$\begin{array}{ll}\\ 16.&\textrm{Jika}\: \: W=\sin 2t\: ,\: \textrm{maka}\: \: \displaystyle \frac{dW}{dt}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos 2t&\textrm{d}.\quad 2t\cos 2t+\sin 2t\\ \textrm{b}.\quad 2\cos 2t\quad &\textrm{e}.\quad \sin 2t-t\cos 2t\\ \textrm{c}.\quad \sin 2t+t\cos 2t\quad \end{array} \end{array}$.
Jawaban: $\begin{array}{ll} &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &W=\sin 2t\\ &W=\sin u\, \quad \textrm{dengan}\: \: u=2t\\ &\displaystyle \frac{dW}{dt}=\displaystyle \frac{dW}{du}.\frac{du}{dt}\\ &\qquad=\cos u.2\\ &\qquad=2\cos u\\ &\qquad=2\cos 2t \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: f(x)=2\cos x-2020\\ &\textrm{Turunan pertama fungsi}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&2\sin x\\ \textrm{b}.&-2\sin x\\ \textrm{c}.&-2\sin x-2020x\\ \textrm{d}.&2\sin ^{2}x\\ \textrm{e}.&2\cos x-2020x \end{array}\\\end{array}$
Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}f(x)&=2\cos x-2020\\ f'(x)&=-2\sin x \end{aligned}\end{array}$
$\begin{array}{ll}\\ 18.&\textrm{Jika}\: \: f'(x)\: \: \textrm{adalah turunan pertama dari}\\ &\textrm{fungsi}\: \: f(x)=\sin ^{7}x\: ,\: \textrm{maka}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\cos^{6} x\\ \textrm{b}.&7\cos^{7} x\\ \textrm{c}.&7\sin^{6} x\cos x\\ \textrm{d}.&7\cos ^{6}x\sin x\\ \textrm{e}.&7\cos ^{6}x\sin ^{6}x \end{array}\\\end{array}$
Jawaban: $\begin{array}{ll}&\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}f(x)&=\sin ^{7}x\\ \textrm{guna}&\textrm{kan formula}:\: y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ f'(x)&=7\sin ^{6}x\left ( \cos x \right )=7\sin ^{6}x\cos x \end{aligned}\end{array}$
$\begin{array}{ll}\\ 19.&\textrm{Turunan pertama fungsi}\: \: g(x)=-5\sin ^{3}x\\ &\textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll} \textrm{a}.&-5\sin ^{2}x\cos x\\ \textrm{b}.&-5\sin ^{2}\cos ^{2}x\\ \textrm{c}.&-15\sin ^{2}x\cos x\\ \textrm{d}.&-15\cos ^{3}x\\ \textrm{e}.&-15\sin ^{4}x \end{array}\\\end{array}$
Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}g(x)&=-5\sin ^{3}x\\ \textrm{guna}&\textrm{kan formula}:\: y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ g'(x)&=-5\left ( 3\sin ^{2}x \right )(\cos x)=-15\sin ^{2}x\cos x \end{aligned}\end{array}$
$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x} \: ,\: \textrm{maka}\: \: {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2} &&\textrm{d}.\quad 2 \\ \textrm{b}.\quad \displaystyle -\frac{1}{2} \quad &\textrm{c}.\quad \displaystyle \frac{3}{4} \quad &\textrm{e}.\quad -2 \end{array}\\\end{array}$.
Jawaban: $\begin{array}{ll}&\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{1+\sin x}{\cos x}\\ {f}\, '(x)&=\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x +\sin ^{2}x}{\cos ^{2}x}\\ &\qquad \textrm{ingat bahwa}\: \: \sin ^{2}x+\cos ^{2}x=1\\ &=\displaystyle \frac{1+\sin x }{\cos ^{2}x}\\ {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )&=\displaystyle \frac{1+\sin \left ( \displaystyle \frac{1}{6}\pi \right )}{\cos ^{2}\left ( \displaystyle \frac{1}{6}\pi \right )}\\&=\displaystyle \frac{1+\displaystyle \frac{1}{2} }{\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{3}{2}\times \frac{4}{3}\\ &=2 \end{aligned} \end{array}$.
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