Contoh Soal 2 Turunan Fungsi Aljabar

 $\begin{array}{ll}\\ 6.&\textrm{Jika}\: \: W=\sin 2t\: ,\: \textrm{maka}\: \: \displaystyle \frac{dW}{dt}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos 2t&\textrm{d}.\quad 2t\cos 2t+\sin 2t\\ \textrm{b}.\quad \color{red}2\cos 2t\quad  &\textrm{e}.\quad \sin 2t-t\cos 2t\\ \textrm{c}.\quad \sin 2t+t\cos 2t\quad \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &W=\sin 2t\\ &W=\sin u\, \quad \textrm{dengan}\: \: u=2t\\ &\displaystyle \frac{dW}{dt}=\displaystyle \frac{dW}{du}.\frac{du}{dt}\\ &\qquad=\cos u.2\\ &\qquad=2\cos u\\ &\qquad=2\cos 2t \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika}\: \: f(x)=\displaystyle \frac{2x+4}{1+\sqrt{x}}\: ,\: \textrm{maka}\: \: \displaystyle {f}\, '(4) =....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\displaystyle \frac{1}{4} &&\textrm{d}.\quad 1 \\ \textrm{b}.\quad \displaystyle \frac{3}{7} \quad &\textrm{c}.\quad \displaystyle \frac{3}{5} \quad &\textrm{e}.\quad 4 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{2x+4}{1+\sqrt{x}}\\ &=\displaystyle \frac{U}{V}\\ {f}\, '(x)&=\displaystyle \frac{{U}\, '.V-U.{V}\, '}{V^{2}}\\ &=\displaystyle \frac{(2)\left ( 1+\sqrt{x} \right )-(2x+4).\left ( 1.\left ( 1+\sqrt{x} \right )^{0} .\displaystyle \frac{1}{2}x^{^{-\frac{1}{2}}} \right )}{\left ( 1+\sqrt{x} \right )^{2}} \\&\textrm{ingat}\: \: \sqrt{x}=x^{^{\frac{1}{2}}} \\ {f}\, '(4) &=\displaystyle \frac{2\left ( 1+\sqrt{4} \right )-(2.4+4).\frac{1}{2}. 4^{^{-\frac{1}{2}}} }{\left ( 1+\sqrt{4} \right )^{2}}\\ &=\displaystyle \frac{2(1+2)-(12).\frac{1}{2}. \frac{1}{2}}{(1+2)^{2}}\\ &\textrm{ingat juga}\: \: 4^{^{-\frac{1}{2}}}=\left ( 2^{2} \right )^{^{-\frac{1}{2}}}=2^{^{-1}}=\displaystyle \frac{1}{2^{1}}=\frac{1}{2}\\ &=\displaystyle \frac{6-3}{9}\\ &=\displaystyle \frac{1}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Jika}\: \: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x} \: ,\: \textrm{maka}\: \: {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2} &&\textrm{d}.\quad \color{red}2 \\ \textrm{b}.\quad \displaystyle -\frac{1}{2} \quad &\textrm{c}.\quad \displaystyle \frac{3}{4} \quad &\textrm{e}.\quad -2 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{1+\sin x}{\cos x}\\ {f}\, '(x)&=\displaystyle \frac{\cos x.\cos x-(1+\sin x).-\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x +\sin ^{2}x}{\cos ^{2}x}\\ &\qquad \textrm{ingat bahwa}\: \: \sin ^{2}x+\cos ^{2}x=1\\ &=\displaystyle \frac{1+\sin x }{\cos ^{2}x}\\ {f}\, '\left ( \displaystyle \frac{1}{6}\pi \right )&=\displaystyle \frac{1+\sin \left ( \displaystyle \frac{1}{6}\pi \right )}{\cos ^{2}\left ( \displaystyle \frac{1}{6}\pi \right )}\\&=\displaystyle \frac{1+\displaystyle \frac{1}{2} }{\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{\displaystyle \frac{3}{2}}{\displaystyle \frac{3}{4}}\\ &=\displaystyle \frac{3}{2}\times \frac{4}{3}\\ &=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: \: f(x)=\displaystyle 3x^{2}-2ax+7\\ &\textrm{dan}\: \: {f}\, '(1)=0 \: ,\: \textrm{maka}\: \: {f}\, '(2)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 1 &&\textrm{d}.\quad \color{red}6 \\ \textrm{b}.\quad \displaystyle 2 \quad &\textrm{c}.\quad \displaystyle 4 \quad &\textrm{e}.\quad 8 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle 3x^{2}-2ax+7\\ {f}\, '(x)&=6x-2a\\ {f}\, '(1)&=0\\ 6(1)-2a&=0\\ 6&=2a\\ 3&=a\\ \textrm{sehingga}\, &\: \\ {f}\, '(x)&=6x-6\\ \textrm{maka}\, ,\: \quad &\\ {f}\, '(2)&=6.2-6\\ &=12-6\\ &=6 \end{aligned}  \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: \: f(x)=\displaystyle (6x-3)^{3}(2x-1)\\ &\textrm{maka}\: \: {f}\, '(1)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 18 &&\textrm{d}.\quad 162 \\\\ \textrm{b}.\quad \displaystyle 24 \quad &\textrm{c}.\quad \displaystyle 54 \quad &\textrm{e}.\quad \color{red}216 \end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle (6x-3)^{3}(2x-1)\\ &=(3.(2x-1))^{3}(2x-1)^{1}\\ &=3^{3}.(2x-1)^{3+1}\\ &=27(2x-1)^{4}\\ {f}\, '(x)&=4.27(2x-1)^{4-1}.2\\ &=216.(2x-1)^{3}\\ {f}\, '(1)&=216.(2.1-1)^{3}\\ &=216.1\\ &=216 \end{aligned} \end{array}$.