CONTOH SOAL 9 TURUNAN FUNGSI

 

$\begin{array}{ll}\\ 41.&\textrm{Persamaan garis singgung pada kurva}\\ &y=3\sin x\: \: \textrm{pada titik yang berabsis}\: \: \displaystyle \frac{\pi }{3}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )-\frac{2\sqrt{2}}{3}\\ \textrm{b}.&y=\displaystyle \frac{2}{3}\left ( x-\frac{\pi }{3} \right )+\frac{2\sqrt{2}}{3}\\ \textrm{c}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{3}}{2}\\ \textrm{d}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )+\frac{3\sqrt{3}}{2}\\ \textrm{e}.&y=\displaystyle \frac{3}{2}\left ( x-\frac{\pi }{3} \right )-\frac{3\sqrt{2}}{2} \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{aligned}&y=3\sin x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=3\sin \left ( \displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\frac{3\sqrt{3}}{2}\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=3\cos x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=3\cos \left ( \displaystyle \frac{\pi }{3} \right )=3\left ( \displaystyle \frac{1}{2} \right )=\frac{3}{2}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=\displaystyle \frac{3}{2}\left ( x-\displaystyle \frac{\pi }{3} \right )+\displaystyle \frac{3\sqrt{3}}{2} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Kurva}\: \: y=\sin x+\cos x\: \: \textrm{untuk}\\ &0\lt x\lt \pi \: \: \textrm{memotong sumbu X}\\ &\textrm{di titik A. Persamaan garis}\\ &\textrm{singgung di titik A adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{b}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{\pi }{2} \right )\\ \textrm{c}.&y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right )\\ \textrm{d}.&y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )\\ \textrm{e}.&y=\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&\textrm{Kurva memotong sumbu X}\\ &\textrm{di titik A, berarti}\: \: y=0\\ &\sin x+\cos x=0\\ &\sin x=-\cos x\\ &\displaystyle \frac{\sin x}{\cos x}=-1\Leftrightarrow \tan x=-1\\ &\tan x=\tan \left ( \displaystyle \frac{3\pi }{4} \right )\Rightarrow x=\displaystyle \frac{3\pi }{4}\\ &\textrm{Jadi, titik A-nya}:\: \: \left ( \displaystyle \frac{3\pi }{4},0 \right )\\ &\textrm{dan nilai gradien}\: \: m=y',\: \: \textrm{yaitu}:\\ &m=\cos x-\sin x\\ &m=\cos \left ( \displaystyle \frac{3\pi }{4} \right )-\sin \left ( \displaystyle \frac{3\pi }{4} \right )\\ &m=-\displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2}=-\sqrt{2}\\ &\textrm{Persamaan garis singgung di A}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) +0\\ &\Leftrightarrow \: y=-\sqrt{2}\left ( x-\displaystyle \frac{3\pi }{4} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec ^{2}x\: \: \textrm{pada titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll} \textrm{a}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{b}.&y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{c}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4\\ \textrm{d}.&y=-8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4\\ \textrm{e}.&y=4\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )-4 \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{b}\\ &\begin{aligned}&y=\sec^{2} x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &y_{0}=\sec^{2} \left ( \displaystyle \frac{\pi }{3} \right )=(2)^{2}=4\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=2\sec^{2} x\tan x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{3}\\ &m=2\sec^{2} \left ( \displaystyle \frac{\pi }{3} \right )\tan \left ( \displaystyle \frac{\pi }{3} \right )\\ &\quad=2(4)\sqrt{3}=8\sqrt{3}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &y=8\sqrt{3}\left ( x-\displaystyle \frac{\pi }{3} \right )+4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&\textrm{Kurva berikut yang memiliki}\\ &\textrm{garis singgung dengan gradien}\\ &4\sqrt{3}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=2\sin x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},\sqrt{3} \right )\\ \textrm{b}.&y=\cos 2x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{12},\frac{1}{2} \right )\\ \textrm{c}.&y=\tan x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \pi ,0 \right )\\ \textrm{d}.&y=2\sec x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{3},2 \right )\\ \textrm{e}.&y=\cot x\: \: \textrm{pada titik}\: \: \left ( \displaystyle \frac{\pi }{4},1 \right ) \end{array}\\ \end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{d}\\ &\begin{array}{|c|l|l|}\hline \textrm{a}&y=2\sin x&m=2\cos \displaystyle \frac{\pi }{3}\\ &y'=2\cos x&m=2.\displaystyle \frac{1}{2}=1\\\hline \textrm{b}&y=\cos 2x&m=-2\sin 2 \left (\displaystyle \frac{\pi }{12} \right )\\ &y'=-2\sin 2x&m=-2.\displaystyle \frac{1}{2}=-1\\\hline \textrm{c}&y=\tan x&m=\sec^{2} \left (\pi \right )\\ &y'=\sec^{2} x&m=(-1)^{2}=1\\\hline \textrm{d}&y=2\sec x&m=2\sec \left ( \displaystyle \frac{\pi }{3} \right )\tan \left (\displaystyle \frac{\pi }{3} \right )\\ &y'=2\sec x\tan x&m=2.2.\sqrt{3}=4\sqrt{3}\\\hline \textrm{e}&y=\cot x&m=-\csc^{2} \left ( \displaystyle \frac{\pi }{4} \right )\\ &y'=-\csc^{2} x&m=-\left ( \sqrt{2} \right )^{2}=-2\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Persamaan garis singgung pada}\\ &\textrm{kurva}\: \: y=\sec x\: \: \textrm{di titik yang}\\ &\textrm{berabsis}\: \: \displaystyle \frac{\pi }{4}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&y=\sqrt{3}x-\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{b}.&y=\sqrt{3}x+\displaystyle \frac{\sqrt{3}\pi }{4}+\sqrt{3}\\ \textrm{c}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{d}.&y=\sqrt{2}x+\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2}\\ \textrm{e}.&y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{3} \end{array}\\\end{array}$

Jawaban: $\begin{array}{ll} &\textrm{Jawab}:\quad \textbf{c}\\ &\begin{aligned}&y=\sec x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &y_{0}=\sec \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}\\ &\textrm{kita cari gradien}\: \: m\: \: \textrm{saat}\: \: y',\: \: \textrm{yaitu}:\\ &m=y'=\sec x\tan x,\: \: \textrm{saat}\: \: x_{0}=\displaystyle \frac{\pi }{4}\\ &m=\sec \left ( \displaystyle \frac{\pi }{4} \right )\tan \left ( \displaystyle \frac{\pi }{4} \right )=\sqrt{2}.1=\sqrt{2}\\ &\textrm{Persamaan garis singgungnya adalah}:\\ &y=m(x-x_{0})+y_{0}\\ &\Leftrightarrow \: y=\sqrt{2}\left ( x-\displaystyle \frac{\pi }{4} \right )+\sqrt{2}\\ &\Leftrightarrow \: y=\sqrt{2}x-\displaystyle \frac{\sqrt{2}\pi }{4}+\sqrt{2} \end{aligned} \end{array}$