\\hline 2x\: \: \: \, \quad +3z&=-2&...(5)\\ \end{array} \\ &\textrm{Dari persamaan}\: \: (4)\&(5)\: \: \textrm{maka},\\ &\color{blue}\begin{array}{lll}\\ -5x-3z&=14&\\ 2x+3z&=-2&+\\\hline -3x&=12&\\ \qquad\quad x&=-4&.....(6)\\ \color{red}\textrm{didapat pula}&z&=2......(7) \end{array}\\ &\textrm{Dari persamaan}\: \: (6)\&(7)\: \: \textrm{didapatkan}\\ &\color{red}\begin{aligned}x+2y+z&=4\\ (-4)+2y+2&=4\\ y&=3 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: xyz=(-4).(3).(2)=-24 \end{aligned} \end{array}$

Lanjutan Materi (6) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI

$\color{blue}\textrm{E. Persamaan Garis Singgung}$

1. Fungsi Aljabar

Perhatikanlah gambar berikut!

Perhatikanlah kurva di atas, yaitu sebuah gambar grafik fungsi kuadrat  $\color{blue}f(x)=x^{2}-2x+1$. Misalkan kita menginginkan garis mana yang merupakan persamaan garis singgung di titik $\color{black}\left ( 2,1 \right )$?

Ada 2 unsur penting dalam menentukan persamaan garis singgung, yaitu:

• titik singgung
• gradien (kemiringan) dari garis singgung itu sendiri, yaitu : $\color{blue}m=\displaystyle \frac{dy}{dx}$

Karena salah satu unsur penentuan persamaan garis singgung telah diketahui, yaitu sebuah titik singgung, langkah berikutnya kita tinggal mencari gradien. Dalam hal ini gradien dari garis singgung diperoleh dengan memasukkan absis seteleh kurva singgung itu diturunkan pertama dan kadang dituliskan dengan notasi  Leibniz  $\color{blue}m=\left ( \displaystyle \frac{dy}{dx} \right )_{\color{black}x=a}$ atau kadang juga dituliskan dengan bentuk notasi $\color{blue}m=\left.\begin{matrix} \displaystyle \frac{dy}{dx} \end{matrix}\right|_{\color{black}x=a}$. Untuk mempermudah, oerhatikanlah kurva di atas, dari keempat garis lurus yang ada, tidak semunya menyinggung. Karena sebagian bahkan berpotongan dengan kurva. Walaupun antara titik potong dan titik singgung sama, tetapi cara mendapatkannya berbeda. Sementara kita fokus pada aplikasi turunan pertama pada suatu kurva. Coba kita perjelas lagi dengan menyertakan persamaan keempat garis lurusnya berikut

Mari kita tentukan persamaan garis singgung kurva di atas dari keempat garis lurus itu, garis yang mana?

Persamaan Garis Singgung kurva dituliskan sebagai: $\color{blue}y=m(x-a)+b$, dengan  $\color{red}(a,b)$  adalah titik singgung. Kada titik singgung juga dituliskan dengan  $\color{red}\left (a,f(a) \right )$.

Sehingga persamaan garis singgung kurva di atas adalah:

$\color{blue}\begin{aligned}f(x)=y&=x^{2}-2x+1=(x-1)^{2}\\ m&=\color{black}2x-2\\ \left (\displaystyle \frac{dy}{dx} \right )_{x=2}&=m=\color{black}2(2)-2=4-2=2\\ \color{red}\textrm{maka}&\: \color{purple}\textrm{persamaan garis singgung kurvanya}\\ y&=m(x-a)+b\\ &=2(x-2)+1\\ &=2x-4+1\\ y&=\color{black}2x-3 \end{aligned}$.

Jadi, garis pada gambar di atas yang merupakan garis singgung kurva yang dimaksud adalah garis $\color{red}g_{3}\: :\: \color{blue}y=2x-3$.

2. Fungsi Trigonometri

Tidak jauh berbeda dengan fungsi aljabra, maka pada fungsi trigonometri berlaku sifat yang sama yang membedakan hanyanya kurvanya serta sumbu X (letak absis).

Sebagai misal kita diberikan sebuah fungsi trigonometri  $\color{blue}f(x)=y=\sin 2x$. Jika dituntut untuk menunjukkan persamaan garis singgung di titik yang berabsis  $\color{blue}\displaystyle \frac{\pi }{2}$, maka kita juga dapat dengan mudah menentukannya.

Perhatikan uraian berikut sebagai pembahasan dari permasalahan di atas.

$\color{blue}\begin{aligned}\textrm{Diketahui}&\: \: x=a=\displaystyle \frac{\pi }{2}\\ \textrm{Kita men}&\textrm{cari titik singgungnya dulu, yaitu}\\ f(a)&=\sin 2\left ( \displaystyle \frac{\pi }{2} \right )=\sin \pi =0,\\ \color{red}\left ( a,f(a) \right )&=\left ( \displaystyle \frac{\pi }{2},0 \right )\\ f(x)=y&=\sin 2x\\ m&=\color{purple}2\cos 2x\quad ......(\textbf{turunan pertama})\\ \left (\displaystyle \frac{dy}{dx} \right )_{x=\frac{\pi }{2}}&=m=\color{black}2\cos 2\left ( \displaystyle \frac{\pi }{2} \right )\\ &=\color{black}2\cos \pi =2.(-1)=-2\\ \color{red}\textrm{maka}&\: \textrm{persamaan garis singgung kurvanya}\\ y&=m(x-a)+b\\ &=-2\left ( x-\displaystyle \frac{\pi }{2} \right )+0\\ &=\color{red}-2x+\pi \end{aligned}$

DAFTAR PUSTAKA

1. Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira.
2. Tampomas, H. 1999. SeribuPena Matematika SMU Kelas 2. Jakarta: ERLANGGA
3. Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI. Jakarta: ERLANGGA.

 

Contoh Soal 5 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 21.&\textrm{Turunan pertama dari fungsi}\\ &g(x)=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\sin ^{2}x}\\ \textrm{c}.&\displaystyle \frac{1}{\sin^{2} x\cos ^{2}x}\\ \textrm{d}.&\displaystyle \frac{-1}{\sin ^{2}x\cos ^{2}x}\\ \textrm{e}.&\displaystyle \sin ^{2}x\cos ^{2}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(x)&=\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\\ &=\frac{\sin ^{2}x+\cos ^{2}x}{\sin x\cos x}=\displaystyle \frac{1}{\sin x\cos x}\\ \color{red}\textrm{maka}&\\ g'(x)&=\displaystyle \frac{0.(\sin x\cos x)-1.\left (\cos ^{2}x -\sin ^{2}x \right )}{(\sin x\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x-\cos ^{2}x}{\sin^{2} x\cos^{2} x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}-\frac{1}{\sin ^{2}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Diketahui}\: \: h(x)=\cos \left ( \displaystyle \frac{3}{x} \right ), \\ &\textrm{maka}\: \: \displaystyle \frac{dh}{dx}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\sin \displaystyle \frac{3}{x}\\ \textrm{b}.&-\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{c}.&-\displaystyle \frac{3}{x}\sin \frac{3}{x}\\ \color{red}\textrm{d}.&\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x}\\ \textrm{e}.&\displaystyle \frac{3}{x}\sin \frac{3}{x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\cos \displaystyle \frac{3}{x}&=-\sin \displaystyle \frac{3}{x}\left ( \displaystyle \frac{0.(x)-3.1}{x^{2}} \right )\\ &=\displaystyle \frac{-(-3)}{x^{2}}\sin \frac{3}{x}\\ &=\displaystyle \frac{3}{x^{2}}\sin \frac{3}{x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Turunan pertama dari}\: \: \tan (\cos x), \\ &\textrm{terhadap}\: \: x\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&-\sec ^{2}(\cos x)\sin x\\ \textrm{b}.&\sec ^{2}(\cos x)\sin x\\ \textrm{c}.&\sec ^{2}(\sin x)\cos x\\ \textrm{d}.&\displaystyle \sin x\\ \textrm{e}.&\displaystyle -\sin x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Misal}&\textrm{kan}\\ y&=\tan x(\cos x)\\ y'&=\sec ^{2}(\cos x)\times (-\sin x)\\ &=-\sec ^{2}(\cos x).\sin x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&(\textbf{UN 2005})\textrm{Turunan pertama dari}\\ &f(x)=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \textrm{b}.&\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\\ \textrm{c}.&-\displaystyle \frac{2}{3}\cos^{.^{-\frac{1}{3}}} \left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ \color{red}\textrm{d}.&-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ \textrm{e}.&\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Misal}&\textrm{kan}\\ f(x)&=\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )}\\ f'(x)&=\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &=\displaystyle \frac{2}{3}\cos ^{.^{-\frac{1}{2}}}\left ( 3x^{2}+5x \right )\times \left ( -\sin \left ( 3x^{2}+5x \right ) \right )\\ &\qquad\qquad\qquad\qquad \times (6x+5)\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{-\frac{1}{3}}}\left ( 3x^{2}+5x \right )\sin \left ( 3x^{2}+5x \right )\\ &=-\displaystyle \frac{2}{3}(6x+5)\cos ^{.^{\frac{2}{3}}}\left ( 3x^{2}+5x \right )\\ &\times \cos^{-1} \left ( 3x^{2}+5x \right )\times \sin \left ( 3x^{2}+5x \right )\\ &=-\displaystyle \frac{2}{3}(6x+5)\tan \left ( 3x^{2}+5x \right )\sqrt[3]{\cos ^{2}\left ( 3x^{2}+5x \right )} \end{aligned} \end{array}$

Contoh Soal 4 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 16.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{1-\cos x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\sin x+\cos x+1}{x^{2}}\\ \textrm{b}.&\displaystyle \frac{x\cos x+\sin x-1}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{x\sin x-\cos x+1}{x^{2}}\\ \color{red}\textrm{d}.&\displaystyle \frac{x\sin x+\cos x-1}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{x\cos x-\sin x+1}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{1-\cos x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=1-\cos x\Rightarrow u'=\sin x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sin x.(x)-(1-\cos x).1}{x^{2}}\\ &=\displaystyle \frac{x\sin x+\cos x-1}{x^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\tan x}{\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1+\cos ^{2}x}{\cos ^{3}x}\\ \textrm{b}.&\displaystyle \frac{1-\cos x}{\cos ^{3}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{1+\sin x}{\cos ^{3}x}\\ \textrm{e}.&\displaystyle \frac{1-\sin ^{2}x}{\cos ^{3}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\tan x}{\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\tan x\Rightarrow u'=\sec ^{2}x\\ v&=\cos x\Rightarrow v'=-\sin x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sec ^{2}x.(\cos x)-(\tan x).(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\sec ^{2}x.\cos x+\tan x\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\cos x+\left ( \displaystyle \frac{\sin x}{\cos x} \right )\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\cos x}+\displaystyle \frac{\sin ^{2}x}{\cos x}}{\cos ^{2}x}\\ &=\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Turunan pertama dari}\: \: g(t)=\displaystyle \frac{\cos t+2t}{\sin t} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2\sin t+2t\cos t-1}{\sin^{2} t}\\ \textrm{b}.&\displaystyle \frac{2\sin t-2t\cos t+1}{\sin^{2} t}\\ \textrm{c}.&\displaystyle \frac{2\sin t+2t\cos t+1}{\sin^{2} t}\\ \color{red}\textrm{d}.&\displaystyle \frac{2\sin t-2t\cos t-1}{\sin^{2} t}\\ \textrm{e}.&\displaystyle \frac{-2\sin t+2t\cos t-1}{\sin^{2} t} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(t)&=\displaystyle \frac{\cos t+2t}{\sin t}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\cos t+2t\Rightarrow u'=-\sin t+2\\ v&=\sin t\Rightarrow v'=\cos t\\ \color{red}\textrm{maka}&\\ g'(t)&=\displaystyle \frac{(-\sin t+2)(\sin t)-(\cos t+2t)(\cos t)}{\sin ^{2}t}\\ &=\displaystyle \frac{-\sin ^{2}t+2\sin t-\cos ^{2}t-2t\cos t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\sin ^{2}t-\cos ^{2}t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\left (\sin ^{2}t+\cos ^{2}t \right )}{\sin ^{2}t}\\ &=\displaystyle \frac{2\sin t-2t\cos t-1}{\sin ^{2}t} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 19.&\textrm{Turunan pertama dari}\: \: h(x)=\displaystyle \frac{\sin x}{\sin x+\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x-\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\sin ^{2}x-\cos ^{2}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1}{(\sin x+\cos x)^{2}}\\ \textrm{d}.&\displaystyle \sin ^{2}x-\cos ^{2}x\\ \textrm{e}.&\displaystyle 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ h(x)&=\displaystyle \frac{\sin x}{\sin x+\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x \Rightarrow u'=\cos x\\ v&=\sin x+\cos x\Rightarrow v'=\cos x-\sin x\\ \color{red}\textrm{maka}&\\ h'(x)&=\displaystyle \frac{\cos x.(\sin x+\cos x)-\sin x.(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\cos x\sin x+\cos ^{2}x-\sin x\cos x+\sin ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x+\cos ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{1}{(\sin x+\cos x)^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\sin x-\cos x}{\tan x}. \: \: \textrm{Nilai}\\ &\textrm{turunan pertama fungsi}\: \: f\: \: \textrm{saat}\: \: x=45^{\circ}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle 1\\ \color{red}\textrm{d}.&\displaystyle \sqrt{2}\\ \textrm{e}.&\displaystyle \sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x-\cos x}{\tan x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x-\cos x \Rightarrow u'=\cos x+\sin x\\ v&=\tan x\Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{(\cos x+\sin x).\tan x-(\sin x-\cos x).\sec ^{2}x}{\tan ^{2}x}\\ f'\left ( 45^{\circ} \right )&=\displaystyle \frac{\left ( \displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2} \right ).1-\left ( \displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2} \right ).\left ( \sqrt{2} \right )^{2}}{1^{2}}\\ &=\displaystyle \frac{\sqrt{2}-0}{1}\\ &=\sqrt{2} \end{aligned} \end{array}$

Contoh Soal 3 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 11.&\textrm{Turunan pertama fungsi}\\ &h(x)=5\sin x\cos x\: \: \textrm{adalah}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\sin 2x\\ \color{red}\textrm{b}.&5\cos 2x\\ \textrm{c}.&5\sin ^{2}x\cos x\\ \textrm{d}.&5\sin ^{2}x\cos^{2} x\\ \textrm{e}.&5\sin 2x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}h(x)=5\sin x\cos x\\ h(x)&=\color{red}\displaystyle \frac{5}{2}\left ( 2\sin x\cos x \right )=\displaystyle \frac{5}{2}\sin 2x\\ h'(x)&=\color{purple}\displaystyle \frac{5}{2}\left ( \cos 2x \right ).(2)\\ &=\color{purple}5\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Turunan pertama fungsi}\\ &k(x)=\cos x\tan x\: \: \textrm{adalah}\: \: k'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin x\cot x+\cos x\sec ^{2}x\\ \color{red}\textrm{b}.&-\sin x\tan x+\cos x\sec ^{2}x\\ \textrm{c}.&\sin x\tan x-\cos x\sec ^{2}x\\ \textrm{d}.&-\displaystyle \frac{1+\sin ^{2}x}{\cos x}\\ \textrm{e}.&\displaystyle \frac{1+\sin ^{2}x}{\cos x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}k(x)=\cos x\tan x\\ \textrm{guna}&\textrm{kan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'v+u.v'\\ u&=\color{black}\cos x \Rightarrow u'=-\sin x\\ v&=\color{black}\tan x \Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ k'(x)&=\left ( -\sin x \right )\tan x+\cos x.\left ( \sec ^{2}x \right )\\ &=-\sin x\tan x+\cos x\sec ^{2}x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Jika diketahui}\: \: f(x)=\left | \tan x \right |,\: \textrm{maka}\: \: \displaystyle \frac{dy}{dx}\\ &\textrm{saat}\: \: x=k,\: \: \textrm{di mana}\: \: \displaystyle \frac{1}{2}\pi <k<\pi\\ & \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin k\\ \textrm{b}.&\cos k\\ \color{red}\textrm{c}.&-\sec ^{2}k\\ \textrm{d}.&\sec ^{2}k\\ \textrm{e}.&\cot k \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}f(x)=\left |\tan x \right |\\ \textrm{saat}&\: \: \color{red}x=k\: \: \color{blue}\textrm{dengan}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \color{black}\textrm{adal}&\color{black}\textrm{ah}:\\ f(x)&=\left | \tan x \right |,\: \: \color{black}\textrm{maka saat}\: \: \color{blue}x=k\\ f(k)&=\left | \tan k \right |=-\tan k,\: \: \color{black}\textrm{karena di}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \displaystyle \frac{dy}{dx}&=f'(k)=-\sec ^{2}k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Turunan pertama}\: \: g(x)=\left | \cos x \right |\\ & \textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\left | \sin x \right |\\ \textrm{b}.&-\sin x\\ \textrm{c}.&\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \color{red}\textrm{d}.&-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{e}.&\left | \sin x \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}g(x)=\left |\cos x \right |=\sqrt{\cos ^{2}x}=\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}\\ g'(x)&=\color{purple}\displaystyle \frac{1}{2}\left ( \cos ^{2}x \right )^{.^{-\frac{1}{2}}}.\left ( 2\cos x \right ).\left ( -\sin x \right )\\ &=\color{purple}\displaystyle \frac{-2\sin x\cos x}{2\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}}\\ &=\color{blue}-\displaystyle \frac{\sin 2x}{2\sqrt{\cos ^{2}x}}\\ &=\color{blue}-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\sin x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\cos x+\sin x}{x^{2}}\\ \color{red}\textrm{b}.&\displaystyle \frac{x\cos x-\sin x}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{-x\cos x-\sin x}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{\cos x-x\sin x}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{\cos x+x\sin x}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x\Rightarrow u'=\cos x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\cos x.(x)-\sin x.1}{x^{2}}\\ &=\displaystyle \frac{x\cos x-\sin x}{x^{2}} \end{aligned} \end{array}$

Contoh Soal 2 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 6.&\textrm{Turunan pertama}\: \: q(x)=\sin ^{2}x+\cos ^{2}x\\ &\textrm{adalah}\: \: q'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cos ^{2}x-\sin ^{2}x\\ \textrm{b}.&2\cos ^{2}x-2\sin ^{2}x\\ \textrm{c}.&\cos x-\sin x\\ \textrm{d}.&2\cos x-2\sin x\\ \color{red}\textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}q(x)&=\sin ^{2}x+\cos ^{2}x\\ \color{red}\textrm{guna}&\color{red}\textrm{kan formula identitas}:\: \color{black}\sin ^{2}x+\cos ^{2}x=1\\ \textrm{Sehi}&\textrm{ngga soal di atas dapat dituliskan menjadi}\\ q(x)&=1,\: \: \textrm{maka}\\ q'(x)&=0\\ \color{purple}\textrm{inga}&\color{purple}\textrm{t bahwa}\: \: \color{black}y=a\Rightarrow \displaystyle \frac{dy}{dx}=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.&-\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dari}&\: \textrm{soal diketahui}:\: \\ f(x)&=\sin \displaystyle \frac{\pi }{3}\\ \textrm{Nila}&\textrm{i dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ \textrm{arti}&\textrm{nya bermakna, berapkah}\: \: f'\left ( x \right )?\\ \color{red}\textrm{maka}&\\ f'\left ( x \right )&=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: f(x)=8x-\sin ^{3}x,\\ &\textrm{maka nilai}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&4x^{2}-3\cos^{2}x \\ \textrm{b}.&8x-3\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&8-3\sin ^{2}x\cos x\\ \textrm{d}.&8+\sin ^{2}x\cos x\\ \textrm{e}.&3\sin ^{2}x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui dari soal}\: f(x)=8x-\sin ^{3}x\\ \color{red}\textrm{maka}&\: \textrm{nilai dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}=f'(x)\\ f'(x)&=8-3\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Turunan pertama fungsi}\: \: f(x)=\sqrt{\sin x},\\ &\textrm{adalah}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2\sqrt{\sin x}} \\ \textrm{b}.&\displaystyle \frac{\cos x}{\sqrt{\sin x}}\\ \color{red}\textrm{c}.&\displaystyle \frac{\cos x}{2\sqrt{\sin x}}\\ \textrm{d}.&-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ \textrm{e}.&\displaystyle \frac{2\cos x}{\sqrt{\sin x}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}f(x)=\sqrt{\sin x}=\sin ^{.^{\frac{1}{2}}}x\\ f'(x)&=\color{purple}\displaystyle \frac{1}{2}\left ( \sin ^{.^{-\frac{1}{2}}}x \right ).(\cos x)\\ &=\color{purple}\displaystyle \frac{\cos x}{2\sin ^{.^{\frac{1}{2}}}x}\\ &=\color{purple}\displaystyle \frac{\cos x}{2\sqrt{\sin x}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: g'(x)\: \: \textrm{adalah turunan pertama}\\ &\textrm{fungsi}\: \: g(x)\: \: \textrm{dengan}\: \: g(x)=5\tan ^{2}x,\\ &\textrm{maka}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&10\cos ^{2}x\sin x\\ \textrm{b}.&10\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&\displaystyle \frac{10\sin x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{10\cos ^{3}x}{\sin x}\\ \textrm{e}.&\displaystyle \frac{10}{\sin ^{2}x-\cos ^{2}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}g(x)=5\tan ^{2}x\\ g'(x)&=\color{purple}5\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )\\ &=\color{purple}10\tan x\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\color{purple}10\left ( \displaystyle \frac{\sin x}{\cos x} \right )\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\color{purple}\displaystyle \frac{10\sin x}{\cos ^{3}x} \end{aligned} \end{array}$

Contoh Soal 1 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: f(x)=2\cos x-2020\\ &\textrm{Turunan pertama fungsi}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&2\sin x\\ \color{red}\textrm{b}.&-2\sin x\\ \textrm{c}.&-2\sin x-2020x\\ \textrm{d}.&2\sin ^{2}x\\ \textrm{e}.&2\cos x-2020x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}f(x)&=2\cos x-2020\\ f'(x)&=-2\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: f'(x)\: \: \textrm{adalah turunan pertama dari}\\ &\textrm{fungsi}\: \: f(x)=\sin ^{7}x\: ,\: \textrm{maka}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\cos^{6} x\\ \textrm{b}.&7\cos^{7} x\\ \color{red}\textrm{c}.&7\sin^{6} x\cos x\\ \textrm{d}.&7\cos ^{6}x\sin x\\ \textrm{e}.&7\cos ^{6}x\sin ^{6}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\sin ^{7}x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ f'(x)&=7\sin ^{6}x\left ( \cos x \right )=7\sin ^{6}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Turunan pertama fungsi}\: \: g(x)=-5\sin ^{3}x\\ &\textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\sin ^{2}x\cos x\\ \textrm{b}.&-5\sin ^{2}\cos ^{2}x\\ \color{red}\textrm{c}.&-15\sin ^{2}x\cos x\\ \textrm{d}.&-15\cos ^{3}x\\ \textrm{e}.&-15\sin ^{4}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}g(x)&=-5\sin ^{3}x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ g'(x)&=-5\left ( 3\sin ^{2}x \right )(\cos x)=-15\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: h(x)=4x^{3}+\sin x+\cos x\\ &\textrm{maka}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&12x^{2}+\cos x-\sin x\\ \textrm{b}.&12x^{2}-\cos x+\sin x\\ \textrm{c}.&4x^{3}-\cos x-\sin x\\ \textrm{d}.&4x^{3}-\sin x-\cos x\\ \textrm{e}.&12x^{3}+\cos x+\sin x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}h(x)&=4x^{3}+\sin x+\cos x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ \textrm{pada}&\: \textrm{fungsi aljabarnya, yaitu}:\color{black}y=4x^{3}\Rightarrow y'=12x^{2}\\ \textrm{seda}&\textrm{ngkan fungsi transendennya mengikuti}\\ \textrm{turu}&\textrm{nan fungsi trigonometri biasa. Sehingga}\\ f'(x)&=12x^{2}+\cos x-\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: p(x)=-\cos ^{4}x,\: \: \textrm{maka nilai}\\ &\textrm{maka}\: \: p'\left ( \displaystyle \frac{\pi }{3} \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{4}\sqrt{3}\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}p(x)&=-\cos ^{4}x\\ \color{red}p'(x)&\color{red}=-4\cos ^{3}x.(-\sin x)=\color{black}4\cos ^{3}x\sin x\\ p'\left ( \displaystyle \frac{\pi }{3} \right )&=4\cos ^{3}\left ( \displaystyle \frac{\pi }{3} \right ).\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ &=4\cos ^{3}60^{\circ}\times \sin 60^{\circ}\\ &=4\left ( \displaystyle \frac{1}{2} \right )^{3}\times \left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{4}{16}\sqrt{3}\\ &=\displaystyle \frac{1}{4}\sqrt{3} \end{aligned} \end{array}$

Lanjutan Materi (5) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

 $\color{blue}\textrm{D. Aturan Rantai Turunan Fungsi Trigonometri}$

Jika fungsi  $y=\left ( f\circ g \right )(x)=f\left ( g(x) \right )=f(u)$  dengan  $u=g(x)$, maka turunan dari fungsi komposisi tersebut adalah:

$\color{blue}\begin{matrix} y'=\left ( f\circ g \right )'(x)=f'\left ( g(x) \right )\times g'(x)\\\\ \color{black}\textbf{atau}\\\\ \displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx} \end{matrix}$

Perluasan dari teorema di atas, adalah berikut:

Diberikan $y=\left ( f\circ g\circ h \right )(x)=h\left (f\left ( g(x) \right ) \right )=f(u)$  dengan  $u=g(v)$  dan  $v=h(x)$, maka turunan pertama dari fungsi komposisi tersebut adalah:

$\color{blue}\begin{matrix} y'=\left ( f\circ g\circ h \right )'(x)=f'\left ( g\left ( h(x) \right ) \right )\times g'\left ( h(x) \right )\times h'(x)\\\\ \color{black}\textbf{atau}\\\\ \displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx} \end{matrix}$

$\LARGE\color{magenta}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukan turunan pertama dari}\\ &f(x)=\sin ^{20}\left ( 8x^{5}+\pi \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}f(x)&=\sin ^{20}\left ( 8x^{5}+\pi \right )=\left ( \sin \left ( 8x^{5}+\pi \right ) \right )^{20}\\ \textrm{Dim}&\textrm{isalkan}\\ y& =u^{20},\: \: \textrm{dengan}\\ u&=\sin \left ( 8x^{5}+\pi \right )\: \: \textrm{serta}\: \: u=\sin v\\ &\textrm{dan}\: \: v=\left ( 8x^{5}+\pi \right ),\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=20u^{19}=20\sin ^{19}\left ( 8x^{5}+\pi \right ),\\ \displaystyle \frac{du}{dv}&=\cos v=\cos \left ( 8x^{5}+\pi \right ),\\ \displaystyle \frac{dv}{dx}&=40x^{4}\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ f'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=20\sin ^{19}\left ( 8x^{5}+\pi \right )\times \cos \left ( 8x^{5}+\pi \right )\times 40x^{4}\\ &=800x^{4}\sin ^{19}\left ( 8x^{5}+\pi \right )\cos \left ( 8x^{5}+\pi \right ) \end{aligned} \\\\ &\color{red}\textbf{atau kalau ingin langsungan saja}\\ &\color{red}\textrm{Tentunya jika Anda sudah lancar adalah}\\\\ &\color{purple}\begin{aligned}f(x)&=\sin ^{20}\left ( 8x^{5}+\pi \right )\\ f'(x)&=20\left ( \sin ^{19}\left ( 8x^{5}+\pi \right ) \right )\times \cos \left ( 8x^{5}+\pi \right )\times \left ( 40x^{4} \right )\\ &=800x^{4}\sin ^{19}\left ( 8x^{5}+\pi \right )\cos \left ( 8x^{5}+\pi \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Tentukan turunan pertama dari}\\ &g(x)=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}g(x)&=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}=\cos ^{.^{\frac{3}{5}}}\left ( x^{2}-\pi \right )\\ \textrm{Dim}&\textrm{isalkan}\\ y& =u^{.^{\frac{3}{5}}},\: \: \textrm{dengan}\\ u&=\cos \left ( x^{2}-\pi \right )\: \: \textrm{serta}\: \: u=\cos v\\ &\textrm{dan}\: \: v=\left ( x^{2}-\pi \right ),\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=\displaystyle \frac{3}{5}u^{.^{-\frac{2}{5}}}=\frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right ),\\ \displaystyle \frac{du}{dv}&=-\sin v=-\sin \left ( x^{2}-\pi \right ),\\ \displaystyle \frac{dv}{dx}&=2x\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ g'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=\displaystyle \frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right )\times \left ( -\sin \left ( x^{2}-\pi \right ) \right )\times (2x)\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\cos ^{.^{\frac{2}{5}}}\left ( x^{2}-\pi \right )}\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\sqrt[5]{\cos^{2} \left ( x^{2}-\pi \right )}} \end{aligned} \\\\ &\color{red}\textbf{atau kalau ingin langsungan saja}\\\\ &\color{purple}\begin{aligned}g(x)&=\sqrt[5]{\cos ^{3}\left ( x^{2}-\pi \right )}=\cos ^{.^{\frac{3}{5}}}\left ( x^{2}-\pi \right )\\ g'(x)&=\displaystyle \frac{3}{5}\cos ^{.^{-\frac{2}{5}}}\left ( x^{2}-\pi \right )\times \left ( -\sin \left ( x^{2}-\pi \right ) \right )\times (2x)\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\cos ^{.^{\frac{2}{5}}}\left ( x^{2}-\pi \right )}\\ &=-\displaystyle \frac{6x\sin \left ( x^{2}-\pi \right )}{5\sqrt[5]{\cos^{2} \left ( x^{2}-\pi \right )}} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Tentukan turunan pertama dari}\\ &h(x)=\cos \left ( \sin x^{2020} \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}h(x)&=\cos \left ( \sin x^{2020} \right )\\ \textrm{Dim}&\textrm{isalkan}\\ y& =\cos \left ( \sin x^{2020} \right )=\cos u,\: \: \textrm{dengan}\\ u&=\sin x^{2020}=\sin v\: ,\: \textrm{serta}\: \: v=x^{2020}\\ \color{black}\textrm{mak}&\color{black}\textrm{a}\\ \displaystyle \frac{dy}{du}&=-\sin u=-\sin \left ( \sin x^{2020} \right ),\\ &\: \: \color{red}\textrm{atau}\: \: \color{black}dy=-\sin u\: \: du\\ \displaystyle \frac{du}{dv}&=\cos v\: \: \color{red}\textrm{atau}\: \: \color{black}du=\cos v\: \: dv\\ \displaystyle \frac{dv}{dx}&=2020x^{2019}\: \: \color{red}\textrm{atau}\: \: \color{black}dv=2020x^{2019}\: \: dx\\ \color{black}\textrm{Seh}&\color{black}\textrm{ingga}\\ h'(x)&=\displaystyle \frac{dy}{dx}\\ &=\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}\\ &=-\sin \left ( \sin x^{2020} \right )\times \cos x^{2020}\times \left ( 2020x^{2019} \right )\\ &=-2020x^{2019}\sin \left ( \sin x^{2020} \right )\cos x^{2020} \end{aligned} \\\\ &\color{black}\textbf{atau}\\ &\color{purple}\begin{aligned}dy&=-\sin u\: \: du\\ &=-\sin u\times \cos v\: \: dv\\ &=-\sin u\times \cos v\times \left ( 2020x^{2019} \right )\: \: dx\\ \displaystyle \frac{dy}{dx}&=-\sin u\times \cos v\times \left ( 2020x^{2019} \right )\\ &=.......(\color{black}\textrm{tinggal dimasukkan}) \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI Berdasarkan Standar Isi 2006. Jakarta: ERLANGGA.

Lanjutan Materi (4) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

 $\color{blue}\textrm{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

Sebelumnya silahkan ingat kembali pada dalil-dalil yang berlaku pada materi turunan fungsi aljabar di kelas XI, maka turunan fungsi trigonometri pun serupa, yaitu:

$\begin{array}{|c|l|l|}\hline \textrm{No}&\color{red}\textrm{Fungsi}&\color{blue}\textrm{Turunan Pertama}\\\hline 1.&y=k.u&y'=k.u'\\\hline 2.&y=u\pm v&y'=u'\pm v'\\\hline 3.&y=u.v&y'=v.u'+u.v'\\\hline 4.&y=k.u^{n}&=n.k.u^{(n-1)}.u'\\\hline 5.&y=\displaystyle \frac{u}{v}&y'=\displaystyle \frac{u'.v-u.v'}{v^{2}}\\\hline \end{array}$

Selanjutnya untuk turunan pertama fungsi di atas semisal fungsi  $y=f(x)$ diturunkan terhadap  $x$, maka turunan pertamnya dapat dituliskan dengan

$\color{blue}y'=\displaystyle \frac{dy}{dx}=f'(x)=\underset{h\rightarrow 0}{\textrm{lim}}\: \frac{f(x+h)-f(x)}{h}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$\color{blue}y''=\displaystyle \frac{d^{2}y}{dx^{2}}=f''(x)\: \: \color{black}\textrm{atau kadang dituliskan}\: \: \displaystyle \color{purple}\frac{df'(x)}{dx}=\frac{d^{2}f}{dx^{2}}$

Selanjutnya perhatikanlah tabel berikut

$\color{blue}\begin{array}{|l|l|}\hline \textrm{Turunan}&\qquad\quad\quad\textrm{Notasi}\\\hline \textrm{Pertama}&y'=f'(x)=\displaystyle \frac{dy}{dx}=\frac{df}{dx}\\ &\\ \textrm{Kedua}&y''=f''(x)=\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d^{2}f}{dx^{2}}\\ &\\ \textrm{Ketiga}&y'''=f'''(x)=\displaystyle \frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}}\\ &\\ \cdots &\cdots \qquad\cdots \qquad\cdots \qquad\cdots \\ \textrm{Ke-n}&y^{n}=f^{n}(x)=\displaystyle \frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}\\\hline \end{array}$

$\LARGE\color{magenta}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah turunan pertama dari}\\ &\begin{array}{ll}\\ \textrm{a}.&y=\sin 2x\\ \textrm{b}.&y=\cos 4x\\ \textrm{c}.&y=\sin^{2} x\\ \textrm{d}.&y=\cos^{4} x\\ \textrm{e}.&y=-\sin x\\ \textrm{f}.&y=-5\cos 2x\\ \textrm{g}.&y=-7\tan x\\ \textrm{h}.&y=3\sin^{3} x\\ \textrm{i}.&y=-10\cos^{5} x\\ \textrm{j}.&y=-4\tan^{2} x\\ \textrm{k}.&y=\sqrt{\cos x}\\ \textrm{l}.&y=2\sin x+5x\\ \textrm{m}.&y=3\cos^{2} x+2x^{2}\\ \textrm{n}.&y=\csc x-2\tan ^{2}x+4x\\ \end{array} \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}\textbf{Jawab}&\\ \textrm{Turun}&\textrm{an pertamanya masing}\\ \textrm{fungsi}&\: \textrm{di atas adalah berikut}:\\ (\textrm{a}).\: \: y&=\sin 2x\\ y'&=2\cos 2x\\ (\textrm{b}).\: \: y&=\cos 4x\\ y'&=-4\sin 4x\\ (\textrm{c}).\: \: y&=\sin^{2} x\\ y'&=2\sin x\cos x,\: \: \color{red}\textrm{atau boleh juga}\\ &=\sin 2x\\ (\textrm{d}).\: \: y&=\cos^{4} x\\ y'&=4\cos ^{3}(-\sin x)=-4\cos ^{3}x.\sin x\\ (\textrm{e}).\: \: y&=-2\sin x\\ y'&=-2\cos x\\ (\textrm{f}).\: \: y&=-5\cos 2x\\ y'&=-5(-\sin 2x.(2))=10x\sin 2x\\ (\textrm{g}).\: \: y&=-7\tan x\\ y'&=-7\sec ^{2}x\\ (\textrm{h}).\: \: y&=3\sin^{3} x\\ y'&=3.\left ( 3\sin ^{2}x \right ).(\cos x)=9\sin ^{2}x\cos x\\ (\textrm{i}).\: \: y&=-10\cos^{5} x\\ y'&=5\left ( -10\cos ^{4}x \right ).(-\sin x)\\ &=50\cos ^{4}x\sin x\\ (\textrm{j}).\: \: y&=-4\tan^{2} x\\ &=2\left ( -4\tan x \right ).\left ( \sec ^{2}x \right )\\ &=-8\tan x\sec ^{2}x\\ (\textrm{k}).\: \: y&=\sqrt{\cos x}=\cos ^{.^{\frac{1}{2}}}x\\ y'&=\displaystyle \frac{1}{2}\left ( \cos ^{.^{-\frac{1}{2}}}x \right ).\left ( -\sin x \right )\\ &=-\displaystyle \frac{1}{2}\cos ^{.^{-\frac{1}{2}}}x\sin x\\ &=-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ (\textrm{l}).\: \: y&=2\sin x+5x\\ y'&=2\cos x+5\\ (\textrm{m}).\: \: y&=3\cos^{2} x+2x^{2}\\ y'&=2\left ( 3\cos x \right ).(-\sin x)+4x\\ &=-6\cos x\sin x+4x,\: \: \color{red}\textrm{atau}\\ &=-3\sin 2x+4x=4x-3\sin 2x\\ (\textrm{n}).\: \: y&=\csc x-2\tan ^{2}x+4x\\ &=-\csc x\cot x-2\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )+4\\ &=-\csc x\cot x-4\tan x\sec ^{2}x+4 \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\\ &\begin{array}{ll}\\ \textrm{a}.&f(x)=\displaystyle \frac{1+\sin x}{\cos x}.\: \: \textrm{Tentukanlah}\: \: f'(x)\\ \textrm{b}.&g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=\displaystyle \frac{\pi }{6}\\ \textrm{c}.&h(x)=\sin x\tan x.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=45^{\circ}\\ \textrm{d}.&k(x)=\sin x+n\cos x\: \: \textrm{dan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0.\\ &\textrm{Tentukanlah nilai}\: \: n \end{array}\\\\ &\color{blue}\textbf{Jawab}: \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{a})\: \: \textrm{dike}&\textrm{tahui}\: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ f'(x)&=\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}\\ &=\sec ^{2}x+\sec x\tan x \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{b})\: \: \textrm{dike}&\textrm{tahui}\: \: g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ g'(x)&=\displaystyle \frac{(\cos x-\sin x)(\cos x)-(\sin x+\cos x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x-\sin x\cos x+\sin x+\sin ^{2}x+\sin x\cos x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1}{\cos ^{2}x}\\ g\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{\cos ^{2}\left ( \displaystyle \frac{\pi }{6} \right )}\\ &=\left (\displaystyle \frac{1}{\cos \left ( \displaystyle \frac{\pi }{6} \right )} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\cos 30^{\circ}} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\displaystyle \frac{1}{2}\sqrt{3}} \right )^{2}\\ &=\left ( \displaystyle \frac{2}{\sqrt{3}} \right )^{2}\\ &=\frac{4}{3}\: \: \color{red}\textrm{Jika Anda tidak terganggu dengan nilai}\\ &\color{red}\textrm{perbandingan trigonometri, Anda bisa langsung saja}\\ &\color{red}\textrm{ke jawabannya, yaitu}\: \: \color{blue}\displaystyle \frac{4}{3} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{c})\: \: \textrm{dike}&\textrm{tahui}\: \: h(x)=\sin x\tan x\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'.v+u.v'\\ h'(x)&=\cos x.(\tan x)+\sin x.\left ( \sec ^{2}x \right )\\ h\left ( 45^{\circ} \right )&=\cos \left ( 45^{\circ} \right )\tan \left ( 45^{\circ} \right )+\sin \left ( 45^{\circ} \right )\sec ^{2}\left ( 45^{\circ} \right )\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right ).1+\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \sqrt{2} \right )^{2}\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}\\ &=\displaystyle \frac{3}{2}\sqrt{2} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{d})\: \: \textrm{dike}&\textrm{tahui}\: \: k(x)=\sin x+n\cos x\\ k'(x)&=\cos x-n\sin x,\: \: \color{red}\textrm{dengan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0\\ k'\left ( \displaystyle \frac{\pi }{6} \right )&=\cos \left ( \displaystyle \frac{\pi }{3} \right )-n\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ 0&=\cos 60^{\circ}-n\sin 60^{\circ}=\displaystyle \frac{1}{2}-n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ \Leftrightarrow \quad&n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\displaystyle \frac{1}{2}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{\sqrt{3}}\times \color{black}\frac{\sqrt{3}}{\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{3}\sqrt{3} \end{aligned}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui fungsi}\: \: y=\displaystyle \frac{1}{2}\sin ^{2}x.\: \: \textrm{Tentukanlah}\\ &\textrm{Turunan pertama, kedua, ketiga, keempat},\\ &\textrm{dan kelima dari fungsi tersebut di atas}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\qquad y&=\displaystyle \frac{1}{2}\sin ^{2}x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan pertama}\\ \displaystyle \frac{dy}{dx}&=2\left ( \displaystyle \frac{1}{2}\sin ^{1}x \right )\times \cos x\\ &=\sin x\cos x=\displaystyle \frac{1}{2}\left ( 2\sin x\cos x \right )=\frac{1}{2}\sin 2x\\ \color{black} \textrm{Turu}&\color{black}\textrm{nan keduanya}\\ \displaystyle \frac{d^{2}y}{dx^{2}}&= \displaystyle \frac{1}{2}(\cos 2x).(2)=\color{red}\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan ketiganya}\\ \displaystyle \frac{d^{3}y}{dx^{3}}&=-\sin 2x.(2)=\color{red}-2\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keempatnya}\\ \displaystyle \frac{d^{4}y}{dx^{4}}&=-2\cos 2x.(2)=\color{red}-4\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan kelimanya}\\ \displaystyle \frac{d^{5}y}{dx^{5}}&=-4(-\sin 2x),(2)=\color{red}8\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keenamnya}\\ \displaystyle \frac{d^{6}y}{dx^{6}}&=8\cos 2x.(2)=\color{red}16\cos 2x \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira
2. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
3. Sembiring, S., Zulkifli, M., Marsito & Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT.
4. Tasari, Aksin, N., Miyanto & Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.
5. Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI Berdasarkan Standar Isi 2006. Jakarta: ERLANGGA.