PAS MATEMATIKA BLOG

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

Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira.
Tampomas, H. 1999. SeribuPena Matematika SMU Kelas 2. Jakarta: ERLANGGA
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

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

Kurnia, N. 2018. Jelajah Matematika 3 SMA Kelas XII Peminatan MIPA. Bogor: Yudhistira
Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
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.
Tasari, Aksin, N., Miyanto & Muklis. 2016. Matematika untuk SMA/MA Kelas XII Peminatan Matematika dan Ilmu-Ilmu Alam. Klaten: INTAN PARIWARA.
Wirodikromo, S. 2007. Matematika Jilid 2 IPA untuk Kelas XI Berdasarkan Standar Isi 2006. Jakarta: ERLANGGA.

Lanjutan Materi (3) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

Selanjutnya saat kita masih kukuh menggu nakan rumus semual, maka saat menentukan turunan pertama fungsi $\tan x$, kita akan ketemu bentuk $\color{blue}\sin (x+h)\cos x$ dan $\color{blue}\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:

$\color{purple}\begin{cases} \sin A\cos B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)+\sin (A-B) \right ) \\\\ \cos A\sin B&=\displaystyle \frac{1}{2}\left ( \sin (A+B)-\sin (A-B) \right ) \end{cases}$

Coba perhatikanlah uraian turunan fungsi tangen berikut:

$\color{purple}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{...+\displaystyle \frac{1}{2}\sin h-...+\displaystyle \frac{1}{2}\sin h}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$

Atau kita juga dapat menggunakan rumus $\color{red}\sin (A-B)=\sin A\cos B-\cos A\sin B$ sebagaimana berikut ini (perhatikanlah proses langkah 5 ke langkah 6):

$\color{blue}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\tan (x+h)-\tan x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)}{\cos (x+h)}-\displaystyle \frac{\sin x}{\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\displaystyle \frac{ \sin (x+h)\cos x}{\cos (x+h)\cos x}-\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin \left ((x+h)-x \right )}{\cos (x+h).\cos x.h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \left ( \displaystyle \frac{\sin h}{h} \right )\left ( \displaystyle \frac{1}{\cos (x+h)\cos x} \right )\\ &=1\times \displaystyle \frac{1}{\cos (x+0).\cos x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}\\ &=\sec ^{2}x \end{aligned}$

Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:

$\color{purple}\begin{aligned}1.\quad &f(x)=\sin x\Rightarrow f'(x)=\cos x\\ 2.\quad &f(x)=\cos x\Rightarrow f'(x)=-\sin x\\ 3.\quad &f(x)=\tan x\Rightarrow f'(x)=\sec ^{2}x\\ 4.\quad &f(x)=\cot x\Rightarrow f'(x)=-\csc ^{2}x\\ 5.\quad &f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x\\ 6.\quad &f(x)=\csc x\Rightarrow f'(x)=-\csc x\cot x \end{aligned}$

Lanjutan Materi (2) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\color{blue}\textrm{B. Turunan Fungsi Trigonometri}$

Fungsi trigonometri di sini adalah suatu fungsi yang mengandung perbandingan trigonometri serta perbandingan trigonometri tersebut bukan merupakan ekponen

Kita ingat sebelumnya untuk menentukan turunan pertama suatu fungsi $f(x)$ yang selanjutnya di dinotasikan dengan $f'(x)$ adalah:

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

Selanjutnya dalam menentukan turunan formula di atas dapat digunakan untuk menentukan turunan pertama fungsi trigonometri, sebagai mana contoh berikut:

Ambil contoh $\color{purple}f(x)=\sin x$, maka kita akan menentukan turuan pertamanya, yaitu:

Pada salah satu langkah di antara langkah di atas ada beberapa rumus yang perlu diingat saat Anda duduk di kelas XI, yaitu penggunaan rumus

$\color{blue}\sin A-\sin B=2\cos \displaystyle \frac{1}{2}(A+B)\sin \displaystyle \frac{1}{2}(A-B)$.

Anda boleh juga menggunakan rumus yang lain. Karena di dalamnya ada $\sin (x+h)$, Anda dapat menggunakan rumus berikut:

$\color{blue}\sin (A+B)=\sin A\cos B+\cos A\sin B$

Coba perhatikan penggunaanya berikut, tapi malah agak panjang dikit jadinya

$\color{purple}\begin{aligned}f'(x)&=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin (x+h)-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\left ( \cos h-1 \right )+\cos x\sin h}{h}\\ &=\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin x\left ( \cos h-1 \right )}{h}+\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos x\sin h}{h}\\ &=\sin x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\cos h-1}{h}+\cos x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{\sin h}{h}\\ &=\sin x.\underset{h\rightarrow 0}{\textrm{Lim}}\: \: \displaystyle \frac{-2\sin ^{2}\displaystyle \frac{1}{2}h}{h}+\cos x.1\\ &=\sin x.0+\cos x\\ &=\cos x \end{aligned}$

Sampai di sini kita akan bisa coba lagi menentukan turunan pertama fungsi kosinus, sebagaimana uraian berikut:

Contoh Soal 6 Fungsi Logaritma (Uraian)

$\begin{array}{l}\\ 26.&\textrm{Diketahui bahwa}\\ & ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\\ &\textrm{maka nilai}\: \: ^{^{44}}\log 66=....\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\frac{1+p+pq}{2+pq} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textbf{(AIME 1984)}\\ &\textrm{Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\\ & \textrm{adalah bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7\\ &\Leftrightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|\\ &\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}....(3)\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|\\ &\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7....(4)\\ &\textrm{saat persamaan}\: \: (3)-(4)\\ &=-\frac{8}{9}.\: ^{^{^{2}}}\log x=\frac{5}{3}-7=-\frac{16}{3}\\ &\color{purple}\textrm{maka}\\ &^{^{^{2}}}\log x=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ &^{^{^{2}}}\log x=6\Leftrightarrow x=2^{6}\Leftrightarrow x=64\\ &\color{purple}\textrm{Pada persamaan 1 selanjutnya}\\ &\frac{1}{3}.\: ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: \frac{1}{3}.6+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: 2+\: ^{^{^{2}}}\log y=5\\ &\Leftrightarrow \: \: ^{^{^{2}}}\log y=5-2=3\Leftrightarrow y=2^{3}=8\\ &\textrm{Jadi},\: \: x.y=64.8=512 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=3\sqrt{5} \end{aligned}\\\hline \end{array}\\ &\color{blue}\begin{array}{|l|}\hline \begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=1\end{aligned}\\\hline \end{array}\\ &\color{purple}\textrm{Pembahasan diserahkan kepada}\\ &\color{purple}\textrm{Pembaca yang budiman untuk poin c} \end{array}$

$\begin{array}{ll}\\ 29.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\\ &\textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}&^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\textrm{diambil}\: \textrm{persamaannya, maka}\\ &\displaystyle \frac{\left ( 1+a^{2} \right )}{25}=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ &\displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\color{purple}\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak}&\cdots \\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 30.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\\ &\textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\color{blue}\textrm{Perhatikanlah bahwa}\\ &\color{blue}\begin{aligned}&\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases} \end{aligned} \\ &\color{blue}\textrm{Selanjutnya}\\ &\color{blue}\textrm{Untuk}\: \: (1-a-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: 2(1-b),\: \textrm{maka}\\ &\color{blue}\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned} \\ &\color{blue}\textrm{Untuk}\: \: \left ( \displaystyle \frac{1-x-y}{2(1-b)} \right ),\: \textrm{maka}\\ &\color{blue}\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\ &\color{blue}\textrm{Jadi},\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}=12^{^{\: ^{12}\log 2}}=2 \end{array}$

$\begin{array}{ll}\\ 31.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\\ & \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\\ & \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\color{blue}\begin{aligned}&2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ &\textrm{maka}\\ &^{x}\log (2y)=\: ^{2x}\log (4y)\\ &\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)...(1)\\ &^{x}\log (2y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)....(2)\\ &^{2x}\log (4y)=\: ^{4x}\log (8yz)\\ &\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)....(3) \end{aligned}\\\\ &\color{blue}\begin{aligned}&\textrm{Perhatikan persamaan}\: \: (2),\: \textrm{yaitu}:\\ &\log (2y)\times \log (4x)=\log x\times \log (8yz)\\ &\log (2y)\times \left (\log (2x)+\log 2 \right )=\log x\times \log (8yz)\\ &\log (2y)\times \log (2x)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\log x\times \log (4y)+\log (2y)\times \log 2=\log x\times \log (8yz)\\ &\quad\textrm{persamaan di atas, persamaan}\: \: (1)\: \: \textrm{disubstitusikan}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: ......(4) \end{aligned}\\\\ &\color{blue}\begin{aligned}&\textrm{Perhatikan juga persamaan}\: \: (3),\: \textrm{yaitu}:\\ &\log (4y)\times \log (4x)=\log (2x)\times \log (8yz)\\ &\left (\log (2y)+\log 2 \right )\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log (2y)\times \log (4x)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\log x\times \log (8yz)+\log 2\times \log (4x)=\log (2x)\times \log (8yz)\\ &\quad\textrm{di atas, persamaan}\: \: (2)\: \: \textrm{disubstitusikan}\\ &\log 2\times \log (4x)=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ &\log 2\times \log (4x)=\log (8yz)\times \log 2\\ &\log 4x=\log (8yz)\\ &4x=8yz\\ &\displaystyle \frac{x}{z}=2y\: ....(5) \end{aligned} \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{dari persamaan}\: \: (4)\: \: \textrm{dan}\: \: (5)\\ &\log (2y)=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log \left ( \displaystyle \frac{x}{z} \right )=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ &\log 2\left ( \log x-\log z \right )=\log x\times \log (2z)\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \left ( \log 2+\log z \right )\\ &\log 2\times \log x-\log 2\times \log z=\log x\times \log 2+\log x\times \log z\\ &-\log 2\times \log z=\log x\times \log z\\ &\log 2^{-1}=\log x\\ &\displaystyle \frac{1}{2}=x\: .....(6) \end{aligned}$

$.\: \: \qquad\color{blue}\begin{array}{|c|c|}\hline \textrm{persamaan}\: \: (2)&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{4}\: .....(7)\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}$

$.\: \: \qquad\color{purple}\begin{aligned}&\textrm{maka nilai untuk}\: \: xy^{5}z\: \: \textrm{adalah}\\ &xy^{5}z=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\\ &\begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ &\textrm{Jadi},\\ &p+q=11+1=12 \end{aligned}$

DAFTAR PUSTAKA

Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.

Contoh Soal 5 Fungsi Logaritma

$\begin{array}{ll}\\ 21.&\textbf{(SPMB '04)}\\ &\textrm{Jika}\: \: a>1\: ,\: \textrm{maka penyelesaian untuk}\\ &\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )=1\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 3\\ \color{red}\textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\left (^{a}\log (2x+1) \right )\left ( ^{3}\log \sqrt{a} \right )&=1\\ \left ( ^{3}\log \sqrt{a} \right )\left (^{a}\log (2x+1) \right )&=1\\ \left ( ^{3}\log a^{.^{^{\frac{1}{2}}}} \right )\left (^{a}\log (2x+1) \right )&=1\\ \displaystyle \frac{1}{2}\left ( ^{3}\log a \right )\left (^{a}\log (2x+1) \right )&=1\\ ^{3}\log (2x+1)&=2\\ 2x+1&=3^{2}\\ 2x&=9-1\\ 2x&=8\\ x&=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textbf{(SPMB '04)}\\ &\textrm{Nilai}\: \: \displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\: \: \textrm{adalah}\: ....\\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\\ \textrm{b}.&\displaystyle 1\\ \color{red}\textrm{c}.&\displaystyle 2\\ \textrm{d}.&\displaystyle 4\\ \textrm{e}.&\displaystyle 5 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&\displaystyle \frac{\left ( ^{5}\log 10 \right )^{2}-\left ( ^{5}\log 2 \right )^{2}}{^{5}\log \sqrt{20}}\\ &=\displaystyle \frac{\left ( ^{5}\log 10+\: ^{5}\log 2 \right )\left ( ^{5}\log 10-\: ^{5}\log 2 \right )}{^{5}\log (20)^{.^{\frac{1}{2}}}}\\ &=\displaystyle \frac{^{5}\log (10.2)\times ^{5}\log \left (\frac{10}{2} \right )}{\displaystyle \frac{1}{2}\times \: ^{5}\log 20}\\ &=2\times \left ( \displaystyle \frac{^{5}\log 20}{^{5}\log 20} \right )\times \: ^{5}\log 5\\ &=2.1.1\\ &=2 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textbf{(SPMB '03)}\\ &\textrm{Jika diketahui bahwa}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle ^{2}\log x=8\\ \textrm{b}.&\displaystyle ^{2}\log x=4\\ \color{red}\textrm{c}.&\displaystyle ^{4}\log x=8\\ \textrm{d}.&\displaystyle ^{4}\log x=16\\ \textrm{e}.&\displaystyle ^{16}\log x=8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 4^{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log 2=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{2^{2}}\log 2^{1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{4}\log \left ( \displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\: ^{2^{2}}\log 2^{-1}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x-\left ( -\displaystyle \frac{1}{2} \right )=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x+\frac{1}{2}=2\\ &\Leftrightarrow \: ^{4}\log \: ^{4}\log x=2-\displaystyle \frac{1}{2}=\frac{3}{2}\\ &\Leftrightarrow \: ^{4}\log x=4^{.\frac{3}{2}}\\ &\Leftrightarrow \: ^{4}\log x=\left (2^{2} \right )^{.^{\frac{3}{2}}}\\ &\Leftrightarrow \: ^{4}\log x=2^{3}\\ &\Leftrightarrow \: ^{4}\log x=8\\ \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textbf{(UMPTN '92)}\\ &\textrm{Jika}\: \: x\: \: \textrm{memenuhi persamaan}\\ &^{4}\log ^{4}\log x-\: ^{4}\log ^{4}\log ^{4}\log 16=2\\ & \textrm{maka nilai}\: \: ^{16}\log x=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle 4\\ \textrm{b}.&\displaystyle 2\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle -2\\ \textrm{e}.&\displaystyle -4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{4}\log \: ^{4}\log x-\: ^{4}\log \: ^{4}\log \: ^{4}\log 16=2\\ &\textrm{menyebabkan}\\ & ^{4}\log x=8\Rightarrow x=4^{8}\\ &(\color{purple}\textrm{lihat pembahasan no.23})\\ &\textrm{maka},\\ &\: ^{16}\log x=\: ^{16}\log 4^{8}=\: ^{4^{2}}\log 4^{8}\\ &=\displaystyle \frac{8}{2}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textbf{(UMPTN '94)}\\ &\textrm{Hasil kali akar-akar persamaan}\\ &^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ & \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&\displaystyle \frac{1}{9}\\ \textrm{b}.&\displaystyle \frac{1}{3}\\ \textrm{c}.&\displaystyle 1\\ \textrm{d}.&\displaystyle 3\\ \textrm{e}.&\displaystyle 9 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}&^{3}\log x^{.\left (^{2+\: ^{3}\log x} \right )}=15\\ &\Leftrightarrow \: \left ({2+\: ^{3}\log x} \right )^{3}\log x-15=0\\ &\Leftrightarrow 2\: ^{3}\log x+\: \left ( ^{3}\log x \right )^{2}-15=0\\ &\Leftrightarrow \: \left ( ^{3}\log x \right )^{2}+2\: ^{3}\log x-15=0\\ &\Leftrightarrow \: \left (^{3}\log x_{1}+5 \right )\left ( ^{3}\log x_{2}-3 \right )=0\\ &\Leftrightarrow \: ^{3}\log x_{1}+5=0\: \: \textrm{atau}\: \: ^{3}\log x_{1}-3=0\\ &\Leftrightarrow \: ^{3}\log x_{1}=-5\: \: \textrm{atau}\: \: ^{3}\log x_{2}=3\\ &\Leftrightarrow \: x_{1}=3^{-5}\: \: \textrm{atau}\: \: x_{2}=3^{3}\\ &\qquad \textrm{maka}\\ &\Leftrightarrow \: x_{1}\times x_{2}=3^{-5}\times 3^{3}=3^{-5+3}=3^{-2}\\ &\Leftrightarrow \qquad =\displaystyle \frac{1}{3^{2}}\\ &\Leftrightarrow \qquad =\frac{1}{9} \end{aligned} \end{array}$