Contoh Soal 3 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 11.&\textrm{Turunan pertama fungsi}\\ &h(x)=5\sin x\cos x\: \: \textrm{adalah}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\sin 2x\\ \color{red}\textrm{b}.&5\cos 2x\\ \textrm{c}.&5\sin ^{2}x\cos x\\ \textrm{d}.&5\sin ^{2}x\cos^{2} x\\ \textrm{e}.&5\sin 2x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}h(x)=5\sin x\cos x\\ h(x)&=\color{red}\displaystyle \frac{5}{2}\left ( 2\sin x\cos x \right )=\displaystyle \frac{5}{2}\sin 2x\\ h'(x)&=\color{purple}\displaystyle \frac{5}{2}\left ( \cos 2x \right ).(2)\\ &=\color{purple}5\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Turunan pertama fungsi}\\ &k(x)=\cos x\tan x\: \: \textrm{adalah}\: \: k'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin x\cot x+\cos x\sec ^{2}x\\ \color{red}\textrm{b}.&-\sin x\tan x+\cos x\sec ^{2}x\\ \textrm{c}.&\sin x\tan x-\cos x\sec ^{2}x\\ \textrm{d}.&-\displaystyle \frac{1+\sin ^{2}x}{\cos x}\\ \textrm{e}.&\displaystyle \frac{1+\sin ^{2}x}{\cos x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}k(x)=\cos x\tan x\\ \textrm{guna}&\textrm{kan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'v+u.v'\\ u&=\color{black}\cos x \Rightarrow u'=-\sin x\\ v&=\color{black}\tan x \Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ k'(x)&=\left ( -\sin x \right )\tan x+\cos x.\left ( \sec ^{2}x \right )\\ &=-\sin x\tan x+\cos x\sec ^{2}x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Jika diketahui}\: \: f(x)=\left | \tan x \right |,\: \textrm{maka}\: \: \displaystyle \frac{dy}{dx}\\ &\textrm{saat}\: \: x=k,\: \: \textrm{di mana}\: \: \displaystyle \frac{1}{2}\pi <k<\pi\\ & \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin k\\ \textrm{b}.&\cos k\\ \color{red}\textrm{c}.&-\sec ^{2}k\\ \textrm{d}.&\sec ^{2}k\\ \textrm{e}.&\cot k \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}f(x)=\left |\tan x \right |\\ \textrm{saat}&\: \: \color{red}x=k\: \: \color{blue}\textrm{dengan}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \color{black}\textrm{adal}&\color{black}\textrm{ah}:\\ f(x)&=\left | \tan x \right |,\: \: \color{black}\textrm{maka saat}\: \: \color{blue}x=k\\ f(k)&=\left | \tan k \right |=-\tan k,\: \: \color{black}\textrm{karena di}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \displaystyle \frac{dy}{dx}&=f'(k)=-\sec ^{2}k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 14.&\textrm{Turunan pertama}\: \: g(x)=\left | \cos x \right |\\ & \textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\left | \sin x \right |\\ \textrm{b}.&-\sin x\\ \textrm{c}.&\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \color{red}\textrm{d}.&-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{e}.&\left | \sin x \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}g(x)=\left |\cos x \right |=\sqrt{\cos ^{2}x}=\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}\\ g'(x)&=\color{purple}\displaystyle \frac{1}{2}\left ( \cos ^{2}x \right )^{.^{-\frac{1}{2}}}.\left ( 2\cos x \right ).\left ( -\sin x \right )\\ &=\color{purple}\displaystyle \frac{-2\sin x\cos x}{2\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}}\\ &=\color{blue}-\displaystyle \frac{\sin 2x}{2\sqrt{\cos ^{2}x}}\\ &=\color{blue}-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 15.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\sin x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\cos x+\sin x}{x^{2}}\\ \color{red}\textrm{b}.&\displaystyle \frac{x\cos x-\sin x}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{-x\cos x-\sin x}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{\cos x-x\sin x}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{\cos x+x\sin x}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x\Rightarrow u'=\cos x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\cos x.(x)-\sin x.1}{x^{2}}\\ &=\displaystyle \frac{x\cos x-\sin x}{x^{2}} \end{aligned} \end{array}$