Lanjutan Materi (8) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

 MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{G. Nilai Stasioner}$

Jika fungsi  $y=f(x)$  kontinu dan diferensiabel di  $x=f^{\prime }(a)=0$ , maka fungsi tersebut mempunyai nilai stasioner di  $x=a$.

$\begin{array}{l}\\ \text{a}.& \text{Suatu fungsi memiliki nilai stasioner}\\ & \text{adalah}{f}^{\prime }(x)=0\text{untuk suatu nilai}x\\ \text{b}& \text{Jika fungsi}f(x)\text{mempunyai nilai}f(a)\\ & \text{di}x=a,\text{maka titik}(a,f(a))\text{adalah}\\ & \mathbf{\text{titik stasioner}}\end{array}$

Selanjutnya titik stasioner disebut juga dengan titik kritis atau titik ekstrim dan titik stasioner ini terbagi dalam 3 macam

• titik maksimum
• titik minimum, dan 
• titik belok

Sebagai ilustrasi pada fungsi trigonometri, perhatikanlah ilustrasi fungsi sinus berikut

$\begin{array}{l}\\ \underset{\begin{array}{c}\Downarrow \\ \overbrace{\begin{array}{ccc}\begin{array}{rl}& \text{Maksimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{maksimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)<0\\ & \text{Pada contoh di atas}\\ & \text{Titik A,C,E}\\ & \end{array}& \begin{array}{rl}& \text{Minimum}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{minimum}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)>0\\ & \text{Pada contoh di atas}\\ & \text{Titik B,D}\\ & \end{array}& \begin{array}{rl}& \text{Belok}\\ & \Downarrow \\ & \text{Nilai}\\ & \text{belok}\\ & =f(a)\\ & \text{titiknya}\\ & =(a,f(a))\\ & \text{atau}\\ & f^{″}(x)=0\\ & \text{Pada contoh di atas}\\ & \text{Titik}(-\pi ,0)\\ & (0^{\circ },0),(\pi ,0),(2\pi ,0)\end{array}\end{array}}\end{array}}{\begin{array}{c}\text{Stasioner}\\ f^{\prime }(x)=0\text{saat}x=a\end{array}}\end{array}$

Sebagai catatan bahwa, nilai maksimum dan minimum yang telah di dapatkan sampai dengan memasukkan titik ujinya adalah sebenarnya titik maksimum atau minimum LOKAL dalam selang yang diberikan. Supaya menjadi nilai maksimum atau minimum mutlak, maka nilai-nilai dari nilai stasioner ini harus dibandingkan dengan nilai-nilai FUNGSI pada titik-titik ujung intervalnya yang diberikan tersebut.

$\boxed{\text{CONTOH SOAL}}$

Pada contoh soal LANJUTAN MATERI (7) lihat di sini tentang fungsi naik fungsi turun

$\begin{array}{l}\\ 1.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga ada dua absis yang memenuhi}\\ & \text{sebagai titik STASIONER},\text{yaitu}\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x=\frac{5\pi }{4}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{4}}\\ & f\left({\displaystyle \frac{\pi }{4}}\right)=\sin \left({\displaystyle \frac{\pi }{4}}\right)-\cos \left({\displaystyle \frac{\pi }{4}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}=\sqrt{2}}\\ & \text{untuk}x={\displaystyle \frac{5\pi }{4}}\\ & f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin \left({\displaystyle \frac{5\pi }{4}}\right)+\cos \left({\displaystyle \frac{5\pi }{4}}\right)\\ & =-{\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}}\\ & \text{Jadi titik stasionernya}:\left({\displaystyle \frac{\pi }{4},2}\right)\mathrm{\&}\left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)\\ & \text{Langkah berikutnya gunakanlah titik}\\ & \text{uji di sekitar nilai stasioner yaitu}:\\ & \begin{array}{ccccccccc}& & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & \pi & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Selanjutnya}\\ & \text{Untuk}{f}^{\prime }(x)=\cos x-\sin x\\ & x=0\Rightarrow f^{\prime }(0)=\cos 0-\sin 0\\ & =1+0=1>0(\text{positif})\\ & x=\pi \Rightarrow f^{\prime }(\pi )=\cos \pi -\sin \pi \\ & =-1+0=-1<0(\text{negatif})\\ & x=0\Rightarrow f^{\prime }(2\pi )=\cos 2\pi -\sin 2\pi \\ & =1+0=1>0(\text{positif})\\ & \begin{array}{|cccccl|}\hline x& 0& {\displaystyle \frac{\pi }{4}}& \pi & {\displaystyle \frac{5\pi }{4}}& 2\pi \\ f^{\prime }(x)& +& 0& -& 0& +\\ & & --& & & \\ \text{Garfik}& /& & \mathrm{\setminus }& & /\\ & & & & \mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}& \\ \hline\end{array}\\ & \text{Dari tabel di atas didapatkan}\\ & \{\begin{array}{ll}\left({\displaystyle \frac{\pi }{4},\sqrt{2}}\right)& \text{titik balik maksimum}\\ \left({\displaystyle \frac{5\pi }{4},-\sqrt{2}}\right)& \text{titik balik minimum}\end{array}\end{array}\end{array}$

$\begin{array}{rl}& \text{Sebagai CATATAN bahwa}:\\ & \text{Nilai ujung intervalnya adalah}:\\ & \{\begin{array}{ll}x=0& \Rightarrow f(0)=\sin 0+\cos 0=0+1=1\\ & \text{titiknya}(0,1)\\ x=2\pi & \Rightarrow f(2\pi )=\sin 2\pi +\cos 2\pi =0+1=1\\ & \text{dan titiknya}(2\pi ,1)\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah semua titik stasioner}\\ & \text{berikut jenisnya dari fungsi}\\ & f(x)=\sin 2x\text{dengan}\\ & 0\le x\le 2\pi \\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin 2x\\ & f^{\prime }(x)=2\cos 2x\\ & \text{Stasioner saat}{f}^{\prime }(x)=0\\ & 2\cos 2x=0\\ & \cos 2x=0\\ & \cos 2x=\cos {\displaystyle \frac{\pi }{2}}\\ & 2x=\pm {\displaystyle \frac{\pi }{2}+k.2\pi }\\ & x=\pm {\displaystyle \frac{\pi }{4}+k.\pi }\\ & \text{saat}k=0,\Rightarrow x={\displaystyle \frac{\pi }{4}}\\ & \text{saat}k=1,\Rightarrow x={\displaystyle \frac{5\pi }{4}\text{dan}x={\displaystyle \frac{3\pi }{4}}}\\ & \text{saat}k=2,\Rightarrow x={\displaystyle \frac{7\pi }{4}}\\ & \text{Nilai stasionernya dari absis di atas}:\\ & \bullet f\left({\displaystyle \frac{\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{3\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{3\pi }{4}}\right)=-1\\ & \bullet f\left({\displaystyle \frac{5\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{5\pi }{4}}\right)=1\\ & \bullet f\left({\displaystyle \frac{7\pi }{4}}\right)=\sin 2\left({\displaystyle \frac{7\pi }{4}}\right)=-1\\ & \\ & \text{SILAHKAN LANJUTKAN SENDIRI}\end{array}\end{array}$

Lanjutan Materi (7) Turunan Pertama Fungsi Trigonometri (Matematika Peminatan Kelas XII)

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI TRIGONOMETRI

$\text{F. Fungsi Naik dan Fungsi Turun}$

Dalam menentukan interval-interval di mana fungsi naik atau turun perhatikan dulu ilustrasi berikut ini

Fungsi di atas adalah fungsi $y=f(x)=\sin x$  untuk  $0<x<\pi$  yang memepunya sumbu simetri di  $x={\displaystyle \frac{\pi }{2}=0,5\pi }$. Semua garis singgung yang berada di sebelah kiri sumbu simetri akan mempunyai nilai positif dan semunya garis singgung yang berada di sebelah kanan sumbu simetri bernilai negatif tetapi garis singgung yang tepat pada sumbu simetri memiliki nilai nol.

Pada bahasan sebelumnya-lihat di sini-telah dijelaskan bahwa gradien suatu garis singgung seperti disinggung di atas merupakan nilai dari turunan fungsi pada titik singgung tersebut.

Perhatikanlah gambar ilustrasi berikut

Untuk  

$\bullet m=f^{\prime }(x)>0(\text{tanda positif})$

$\bullet m=f^{\prime }(x)=0$

$\bullet m=f^{\prime }(x)<0(\text{tanda negatif})$

Selanjutnya perhatikan tabel berikut

$\begin{array}{|cll|}\hline \text{Interval}& \text{Nilai}& \text{Keterangan}\\ x<{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)>0& \text{fungsi}f\text{naik}\\ x={\displaystyle \frac{\pi }{2}}& f^{\prime }(x)=0& \text{tidak naik/turun}\\ x>{\displaystyle \frac{\pi }{2}}& f^{\prime }(x)<0& \text{fungsi}f\text{turun}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)=\sin x+\cos x\text{dengan}\\ & 0<x<2\pi \\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\\ & f(x)=\sin x+\cos x\\ & f^{\prime }(x)=\cos x-\sin x\\ & \text{Saat}{f}^{\prime }(x)=0,\\ & f^{\prime }(x)=\cos x-\sin x=0\cos x=\sin x\\ & \cos x=\cos \left({\displaystyle \frac{\pi }{2}-x}\right)\\ & x=\pm \left({\displaystyle \frac{\pi }{2}-x}\right)+k.2\pi \\ & \{\begin{array}{ll}x+x& ={\displaystyle \frac{\pi }{2}+k.2\pi ,\text{atau}}\\ x-x& =-{\displaystyle \frac{\pi }{2}+k.2\pi }\end{array}\\ & \text{maka}\\ & \{\begin{array}{ll}x& ={\displaystyle \frac{\pi }{4}+k.\pi ,\text{atau}}\\ 0& =-{\displaystyle \frac{\pi }{2}+k.2\pi (\text{tidak memenuhi})}\end{array}\\ & \text{Sehingga nilai}x\text{yang memenuhi}:\\ & x={\displaystyle \frac{\pi }{4}\text{dan}x={\displaystyle \frac{5}{4}\pi }}\\ & \begin{array}{c}\\ & & & & & & & \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Pilih titik uji bebas, misalkan}\\ & x={\displaystyle \frac{\pi }{6},x=\frac{\pi }{3},\text{dan}x={\displaystyle \frac{3\pi }{2}}}\\ & \text{untuk}x={\displaystyle \frac{\pi }{6}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{6}}\right)-\sin \left({\displaystyle \frac{\pi }{6}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}(\text{positif})}\\ & \text{untuk}x={\displaystyle \frac{\pi }{3}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{\pi }{3}}\right)-\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & ={\displaystyle \frac{1}{2}-\frac{1}{2}\sqrt{3}(\text{negatif})}\\ & \text{dan untuk}x={\displaystyle \frac{3\pi }{2}}\\ & f^{\prime }(x)=\cos \left({\displaystyle \frac{3\pi }{2}}\right)-\sin \left({\displaystyle \frac{3\pi }{2}}\right)\\ & =0-(-1)=1(\text{positif})\\ & \begin{array}{c}\\ & & & & & & & \\ & ++& & -& -& & ++& \\ 0& & {\displaystyle \frac{\pi }{4}}& & & {\displaystyle \frac{5\pi }{4}}& & 2\pi \end{array}\\ & \text{Berdasarkan garis bilangan di atas}\\ & \bullet f\text{naik pada}:0<x<{\displaystyle \frac{\pi }{4}}\\ & \text{atau}{\displaystyle \frac{5\pi }{4}<x<2\pi }\\ & \bullet f\text{turun pada}:{\displaystyle \frac{\pi }{4}<x<{\displaystyle \frac{5\pi }{4}}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Tentukanlah interval ketika fungsi}\\ & f(x)={\cos }^{2}x\text{dengan}\\ & 0^{\circ }<x<{360}^{\circ }\\ & \text{a}.\text{naik}\\ & \text{b}.\text{turun}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& f(x)={\cos }^{2}x\\ & f^{\prime }(x)=2\cos x(-\sin x)=-2\sin x\cos x\\ & =-\sin 2x\\ & \text{Saat}{f}^{\prime }(x)=0\\ & -\sin 2x=0\\ & \sin 2x=0\\ & \sin 2x=\sin 0^{\circ }\\ & \begin{array}{ll}\begin{array}{rl}2x& =0^{\circ }+k{.360}^{\circ }\\ x& =0^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_1={180}^{\circ }\\ k& =1\Rightarrow x_2={360}^{\circ }\end{array}& \begin{array}{rl}2x& ={180}^{\circ }+k{.360}^{\circ }\\ x& ={90}^{\circ }+k{.180}^{\circ }\\ k& =0\Rightarrow x_3={90}^{\circ }\\ k& =1\Rightarrow x_4={270}^{\circ }\end{array}\end{array}\\ & \text{Lalau kita buat diagram nilai}{f}^{\prime }(x)\text{nya}\\ & \begin{array}{c}\\ & & & & & & & & & & & \\ & --& & ++& & & --& & & ++& \\ 0^{\circ }& & {90}^{\circ }& & & {180}^{\circ }& & & {270}^{\circ }& & {360}^{\circ }\end{array}\\ & \text{Berdasar garis bilangan di atas}\\ & (\text{untuk mengecek gunakan titik uji})\\ & \text{maka fungsi}f(x)\\ & \bullet \text{naik}{90}^{\circ }<x<{180}^{\circ }\text{dan}{270}^{\circ }<x<{360}^{\circ }\\ & \bullet \text{turun}{0}^{\circ }<x<{90}^{\circ }\text{dan}{180}^{\circ }<x<{270}^{\circ }\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA

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

MASALAH YANG MELIBATKAN TURUNAN PERTAMA FUNGSI

$\text{E. Persamaan Garis Singgung}$

1. Fungsi Aljabar

Perhatikanlah gambar berikut!

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

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

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

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  $m={\left({\displaystyle \frac{dy}{dx}}\right)}_{x=a}$ atau kadang juga dituliskan dengan bentuk notasi $m={\begin{array}{c}{\displaystyle \frac{dy}{dx}}\end{array}|}_{x=a}$. 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: $y=m(x-a)+b$, dengan  $(a,b)$  adalah titik singgung. Kada titik singgung juga dituliskan dengan  $(a,f(a))$.

Sehingga persamaan garis singgung kurva di atas adalah:

$\begin{array}{rl}f(x)=y& =x^2-2x+1=(x-1{)}^2\\ m& =2x-2\\ {\left({\displaystyle \frac{dy}{dx}}\right)}_{x=2}& =m=2(2)-2=4-2=2\\ \text{maka}& \text{persamaan garis singgung kurvanya}\\ y& =m(x-a)+b\\ & =2(x-2)+1\\ & =2x-4+1\\ y& =2x-3\end{array}$.

Jadi, garis pada gambar di atas yang merupakan garis singgung kurva yang dimaksud adalah garis $g_3:y=2x-3$.

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

Perhatikan uraian berikut sebagai pembahasan dari permasalahan di atas.

$\begin{array}{rl}\text{Diketahui}& x=a={\displaystyle \frac{\pi }{2}}\\ \text{Kita men}& \text{cari titik singgungnya dulu, yaitu}\\ f(a)& =\sin 2\left({\displaystyle \frac{\pi }{2}}\right)=\sin \pi =0,\\ (a,f(a))& =\left({\displaystyle \frac{\pi }{2},0}\right)\\ f(x)=y& =\sin 2x\\ m& =2\cos 2x......(\mathbf{\text{turunan pertama}})\\ {\left({\displaystyle \frac{dy}{dx}}\right)}_{x=\frac{\pi }{2}}& =m=2\cos 2\left({\displaystyle \frac{\pi }{2}}\right)\\ & =2\cos \pi =2.(-1)=-2\\ \text{maka}& \text{persamaan garis singgung kurvanya}\\ y& =m(x-a)+b\\ & =-2(x-{\displaystyle \frac{\pi }{2}})+0\\ & =-2x+\pi \end{array}$

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)

$\begin{array}{l}\\ 21.& \text{Turunan pertama dari fungsi}\\ & g(x)={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\\ \text{b}.& {\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}}\\ \text{c}.& {\displaystyle \frac{1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{d}.& {\displaystyle \frac{-1}{{\sin }^{2}x{\cos }^{2}x}}\\ \text{e}.& {\displaystyle {\sin }^{2}x{\cos }^{2}x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ g(x)& ={\displaystyle \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}\\ & =\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}={\displaystyle \frac{1}{\sin x\cos x}}\\ \text{maka}& \\ g^{\prime }(x)& ={\displaystyle \frac{0.(\sin x\cos x)-1.({\cos }^{2}x-{\sin }^{2}x)}{(\sin x\cos x{)}^2}}\\ & ={\displaystyle \frac{{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}-\frac{1}{{\sin }^{2}x}}\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Diketahui}h(x)=\cos \left({\displaystyle \frac{3}{x}}\right),\\ & \text{maka}{\displaystyle \frac{dh}{dx}}\\ & \begin{array}{l}\\ \text{a}.& -3\sin {\displaystyle \frac{3}{x}}\\ \text{b}.& -{\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{c}.& -{\displaystyle \frac{3}{x}\sin \frac{3}{x}}\\ \text{d}.& {\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\\ \text{e}.& {\displaystyle \frac{3}{x}\sin \frac{3}{x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos {\displaystyle \frac{3}{x}}& =-\sin {\displaystyle \frac{3}{x}\left({\displaystyle \frac{0.(x)-3.1}{x^2}}\right)}\\ & ={\displaystyle \frac{-(-3)}{x^2}\sin \frac{3}{x}}\\ & ={\displaystyle \frac{3}{x^2}\sin \frac{3}{x}}\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Turunan pertama dari}\tan (\cos x),\\ & \text{terhadap}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\sec }^2(\cos x)\sin x\\ \text{b}.& {\sec }^2(\cos x)\sin x\\ \text{c}.& {\sec }^2(\sin x)\cos x\\ \text{d}.& {\displaystyle \sin x}\\ \text{e}.& {\displaystyle -\sin x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ y& =\tan x(\cos x)\\ y^{\prime }& ={\sec }^2(\cos x)\times (-\sin x)\\ & =-{\sec }^2(\cos x).\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 24.& (\mathbf{\text{UN 2005}})\text{Turunan pertama dari}\\ & f(x)=\sqrt[3]{{\cos }^2(3x^2+5x)}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{b}.& {\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)}\\ \text{c}.& -{\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ \text{d}.& -{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\\ \text{e}.& {\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Misal}& \text{kan}\\ f(x)& =\sqrt[3]{{\cos }^2(3x^2+5x)}\\ f^{\prime }(x)& ={\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)\\ & ={\displaystyle \frac{2}{3}{\cos }^{{.}^{-\frac{1}{2}}}(3x^2+5x)\times (-\sin (3x^2+5x))}\\ & \times (6x+5)\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{-\frac{1}{3}}}(3x^2+5x)\sin (3x^2+5x)}\\ & =-{\displaystyle \frac{2}{3}(6x+5){\cos }^{{.}^{\frac{2}{3}}}(3x^2+5x)}\\ & \times {\cos }^{-1}(3x^2+5x)\times \sin (3x^2+5x)\\ & =-{\displaystyle \frac{2}{3}(6x+5)\tan (3x^2+5x)\sqrt[3]{{\cos }^2(3x^2+5x)}}\end{array}\end{array}$

Contoh Soal 4 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 16.& \text{Turunan pertama dari}f(x)={\displaystyle \frac{1-\cos x}{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{x\sin x+\cos x+1}{x^2}}\\ \text{b}.& {\displaystyle \frac{x\cos x+\sin x-1}{x^2}}\\ \text{c}.& {\displaystyle \frac{x\sin x-\cos x+1}{x^2}}\\ \text{d}.& {\displaystyle \frac{x\sin x+\cos x-1}{x^2}}\\ \text{e}.& {\displaystyle \frac{x\cos x-\sin x+1}{x^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ f(x)& ={\displaystyle \frac{1-\cos x}{x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =1-\cos x\Rightarrow u^{\prime }=\sin x\\ v& =x\Rightarrow v^{\prime }=1\\ \text{maka}& \\ f^{\prime }(x)& ={\displaystyle \frac{\sin x.(x)-(1-\cos x).1}{x^2}}\\ & ={\displaystyle \frac{x\sin x+\cos x-1}{x^2}}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Turunan pertama dari}f(x)={\displaystyle \frac{\tan x}{\cos x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1+{\cos }^{2}x}{{\cos }^{3}x}}\\ \text{b}.& {\displaystyle \frac{1-\cos x}{{\cos }^{3}x}}\\ \text{c}.& {\displaystyle \frac{1+{\sin }^{2}x}{{\cos }^{3}x}}\\ \text{d}.& {\displaystyle \frac{1+\sin x}{{\cos }^{3}x}}\\ \text{e}.& {\displaystyle \frac{1-{\sin }^{2}x}{{\cos }^{3}x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ f(x)& ={\displaystyle \frac{\tan x}{\cos x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\tan x\Rightarrow u^{\prime }={\sec }^{2}x\\ v& =\cos x\Rightarrow v^{\prime }=-\sin x\\ \text{maka}& \\ f^{\prime }(x)& ={\displaystyle \frac{{\sec }^{2}x.(\cos x)-(\tan x).(-\sin x)}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\sec }^{2}x.\cos x+\tan x\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\cos x+\left({\displaystyle \frac{\sin x}{\cos x}}\right)\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{\cos x}+{\displaystyle \frac{{\sin }^{2}x}{\cos x}}}}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1+{\sin }^{2}x}{{\cos }^{3}x}}\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Turunan pertama dari}g(t)={\displaystyle \frac{\cos t+2t}{\sin t}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2\sin t+2t\cos t-1}{{\sin }^{2}t}}\\ \text{b}.& {\displaystyle \frac{2\sin t-2t\cos t+1}{{\sin }^{2}t}}\\ \text{c}.& {\displaystyle \frac{2\sin t+2t\cos t+1}{{\sin }^{2}t}}\\ \text{d}.& {\displaystyle \frac{2\sin t-2t\cos t-1}{{\sin }^{2}t}}\\ \text{e}.& {\displaystyle \frac{-2\sin t+2t\cos t-1}{{\sin }^{2}t}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ g(t)& ={\displaystyle \frac{\cos t+2t}{\sin t}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\cos t+2t\Rightarrow u^{\prime }=-\sin t+2\\ v& =\sin t\Rightarrow v^{\prime }=\cos t\\ \text{maka}& \\ g^{\prime }(t)& ={\displaystyle \frac{(-\sin t+2)(\sin t)-(\cos t+2t)(\cos t)}{{\sin }^{2}t}}\\ & ={\displaystyle \frac{-{\sin }^{2}t+2\sin t-{\cos }^{2}t-2t\cos t}{{\sin }^{2}t}}\\ & ={\displaystyle \frac{t+2\sin t-2t\cos t-{\sin }^{2}t-{\cos }^{2}t}{{\sin }^{2}t}}\\ & ={\displaystyle \frac{t+2\sin t-2t\cos t-({\sin }^{2}t+{\cos }^{2}t)}{{\sin }^{2}t}}\\ & ={\displaystyle \frac{2\sin t-2t\cos t-1}{{\sin }^{2}t}}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Turunan pertama dari}h(x)={\displaystyle \frac{\sin x}{\sin x+\cos x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{{\cos }^{2}x-{\sin }^{2}x}}\\ \text{b}.& {\displaystyle \frac{1}{{\sin }^{2}x-{\cos }^{2}x}}\\ \text{c}.& {\displaystyle \frac{1}{(\sin x+\cos x{)}^2}}\\ \text{d}.& {\displaystyle {\sin }^{2}x-{\cos }^{2}x}\\ \text{e}.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ h(x)& ={\displaystyle \frac{\sin x}{\sin x+\cos x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\sin x\Rightarrow u^{\prime }=\cos x\\ v& =\sin x+\cos x\Rightarrow v^{\prime }=\cos x-\sin x\\ \text{maka}& \\ h^{\prime }(x)& ={\displaystyle \frac{\cos x.(\sin x+\cos x)-\sin x.(\cos x-\sin x)}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{\cos x\sin x+{\cos }^{2}x-\sin x\cos x+{\sin }^{2}x}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{{\sin }^{2}x+{\cos }^{2}x}{(\sin x+\cos x{)}^2}}\\ & ={\displaystyle \frac{1}{(\sin x+\cos x{)}^2}}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahui}f(x)={\displaystyle \frac{\sin x-\cos x}{\tan x}.\text{Nilai}}\\ & \text{turunan pertama fungsi}f\text{saat}x={45}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle \sqrt{2}}\\ \text{e}.& {\displaystyle \sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ f(x)& ={\displaystyle \frac{\sin x-\cos x}{\tan x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\sin x-\cos x\Rightarrow u^{\prime }=\cos x+\sin x\\ v& =\tan x\Rightarrow v^{\prime }={\sec }^{2}x\\ \text{maka}& \\ f^{\prime }(x)& ={\displaystyle \frac{(\cos x+\sin x).\tan x-(\sin x-\cos x).{\sec }^{2}x}{{\tan }^{2}x}}\\ f^{\prime }\left({45}^{\circ }\right)& ={\displaystyle \frac{\left({\displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2}}\right).1-\left({\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}}\right).{\left(\sqrt{2}\right)}^2}{1^2}}\\ & ={\displaystyle \frac{\sqrt{2}-0}{1}}\\ & =\sqrt{2}\end{array}\end{array}$

Contoh Soal 3 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 11.& \text{Turunan pertama fungsi}\\ & h(x)=5\sin x\cos x\text{adalah}{h}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 5\sin 2x\\ \text{b}.& 5\cos 2x\\ \text{c}.& 5{\sin }^{2}x\cos x\\ \text{d}.& 5{\sin }^{2}x{\cos }^{2}x\\ \text{e}.& 5\sin 2x\cos x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}h(x)=5\sin x\cos x\\ h(x)& ={\displaystyle \frac{5}{2}(2\sin x\cos x)={\displaystyle \frac{5}{2}\sin 2x}}\\ h^{\prime }(x)& ={\displaystyle \frac{5}{2}(\cos 2x).(2)}\\ & =5\cos 2x\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Turunan pertama fungsi}\\ & k(x)=\cos x\tan x\text{adalah}{k}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -\sin x\cot x+\cos x{\sec }^{2}x\\ \text{b}.& -\sin x\tan x+\cos x{\sec }^{2}x\\ \text{c}.& \sin x\tan x-\cos x{\sec }^{2}x\\ \text{d}.& -{\displaystyle \frac{1+{\sin }^{2}x}{\cos x}}\\ \text{e}.& {\displaystyle \frac{1+{\sin }^{2}x}{\cos x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}k(x)=\cos x\tan x\\ \text{guna}& \text{kan formula}y=u.v\Rightarrow y^{\prime }=u^{\prime }v+u.v^{\prime }\\ u& =\cos x\Rightarrow u^{\prime }=-\sin x\\ v& =\tan x\Rightarrow v^{\prime }={\sec }^{2}x\\ \text{maka}& \\ k^{\prime }(x)& =(-\sin x)\tan x+\cos x.({\sec }^{2}x)\\ & =-\sin x\tan x+\cos x{\sec }^{2}x\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jika diketahui}f(x)=|\tan x|,\text{maka}{\displaystyle \frac{dy}{dx}}\\ & \text{saat}x=k,\text{di mana}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -\sin k\\ \text{b}.& \cos k\\ \text{c}.& -{\sec }^{2}k\\ \text{d}.& {\sec }^{2}k\\ \text{e}.& \cot k\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}f(x)=|\tan x|\\ \text{saat}& x=k\text{dengan}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ \text{adal}& \text{ah}:\\ f(x)& =|\tan x|,\text{maka saat}x=k\\ f(k)& =|\tan k|=-\tan k,\text{karena di}{\displaystyle \frac{1}{2}\pi <k<\pi }\\ {\displaystyle \frac{dy}{dx}}& =f^{\prime }(k)=-{\sec }^{2}k\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Turunan pertama}g(x)=|\cos x|\\ & \text{adalah}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -|\sin x|\\ \text{b}.& -\sin x\\ \text{c}.& {\displaystyle \frac{\sin 2x}{2|\cos x|}}\\ \text{d}.& -{\displaystyle \frac{\sin 2x}{2|\cos x|}}\\ \text{e}.& |\sin x|\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}g(x)=|\cos x|=\sqrt{{\cos }^{2}x}={({\cos }^{2}x)}^{{.}^{\frac{1}{2}}}\\ g^{\prime }(x)& ={\displaystyle \frac{1}{2}{({\cos }^{2}x)}^{{.}^{-\frac{1}{2}}}.(2\cos x).(-\sin x)}\\ & ={\displaystyle \frac{-2\sin x\cos x}{2{({\cos }^{2}x)}^{{.}^{\frac{1}{2}}}}}\\ & =-{\displaystyle \frac{\sin 2x}{2\sqrt{{\cos }^{2}x}}}\\ & =-{\displaystyle \frac{\sin 2x}{2|\cos x|}}\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Turunan pertama dari}f(x)={\displaystyle \frac{\sin x}{x}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{x\cos x+\sin x}{x^2}}\\ \text{b}.& {\displaystyle \frac{x\cos x-\sin x}{x^2}}\\ \text{c}.& {\displaystyle \frac{-x\cos x-\sin x}{x^2}}\\ \text{d}.& {\displaystyle \frac{\cos x-x\sin x}{x^2}}\\ \text{e}.& {\displaystyle \frac{\cos x+x\sin x}{x^2}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Diket}& \text{ahui}\\ f(x)& ={\displaystyle \frac{\sin x}{x}}\\ \text{Guna}& \text{kan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u.v^{\prime }}{v^2}}}\\ u& =\sin x\Rightarrow u^{\prime }=\cos x\\ v& =x\Rightarrow v^{\prime }=1\\ \text{maka}& \\ f^{\prime }(x)& ={\displaystyle \frac{\cos x.(x)-\sin x.1}{x^2}}\\ & ={\displaystyle \frac{x\cos x-\sin x}{x^2}}\end{array}\end{array}$

Contoh Soal 2 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 6.& \text{Turunan pertama}q(x)={\sin }^{2}x+{\cos }^{2}x\\ & \text{adalah}{q}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\cos }^{2}x-{\sin }^{2}x\\ \text{b}.& 2{\cos }^{2}x-2{\sin }^{2}x\\ \text{c}.& \cos x-\sin x\\ \text{d}.& 2\cos x-2\sin x\\ \text{e}.& 0\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}q(x)& ={\sin }^{2}x+{\cos }^{2}x\\ \text{guna}& \text{kan formula identitas}:{\sin }^{2}x+{\cos }^{2}x=1\\ \text{Sehi}& \text{ngga soal di atas dapat dituliskan menjadi}\\ q(x)& =1,\text{maka}\\ q^{\prime }(x)& =0\\ \text{inga}& \text{t bahwa}y=a\Rightarrow {\displaystyle \frac{dy}{dx}=0}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.& -{\displaystyle \frac{1}{2}}\\ \text{c}.& 0\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dari}& \text{soal diketahui}:\\ f(x)& =\sin {\displaystyle \frac{\pi }{3}}\\ \text{Nila}& \text{i dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{\sin \left({\displaystyle \frac{\pi }{3}+h}\right)-\sin {\displaystyle \frac{\pi }{3}}}{h}}\\ \text{arti}& \text{nya bermakna, berapkah}{f}^{\prime }\left(x\right)?\\ \text{maka}& \\ f^{\prime }\left(x\right)& =0\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jika}f(x)=8x-{\sin }^{3}x,\\ & \text{maka nilai}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 4x^2-3{\cos }^{2}x\\ \text{b}.& 8x-3{\sin }^{2}x\cos x\\ \text{c}.& 8-3{\sin }^{2}x\cos x\\ \text{d}.& 8+{\sin }^{2}x\cos x\\ \text{e}.& 3{\sin }^{2}x\cos x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui dari soal}f(x)=8x-{\sin }^{3}x\\ \text{maka}& \text{nilai dari}\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}=f^{\prime }(x)}\\ f^{\prime }(x)& =8-3{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Turunan pertama fungsi}f(x)=\sqrt{\sin x},\\ & \text{adalah}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2\sqrt{\sin x}}}\\ \text{b}.& {\displaystyle \frac{\cos x}{\sqrt{\sin x}}}\\ \text{c}.& {\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\\ \text{d}.& -{\displaystyle \frac{\sin x}{2\sqrt{\cos x}}}\\ \text{e}.& {\displaystyle \frac{2\cos x}{\sqrt{\sin x}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}f(x)=\sqrt{\sin x}={\sin }^{{.}^{\frac{1}{2}}}x\\ f^{\prime }(x)& ={\displaystyle \frac{1}{2}({\sin }^{{.}^{-\frac{1}{2}}}x).(\cos x)}\\ & ={\displaystyle \frac{\cos x}{2{\sin }^{{.}^{\frac{1}{2}}}x}}\\ & ={\displaystyle \frac{\cos x}{2\sqrt{\sin x}}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}{g}^{\prime }(x)\text{adalah turunan pertama}\\ & \text{fungsi}g(x)\text{dengan}g(x)=5{\tan }^{2}x,\\ & \text{maka}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 10{\cos }^{2}x\sin x\\ \text{b}.& 10{\sin }^{2}x\cos x\\ \text{c}.& {\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\\ \text{d}.& {\displaystyle \frac{10{\cos }^{3}x}{\sin x}}\\ \text{e}.& {\displaystyle \frac{10}{{\sin }^{2}x-{\cos }^{2}x}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Dike}& \text{tahui}g(x)=5{\tan }^{2}x\\ g^{\prime }(x)& =5(2\tan x).({\sec }^{2}x)\\ & =10\tan x\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & =10\left({\displaystyle \frac{\sin x}{\cos x}}\right)\times \left({\displaystyle \frac{1}{{\cos }^{2}x}}\right)\\ & ={\displaystyle \frac{10\sin x}{{\cos }^{3}x}}\end{array}\end{array}$

Contoh Soal 1 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{l}\\ 1.& \text{Diketahui}f(x)=2\cos x-2020\\ & \text{Turunan pertama fungsi}f(x)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2\sin x\\ \text{b}.& -2\sin x\\ \text{c}.& -2\sin x-2020x\\ \text{d}.& 2{\sin }^{2}x\\ \text{e}.& 2\cos x-2020x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(x)& =2\cos x-2020\\ f^{\prime }(x)& =-2\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Jika}{f}^{\prime }(x)\text{adalah turunan pertama dari}\\ & \text{fungsi}f(x)={\sin }^{7}x,\text{maka}{f}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 7{\cos }^{6}x\\ \text{b}.& 7{\cos }^{7}x\\ \text{c}.& 7{\sin }^{6}x\cos x\\ \text{d}.& 7{\cos }^{6}x\sin x\\ \text{e}.& 7{\cos }^{6}x{\sin }^{6}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& ={\sin }^{7}x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ f^{\prime }(x)& =7{\sin }^{6}x(\cos x)=7{\sin }^{6}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Turunan pertama fungsi}g(x)=-5{\sin }^{3}x\\ & \text{adalah}{g}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& -5{\sin }^{2}x\cos x\\ \text{b}.& -5{\sin }^2{\cos }^{2}x\\ \text{c}.& -15{\sin }^{2}x\cos x\\ \text{d}.& -15{\cos }^{3}x\\ \text{e}.& -15{\sin }^{4}x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}g(x)& =-5{\sin }^{3}x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ g^{\prime }(x)& =-5(3{\sin }^{2}x)(\cos x)=-15{\sin }^{2}x\cos x\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Jika}h(x)=4x^3+\sin x+\cos x\\ & \text{maka}{h}^{\prime }(x)=....\\ & \begin{array}{l}\\ \text{a}.& 12x^2+\cos x-\sin x\\ \text{b}.& 12x^2-\cos x+\sin x\\ \text{c}.& 4x^3-\cos x-\sin x\\ \text{d}.& 4x^3-\sin x-\cos x\\ \text{e}.& 12x^3+\cos x+\sin x\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}h(x)& =4x^3+\sin x+\cos x\\ \text{guna}& \text{kan formula}:y=a.u^n\Rightarrow y^{\prime }=n.a.u^{n-1}.u^{\prime }\\ \text{pada}& \text{fungsi aljabarnya, yaitu}:y=4x^3\Rightarrow y^{\prime }=12x^2\\ \text{seda}& \text{ngkan fungsi transendennya mengikuti}\\ \text{turu}& \text{nan fungsi trigonometri biasa. Sehingga}\\ f^{\prime }(x)& =12x^2+\cos x-\sin x\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika}p(x)=-{\cos }^{4}x,\text{maka nilai}\\ & \text{maka}{p}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& \sqrt{3}\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{4}\sqrt{3}}\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}p(x)& =-{\cos }^{4}x\\ p^{\prime }(x)& =-4{\cos }^{3}x.(-\sin x)=4{\cos }^{3}x\sin x\\ p^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)& =4{\cos }^3\left({\displaystyle \frac{\pi }{3}}\right).\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ & =4{\cos }^3{60}^{\circ }\times \sin {60}^{\circ }\\ & =4{\left({\displaystyle \frac{1}{2}}\right)}^3\times \left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{4}{16}\sqrt{3}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}}\end{array}\end{array}$

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

 $\text{D. Aturan Rantai Turunan Fungsi Trigonometri}$

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

$\begin{array}{c}{y}^{\prime }={(f\circ g)}^{\prime }(x)=f^{\prime }(g(x))\times g^{\prime }(x)\\ \\ \mathbf{\text{atau}}\\ \\ {\displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}}\end{array}$

Perluasan dari teorema di atas, adalah berikut:

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

$\begin{array}{c}{y}^{\prime }={(f\circ g\circ h)}^{\prime }(x)=f^{\prime }\left(g(h(x))\right)\times g^{\prime }(h(x))\times h^{\prime }(x)\\ \\ \mathbf{\text{atau}}\\ \\ {\displaystyle \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

$\begin{array}{l}\\ 2.& \text{Tentukan turunan pertama dari}\\ & g(x)=\sqrt[5]{{\cos }^3(x^2-\pi )}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}g(x)& =\sqrt[5]{{\cos }^3(x^2-\pi )}={\cos }^{{.}^{\frac{3}{5}}}(x^2-\pi )\\ \text{Dim}& \text{isalkan}\\ y& =u^{{.}^{\frac{3}{5}}},\text{dengan}\\ u& =\cos (x^2-\pi )\text{serta}u=\cos v\\ & \text{dan}v=(x^2-\pi ),\\ \text{mak}& \text{a}\\ {\displaystyle \frac{dy}{du}}& ={\displaystyle \frac{3}{5}{u}^{{.}^{-\frac{2}{5}}}=\frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi ),}\\ {\displaystyle \frac{du}{dv}}& =-\sin v=-\sin (x^2-\pi ),\\ {\displaystyle \frac{dv}{dx}}& =2x\\ \text{Seh}& \text{ingga}\\ g^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ & ={\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\\ & ={\displaystyle \frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi )\times (-\sin (x^2-\pi ))\times (2x)}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5{\cos }^{{.}^{\frac{2}{5}}}(x^2-\pi )}}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5\sqrt[5]{{\cos }^2(x^2-\pi )}}}\end{array}\\ \\ & \mathbf{\text{atau kalau ingin langsungan saja}}\\ \\ & \begin{array}{rl}g(x)& =\sqrt[5]{{\cos }^3(x^2-\pi )}={\cos }^{{.}^{\frac{3}{5}}}(x^2-\pi )\\ g^{\prime }(x)& ={\displaystyle \frac{3}{5}{\cos }^{{.}^{-\frac{2}{5}}}(x^2-\pi )\times (-\sin (x^2-\pi ))\times (2x)}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5{\cos }^{{.}^{\frac{2}{5}}}(x^2-\pi )}}\\ & =-{\displaystyle \frac{6x\sin (x^2-\pi )}{5\sqrt[5]{{\cos }^2(x^2-\pi )}}}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Tentukan turunan pertama dari}\\ & h(x)=\cos (\sin x^{2020})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}h(x)& =\cos (\sin x^{2020})\\ \text{Dim}& \text{isalkan}\\ y& =\cos (\sin x^{2020})=\cos u,\text{dengan}\\ u& =\sin x^{2020}=\sin v,\text{serta}v=x^{2020}\\ \text{mak}& \text{a}\\ {\displaystyle \frac{dy}{du}}& =-\sin u=-\sin (\sin x^{2020}),\\ & \text{atau}dy=-\sin udu\\ {\displaystyle \frac{du}{dv}}& =\cos v\text{atau}du=\cos vdv\\ {\displaystyle \frac{dv}{dx}}& =2020x^{2019}\text{atau}dv=2020x^{2019}dx\\ \text{Seh}& \text{ingga}\\ h^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ & ={\displaystyle \frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}}\\ & =-\sin (\sin x^{2020})\times \cos x^{2020}\times \left(2020x^{2019}\right)\\ & =-2020x^{2019}\sin (\sin x^{2020})\cos x^{2020}\end{array}\\ \\ & \mathbf{\text{atau}}\\ & \begin{array}{rl}dy& =-\sin udu\\ & =-\sin u\times \cos vdv\\ & =-\sin u\times \cos v\times \left(2020x^{2019}\right)dx\\ {\displaystyle \frac{dy}{dx}}& =-\sin u\times \cos v\times \left(2020x^{2019}\right)\\ & =.......(\text{tinggal dimasukkan})\end{array}\end{array}$

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)

 $\text{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

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

$\begin{array}{|cll|}\hline \text{No}& \text{Fungsi}& \text{Turunan Pertama}\\ 1.& y=k.u& y^{\prime }=k.u^{\prime }\\ 2.& y=u\pm v& y^{\prime }=u^{\prime }\pm v^{\prime }\\ 3.& y=u.v& y^{\prime }=v.u^{\prime }+u.v^{\prime }\\ 4.& y=k.u^n& =n.k.u^{(n-1)}.u^{\prime }\\ 5.& y={\displaystyle \frac{u}{v}}& y^{\prime }={\displaystyle \frac{u^{\prime }.v-u.v^{\prime }}{v^2}}\\ \hline\end{array}$

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

$y^{\prime }={\displaystyle \frac{dy}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$y^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}=f^{″}(x)\text{atau kadang dituliskan}{\displaystyle \frac{d{f}^{\prime }(x)}{dx}=\frac{d^{2}f}{d{x}^2}}}$

Selanjutnya perhatikanlah tabel berikut

$\begin{array}{|ll|}\hline \text{Turunan}& \text{Notasi}\\ \text{Pertama}& y^{\prime }=f^{\prime }(x)={\displaystyle \frac{dy}{dx}=\frac{df}{dx}}\\ & \\ \text{Kedua}& y^{″}=f^{″}(x)={\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{d^{2}f}{d{x}^2}}\\ & \\ \text{Ketiga}& y^{‴}=f^{‴}(x)={\displaystyle \frac{d^{3}y}{d{x}^3}=\frac{d^{3}f}{d{x}^3}}\\ & \\ \cdots & \cdots \cdots \cdots \cdots \\ \text{Ke-n}& y^n=f^n(x)={\displaystyle \frac{d^{n}y}{d{x}^n}=\frac{d^{n}f}{d{x}^n}}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

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

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

$.\begin{array}{rl}2.(\text{a})\text{dike}& \text{tahui}f(x)={\displaystyle \frac{1+\sin x}{\cos x}}\\ \text{gun}& \text{akan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}\\ f^{\prime }(x)& ={\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+{\sin }^{2}x+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}}\\ & ={\sec }^{2}x+\sec x\tan x\end{array}$

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

$.\begin{array}{rl}2.(\text{c})\text{dike}& \text{tahui}h(x)=\sin x\tan x\\ \text{gun}& \text{akan formula}y=u.v\Rightarrow y^{\prime }=u^{\prime }.v+u.v^{\prime }\\ h^{\prime }(x)& =\cos x.(\tan x)+\sin x.({\sec }^{2}x)\\ h\left({45}^{\circ }\right)& =\cos \left({45}^{\circ }\right)\tan \left({45}^{\circ }\right)+\sin \left({45}^{\circ }\right){\sec }^2\left({45}^{\circ }\right)\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right).1+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right){\left(\sqrt{2}\right)}^2\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}}\\ & ={\displaystyle \frac{3}{2}\sqrt{2}}\end{array}$

$.\begin{array}{rl}2.(\text{d})\text{dike}& \text{tahui}k(x)=\sin x+n\cos x\\ k^{\prime }(x)& =\cos x-n\sin x,\text{dengan}{k}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=0\\ k^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& =\cos \left({\displaystyle \frac{\pi }{3}}\right)-n\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ 0& =\cos {60}^{\circ }-n\sin {60}^{\circ }={\displaystyle \frac{1}{2}-n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ \iff & n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)={\displaystyle \frac{1}{2}}\\ \iff & n={\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ \iff & n={\displaystyle \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ \iff & n={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}$

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

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.

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 $\sin (x+h)\cos x$  dan $\cos (x+h)\sin x$, maka saat ketemu bentuk itu kita gunakan rumus:

$\{\begin{array}{ll}\sin A\cos B& ={\displaystyle \frac{1}{2}(\sin (A+B)+\sin (A-B))}\\ \\ \cos A\sin B& ={\displaystyle \frac{1}{2}(\sin (A+B)-\sin (A-B))}\end{array}$

Coba perhatikanlah uraian turunan fungsi tangen berikut:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{...+{\displaystyle \frac{1}{2}\sin h-...+{\displaystyle \frac{1}{2}\sin h}}}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\tan (x+h)-\tan x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)}{\cos (x+h)}-{\displaystyle \frac{\sin x}{\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (x+h)\cos x}{\cos (x+h)\cos x}-{\displaystyle \frac{\cos (x+h)\sin x}{\cos (x+h)\cos x}}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)\cos x-\cos (x+h)\sin x}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin ((x+h)-x)}{\cos (x+h).\cos x.h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h.\cos (x+h).\cos x}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \left({\displaystyle \frac{\sin h}{h}}\right)\left({\displaystyle \frac{1}{\cos (x+h)\cos x}}\right)}\\ & =1\times {\displaystyle \frac{1}{\cos (x+0).\cos x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}}\\ & ={\sec }^{2}x\end{array}$

Berikut hasil turunan pertama untuk fungsi trigonometri yang perlu diingat:

$\begin{array}{rl}1.& f(x)=\sin x\Rightarrow f^{\prime }(x)=\cos x\\ 2.& f(x)=\cos x\Rightarrow f^{\prime }(x)=-\sin x\\ 3.& f(x)=\tan x\Rightarrow f^{\prime }(x)={\sec }^{2}x\\ 4.& f(x)=\cot x\Rightarrow f^{\prime }(x)=-{\csc }^{2}x\\ 5.& f(x)=\sec x\Rightarrow f^{\prime }(x)=\sec x\tan x\\ 6.& f(x)=\csc x\Rightarrow f^{\prime }(x)=-\csc x\cot x\end{array}$

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

$\text{B. Turunan Fungsi Trigonometri}$

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  $f^{\prime }(x)$ adalah:

$f^{\prime }(x)=\underset{h\to 0}{\text{lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}$

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

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2x+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2x+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2x+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2x)}\\ & =\cos x\end{array}$

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

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

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

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

Coba perhatikan penggunaanya berikut, tapi malah agak panjang dikit jadinya

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (x+h)-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x\cos h+\cos x\sin h-\sin x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x(\cos h-1)+\cos x\sin h}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin x(\cos h-1)}{h}+\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos x\sin h}{h}}}\\ & =\sin x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos h-1}{h}+\cos x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h}}}\\ & =\sin x.\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2{\sin }^2{\displaystyle \frac{1}{2}h}}{h}+\cos x.1}\\ & =\sin x.0+\cos x\\ & =\cos x\end{array}$

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

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\cos (x+h)-\cos x}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{-2\sin {\displaystyle \frac{1}{2}(2x+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h).\frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle -2\sin {\displaystyle \frac{1}{2}(2x+h)\times \frac{1}{2}}}\\ & =-2\sin {\displaystyle \frac{1}{2}(2x+0)\times \frac{1}{2}}\\ & =-\sin {\displaystyle \frac{1}{2}(2x)}\\ & =-\sin x\end{array}$

Contoh Soal Turunan Fungsi Trigonomometri (Bagian 1)

$\begin{array}{l}\\ 1.& \text{Sebuah partikel bergerak menurut}f(t)=3\sin {\displaystyle \frac{1}{6}t}\\ & \text{(dalam cm)}.\text{Posisi partikel saat}t=5\pi \text{detik adalah}\\ & ....\text{cm}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{3}{2}\sqrt{3}}\\ \text{b}.& -{\displaystyle \frac{3}{2}}\\ \text{c}.& {\displaystyle \frac{3}{2}}\\ \text{d}.& {\displaystyle \frac{3}{2}\sqrt{2}}\\ \text{e}.& {\displaystyle \frac{3}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(t)& =3\sin {\displaystyle \frac{1}{6}t}\\ \text{posi}& \text{si benda saat}t=5\pi \text{detik adalah}:\\ f(5\pi )& =3\sin {\displaystyle \frac{1}{6}(5\pi )}\\ & =3\sin {\displaystyle \frac{5}{6}\pi }\\ & =3\sin {150}^{\circ }\\ & =3\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{3}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Sebuah objek bergerak melingkar mengikuti rumus}\\ & f(t)=-2\sin t\text{pada waktu}t.\text{Fungsi kecepatan}\\ & \text{dari benda tersebut adalah}....\text{cm}\\ & \begin{array}{l}\\ \text{a}.& -2\cos 2t\\ \text{b}.& -2\cos t\\ \text{c}.& -\cos 2t\\ \text{d}.& \cos t\\ \text{e}.& 2\cos t\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}f(t)& =-2\sin t\\ \text{fun}& \text{gsi kecepatannya adalah}:\\ v(t)& =f^{\prime }(t)=-2\cos t\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Kedudukan sebuah partikel yang bergerak}\\ & \text{mengikuti model}f(t)=\cos 5t\text{(dalam cm)}\\ & \text{pada saat}t=\pi \text{detik adalah}....\\ & \begin{array}{l}\\ \text{a}.& \text{1 cm di bawah sumbu horizontal}\\ \text{b}.& \text{1 cm di atas sumbu horizontal}\\ \text{c}.& \text{5 cm di bawah sumbu horizontal}\\ \text{d}.& \text{5 cm di atas sumbu horizontal}\\ \text{e}.& \text{tepat pada sumbu horizontal}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}f(t)& =\cos 5t\\ \text{kedu}& \text{dukannya adalah}:\\ f\left(\pi \right)& =\cos 5(\pi )\\ & =\cos \pi \\ & =-1\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Kedudukan sebuah partikel yang bergerak}\\ & \text{mengikuti rumus}f(t)=2\sin t+\cos {\displaystyle \frac{1}{2}t}\\ & \text{Selang waktu berikut yang menunjukkan}\\ & \text{posisi partikel di bawah sumbu X dan}\\ & \text{bergerak naik adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0<t<{\displaystyle \frac{1}{2}\pi }\\ \text{b}.& {\displaystyle \frac{1}{2}\pi <t<\pi }\\ \text{c}.& {\displaystyle \frac{1}{2}\pi <t<{\displaystyle \frac{3}{2}\pi }}\\ \text{d}.& \pi <t<{\displaystyle \frac{3}{2}\pi }\\ \text{e}.& {\displaystyle \frac{3}{2}\pi <t<2\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}f(t)& =2\sin t+\cos {\displaystyle \frac{1}{2}t}\\ \text{kedu}& \text{dukannya adalah}:\end{array}\end{array}$

Lanjutan Materi Turunan Fungsi Trigonometri (Matematika Peminatan Kelas XII)

$\begin{array}{l}\\ 3.& \text{Diketahui suatu gelombang bergerak teratur}\\ & \text{sebagaimana gambar berikut}\end{array}$

$.\begin{array}{l}\\ & \text{Gelombang tersebut pada waktu}t\text{detik mengikuti}\\ & \text{rumus}y=f(t)=2\sin {\displaystyle \frac{1}{2}\pi t.\text{Dan diketahui pula cepat}}\\ & \text{rambat gelomnya dapat dinyatakan dalam}\\ & v_t=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}\text{dengan}h\ne 0,\text{tentukanlah}:}\\ & \text{a}.\text{posisi gelombang pada ketika}t=1,5\text{detik}\\ & \text{b}.\text{rumus cepat rambat gelombang pada saat}t\\ & \text{c}.\text{cepat rambat gelombang saat}t=2{\displaystyle \frac{1}{2}\text{detik}}\end{array}$

$.\begin{array}{rl}& \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \text{Posisi gelombang saat}t=1,5={\displaystyle \frac{3}{2}\text{detik}}\\ & f(1,5)=2\sin {\displaystyle \frac{1}{2}\pi \left(\frac{3}{2}\right)}\\ & f(1,5)=2\sin {\displaystyle \frac{3}{4}\pi =2\sin {135}^{\circ }=2\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)=\sqrt{2}}\\ \text{b}.& \text{Cepat rambat gelombang saat}t\text{detik}\\ & v_t=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\sin {\displaystyle \frac{1}{2}\pi (t+h)-2\sin {\displaystyle \frac{1}{2}\pi t}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2.2\cos \left({\displaystyle \frac{1}{2}t+\frac{1}{4}h}\right).\sin {\displaystyle \frac{1}{4}\pi h}}{h}}\\ & =4\times {\displaystyle \frac{1}{4}\pi \times \cos {\displaystyle \frac{1}{2}\pi t}}\\ & =\pi \cos {\displaystyle \frac{1}{2}\pi t}\\ \text{c}.& \text{Cepat rambat saat}t=2{\displaystyle \frac{1}{2}=\frac{5}{2}\text{detik}}\\ & v_t=\pi \cos {\displaystyle \frac{1}{2}\pi t}\\ & =\pi \cos {\displaystyle \frac{1}{2}\pi \left({\displaystyle \frac{5}{2}}\right)}\\ & =\pi \cos {\displaystyle \frac{5}{4}\pi }\\ & =\pi (-{\displaystyle \frac{1}{2}\sqrt{2}})\\ & =-{\displaystyle \frac{1}{2}\pi \sqrt{2}}\end{array}\end{array}$

Turunan Fungsi Trigonometri (Matematika Peminatan Kelas XII)

 $\mathbf{\text{A. Pendahuluan}}$

Konsep turunan fungsi pada awalnya digunakan dalam bidang kususnya Matematika dan fisika, dalam hal hal ini kita berikan contohnya adalah laju perubahan kecepatan.

Coba perhatikanlah, misal pada kasus gerak jatuh bebas suatu benda yang dinyaatakan dengan  $h={\displaystyle \frac{1}{2}g{t}^2}$  dengan  $h$  adalah tinggi benda dengan percepatan grafitasinya adalah  $g=10m/s^2$ dan  $t$  adalah waktu tempuh.

Misalkan suatu benda jatuh dari ketinggian 125 meter dari permukaan tanah dengan percepatan grafitasinya adalah $g=10m/s^2$, maka waktu yang dibutuhkan benda tersebut untuk sampai ke tanah adalah:

$\begin{array}{rl}h& ={\displaystyle \frac{1}{2}g{t}^2}\\ 125& =\frac{1}{2}(10)t^2\\ 25& =t^2\\ 5& =t\end{array}$

Dari kejadian di atas dapat kita dapatkan kecepatan rata-ratanya yaitu: perubahan tinggi dibagi perubahan waktu terjadinya, atau misal dituliskan

$△v={\displaystyle \frac{△y}{△t}={\displaystyle \frac{y_n-y_1}{t_n-t_1}}}$

Sehingga kecepatan rata-ratanya adalah :  ${\displaystyle \frac{125}{5}=25m/s^2}$

Misalkan $f(t)$ untuk fungsi yang menujukkan posisi benda yang terjatuh dalam  $t$ dengan $f(t)=5t^2$, maka kecepatan rata-ratanya kita dapat menghitungnya untuk beberapa selang termasuk kita dapat menghitung kecepatan sesaatnya.

Coba perhatikanlah tabel berikut:

$\begin{array}{|ll|}\hline \{\begin{array}{ll}f(4)& ={5.4}^2=80\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{80-45}{4-3}}\\ & ={\displaystyle \frac{35}{1}=35}\end{array}\\ \{\begin{array}{ll}f(3,5)& =5.(3,5{)}^2=61,25\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{61,25-45}{3,5-3}}\\ & ={\displaystyle \frac{16,25}{0,5}=32,5}\end{array}\\ \{\begin{array}{ll}f(3,25)& =5.(3,25{)}^2=\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{52,8125-45}{3,25-3}}\\ & ={\displaystyle \frac{7,8125}{0,25}=31,25}\end{array}\\ \{\begin{array}{ll}f(3,1)& =5.(3,1{)}^2=48,05\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{48,05-45}{3,1-3}}\\ & ={\displaystyle \frac{3,05}{0,1}=30,5}\end{array}\\ \{\begin{array}{ll}f(3,1)& =5.(3,01{)}^2=45,3005\\ f(3)& ={5.3}^2=45\end{array}& \begin{array}{rl}△v& ={\displaystyle \frac{45,3005-45}{3,01-3}}\\ & ={\displaystyle \frac{0,3005}{0,01}=30,05}\end{array}\\ \hline\end{array}$

Dari ilsutrasi tabel di atas jika selisih waktu diperkecil terus menerus sampai mendekati nol, maka kecepatan sesaatnya akan mendekati nilai 30.

Sehingga kecepatan ketika $t=3$ ditentukan sebagai laju perubahan jarak terhadap waktu yang dibutuhkan dapat dituliskan dengan:

$\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(a+h)-f(a)}{h}}$

Selanjutnya jika benda jatuh yang memenuhi kasus di atas, jika dihitung dengan pendekatan ini saat  $t=3$  adalah:

$\begin{array}{rl}\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t+h{)}^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5(t^2+2th+h^2)-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{5t^2+10th+5h^2-5t^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{10th+5h^2}{h}}\\ & =\underset{h\to 0}{\text{Lim}}10t+5h\\ & =10t\end{array}$

Dari saat  $t=3$  kecepatan sesaatnya adalah $10t=10(3)=30m/s^2$.

Secara matematis, perubahan laju terhadap suatu fungsi di  $x=a$ selanjutnya dinotasikan dengan $f^{\prime }(x)$ dan didefiniskan dengan:

$\boxed{{\displaystyle f^{\prime }(x)=\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(x+h)-f(x)}{h}}}}$

Bentuk di atas dinamakan dengan derivatif atau turunan pertama pada fungsi  $f(x)$  dan dinotasikan dengan  $f^{\prime }(x)$ dan proses pencarian derivatif ini dinamakan differensial.

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah kecepatan jika diketahui}f(t)=\sin t\\ & \text{saat}t\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}{f}^{\prime }(t)=v(t)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin (t+h)-\sin t}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{2\cos {\displaystyle \frac{1}{2}(2t+h)\sin {\displaystyle \frac{1}{2}h}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}2\cos {\displaystyle \frac{1}{2}(2t+h).{\displaystyle \frac{\sin {\displaystyle \frac{1}{2}h}}{h}}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle 2\cos {\displaystyle \frac{1}{2}(2t+h)\times {\displaystyle \frac{1}{2}}}}\\ & =2\cos {\displaystyle \frac{1}{2}(2t+0)\times {\displaystyle \frac{1}{2}}}\\ & =\cos {\displaystyle \frac{1}{2}(2t)}\\ & =\cos t\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Diketahui sebuah bola bergerak melingkar beraturan}\\ & \text{dengan persamaan}f(t)=2\sin 2t.\text{Tentukanlah}\\ & \text{kecepatan bola saat}t={\displaystyle \frac{1}{12}\pi }\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}v(t)& =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{f(t+h)-f(t)}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4\sin 2(t+h)-2\sin 2t}{h}}\\ & =\underset{h\to 0}{\text{Lim}}{\displaystyle \frac{4\cos {\displaystyle \frac{1}{2}(4t+2h)\sin {\displaystyle \frac{1}{2}(2h)}}}{h}}\\ & =\underset{h\to 0}{\text{Lim}}4\cos {\displaystyle \frac{1}{2}(4t+2h)\times \underset{h\to 0}{\text{Lim}}{\displaystyle \frac{\sin h}{h}}}\\ & =4\cos {\displaystyle \frac{1}{2}(4t)}\\ & =4\cos 2t\\ v\left({\displaystyle \frac{1}{12}\pi }\right)& =4\cos 2\left({\displaystyle \frac{1}{12}\pi }\right)\\ & =4\cos {\displaystyle \frac{1}{6}\pi }\\ & =4\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & =2\sqrt{3}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Noormandiri, B. K. 2004. Matematika SMA Jilid $2^A$ Berdasarkan Kurikulum 2004. Jakarta: ERLANGGA.
2. Noormandiri, B. K. 2017. Matematika Jilid 3 untuk SMA/MA Kelas XII Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.