Contoh Soal 2 Turunan Fungsi Trigonometri (Bagian 1)

$\begin{array}{ll}\\ 6.&\textrm{Turunan pertama}\: \: q(x)=\sin ^{2}x+\cos ^{2}x\\ &\textrm{adalah}\: \: q'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cos ^{2}x-\sin ^{2}x\\ \textrm{b}.&2\cos ^{2}x-2\sin ^{2}x\\ \textrm{c}.&\cos x-\sin x\\ \textrm{d}.&2\cos x-2\sin x\\ \color{red}\textrm{e}.&0 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}q(x)&=\sin ^{2}x+\cos ^{2}x\\ \color{red}\textrm{guna}&\color{red}\textrm{kan formula identitas}:\: \color{black}\sin ^{2}x+\cos ^{2}x=1\\ \textrm{Sehi}&\textrm{ngga soal di atas dapat dituliskan menjadi}\\ q(x)&=1,\: \: \textrm{maka}\\ q'(x)&=0\\ \color{purple}\textrm{inga}&\color{purple}\textrm{t bahwa}\: \: \color{black}y=a\Rightarrow \displaystyle \frac{dy}{dx}=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Nilai dari}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.&-\displaystyle \frac{1}{2}\\ \color{red}\textrm{c}.&0\\ \textrm{d}.&\displaystyle \frac{1}{2}\\ \textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dari}&\: \textrm{soal diketahui}:\: \\ f(x)&=\sin \displaystyle \frac{\pi }{3}\\ \textrm{Nila}&\textrm{i dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin \left (\displaystyle \frac{\pi }{3}+h \right )-\sin \displaystyle \frac{\pi }{3}}{h}\\ \textrm{arti}&\textrm{nya bermakna, berapkah}\: \: f'\left ( x \right )?\\ \color{red}\textrm{maka}&\\ f'\left ( x \right )&=0 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: f(x)=8x-\sin ^{3}x,\\ &\textrm{maka nilai}\: \: \underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&4x^{2}-3\cos^{2}x \\ \textrm{b}.&8x-3\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&8-3\sin ^{2}x\cos x\\ \textrm{d}.&8+\sin ^{2}x\cos x\\ \textrm{e}.&3\sin ^{2}x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui dari soal}\: f(x)=8x-\sin ^{3}x\\ \color{red}\textrm{maka}&\: \textrm{nilai dari}\: \: \color{purple}\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{f(x+h)-f(x)}{h}=f'(x)\\ f'(x)&=8-3\sin ^{2}x\cos x \end{aligned}\end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Turunan pertama fungsi}\: \: f(x)=\sqrt{\sin x},\\ &\textrm{adalah}\: \: f'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2\sqrt{\sin x}} \\ \textrm{b}.&\displaystyle \frac{\cos x}{\sqrt{\sin x}}\\ \color{red}\textrm{c}.&\displaystyle \frac{\cos x}{2\sqrt{\sin x}}\\ \textrm{d}.&-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ \textrm{e}.&\displaystyle \frac{2\cos x}{\sqrt{\sin x}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}f(x)=\sqrt{\sin x}=\sin ^{.^{\frac{1}{2}}}x\\ f'(x)&=\color{purple}\displaystyle \frac{1}{2}\left ( \sin ^{.^{-\frac{1}{2}}}x \right ).(\cos x)\\ &=\color{purple}\displaystyle \frac{\cos x}{2\sin ^{.^{\frac{1}{2}}}x}\\ &=\color{purple}\displaystyle \frac{\cos x}{2\sqrt{\sin x}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: g'(x)\: \: \textrm{adalah turunan pertama}\\ &\textrm{fungsi}\: \: g(x)\: \: \textrm{dengan}\: \: g(x)=5\tan ^{2}x,\\ &\textrm{maka}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&10\cos ^{2}x\sin x\\ \textrm{b}.&10\sin ^{2}x\cos x\\ \color{red}\textrm{c}.&\displaystyle \frac{10\sin x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{10\cos ^{3}x}{\sin x}\\ \textrm{e}.&\displaystyle \frac{10}{\sin ^{2}x-\cos ^{2}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}g(x)=5\tan ^{2}x\\ g'(x)&=\color{purple}5\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )\\ &=\color{purple}10\tan x\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\color{purple}10\left ( \displaystyle \frac{\sin x}{\cos x} \right )\times \left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\\ &=\color{purple}\displaystyle \frac{10\sin x}{\cos ^{3}x} \end{aligned} \end{array}$