Contoh Soal 4 Turunan Fungsi Trigonometri (Bagian 1)

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

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

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

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

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