Contoh Soal 3 Turunan Fungsi Trigonometri (Bagian 1)

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

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

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

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

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