Contoh Soal 1 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: f(x)=2\cos x-2020\\ &\textrm{Turunan pertama fungsi}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&2\sin x\\ \color{red}\textrm{b}.&-2\sin x\\ \textrm{c}.&-2\sin x-2020x\\ \textrm{d}.&2\sin ^{2}x\\ \textrm{e}.&2\cos x-2020x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}f(x)&=2\cos x-2020\\ f'(x)&=-2\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Jika}\: \: f'(x)\: \: \textrm{adalah turunan pertama dari}\\ &\textrm{fungsi}\: \: f(x)=\sin ^{7}x\: ,\: \textrm{maka}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&7\cos^{6} x\\ \textrm{b}.&7\cos^{7} x\\ \color{red}\textrm{c}.&7\sin^{6} x\cos x\\ \textrm{d}.&7\cos ^{6}x\sin x\\ \textrm{e}.&7\cos ^{6}x\sin ^{6}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}f(x)&=\sin ^{7}x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ f'(x)&=7\sin ^{6}x\left ( \cos x \right )=7\sin ^{6}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Turunan pertama fungsi}\: \: g(x)=-5\sin ^{3}x\\ &\textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-5\sin ^{2}x\cos x\\ \textrm{b}.&-5\sin ^{2}\cos ^{2}x\\ \color{red}\textrm{c}.&-15\sin ^{2}x\cos x\\ \textrm{d}.&-15\cos ^{3}x\\ \textrm{e}.&-15\sin ^{4}x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}g(x)&=-5\sin ^{3}x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ g'(x)&=-5\left ( 3\sin ^{2}x \right )(\cos x)=-15\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: h(x)=4x^{3}+\sin x+\cos x\\ &\textrm{maka}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&12x^{2}+\cos x-\sin x\\ \textrm{b}.&12x^{2}-\cos x+\sin x\\ \textrm{c}.&4x^{3}-\cos x-\sin x\\ \textrm{d}.&4x^{3}-\sin x-\cos x\\ \textrm{e}.&12x^{3}+\cos x+\sin x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}h(x)&=4x^{3}+\sin x+\cos x\\ \textrm{guna}&\textrm{kan formula}:\: \color{red}y=a.u^{n}\Rightarrow y'=n.a.u^{n-1}.u'\\ \textrm{pada}&\: \textrm{fungsi aljabarnya, yaitu}:\color{black}y=4x^{3}\Rightarrow y'=12x^{2}\\ \textrm{seda}&\textrm{ngkan fungsi transendennya mengikuti}\\ \textrm{turu}&\textrm{nan fungsi trigonometri biasa. Sehingga}\\ f'(x)&=12x^{2}+\cos x-\sin x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Jika}\: \: p(x)=-\cos ^{4}x,\: \: \textrm{maka nilai}\\ &\textrm{maka}\: \: p'\left ( \displaystyle \frac{\pi }{3} \right )=....\\ &\begin{array}{llll}\\ \textrm{a}.&0\\ \textrm{b}.&\sqrt{3}\\ \textrm{c}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \color{red}\textrm{d}.&\displaystyle \frac{1}{4}\sqrt{3}\\ \textrm{e}.&1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}p(x)&=-\cos ^{4}x\\ \color{red}p'(x)&\color{red}=-4\cos ^{3}x.(-\sin x)=\color{black}4\cos ^{3}x\sin x\\ p'\left ( \displaystyle \frac{\pi }{3} \right )&=4\cos ^{3}\left ( \displaystyle \frac{\pi }{3} \right ).\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ &=4\cos ^{3}60^{\circ}\times \sin 60^{\circ}\\ &=4\left ( \displaystyle \frac{1}{2} \right )^{3}\times \left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{4}{16}\sqrt{3}\\ &=\displaystyle \frac{1}{4}\sqrt{3} \end{aligned} \end{array}$