Contoh Soal 2 Turunan Fungsi Trigonometri (Bagian 1)

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

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

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

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

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