Latihan Soal 5 Persiapan PAS Gasal Matematika Peminatan Kelas XII (Limit dan Turunan Fungsi Trigonometri)

 $\begin{array}{ll}\\ 41.&\textrm{Turunan pertama fungsi}\\ &h(x)=5\sin x\cos x\: \: \textrm{adalah}\: \: h'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&5\sin 2x\\ \color{red}\textrm{b}.&5\cos 2x\\ \textrm{c}.&5\sin ^{2}x\cos x\\ \textrm{d}.&5\sin ^{2}x\cos^{2} x\\ \textrm{e}.&5\sin 2x\cos x \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: h(x)=5\sin x\cos x\\ h(x)&=\color{red}\displaystyle \frac{5}{2}\left ( 2\sin x\cos x \right )=\displaystyle \frac{5}{2}\sin 2x\\ h'(x)&=\color{purple}\displaystyle \frac{5}{2}\left ( \cos 2x \right ).(2)\\ &=\color{red}5\cos 2x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 42.&\textrm{Turunan pertama fungsi}\\ &k(x)=\cos x\tan x\: \: \textrm{adalah}\: \: k'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin x\cot x+\cos x\sec ^{2}x\\ \color{red}\textrm{b}.&-\sin x\tan x+\cos x\sec ^{2}x\\ \textrm{c}.&\sin x\tan x-\cos x\sec ^{2}x\\ \textrm{d}.&-\displaystyle \frac{1+\sin ^{2}x}{\cos x}\\ \textrm{e}.&\displaystyle \frac{1+\sin ^{2}x}{\cos x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \color{black}k(x)=\cos x\tan x\\ \textrm{guna}&\textrm{kan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'v+u.v'\\ u&=\color{black}\cos x \Rightarrow u'=-\sin x\\ v&=\color{black}\tan x \Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ k'(x)&=\left ( -\sin x \right )\tan x+\cos x.\left ( \sec ^{2}x \right )\\ &=\color{red}-\sin x\tan x+\cos x\sec ^{2}x \end{aligned} \end{array}$

$\begin{array}{ll}\\ 43.&\textrm{Jika diketahui}\: \: f(x)=\left | \tan x \right |,\: \textrm{maka}\: \: \displaystyle \frac{dy}{dx}\\ &\textrm{saat}\: \: x=k,\: \: \textrm{di mana}\: \: \displaystyle \frac{1}{2}\pi <k<\pi\\ & \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sin k\\ \textrm{b}.&\cos k\\ \color{red}\textrm{c}.&-\sec ^{2}k\\ \textrm{d}.&\sec ^{2}k\\ \textrm{e}.&\cot k \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}f(x)=\left |\tan x \right |\\ \textrm{saat}&\: \: \color{red}x=k\: \: \color{blue}\textrm{dengan}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \color{black}\textrm{adal}&\color{black}\textrm{ah}:\\ f(x)&=\left | \tan x \right |,\: \: \color{black}\textrm{maka saat}\: \: \color{blue}x=k\\ f(k)&=\left | \tan k \right |=-\tan k,\: \: \color{black}\textrm{karena di}\: \: \color{red}\displaystyle \frac{1}{2}\pi <k<\pi\\ \displaystyle \frac{dy}{dx}&=f'(k)=\color{red}-\sec ^{2}k \end{aligned} \end{array}$

$\begin{array}{ll}\\ 44.&\textrm{Turunan pertama}\: \: g(x)=\left | \cos x \right |\\ & \textrm{adalah}\: \: g'(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\left | \sin x \right |\\ \textrm{b}.&-\sin x\\ \textrm{c}.&\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \color{red}\textrm{d}.&-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |}\\ \textrm{e}.&\left | \sin x \right | \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Dike}&\textrm{tahui}\: \: \color{black}g(x)=\left |\cos x \right |=\sqrt{\cos ^{2}x}=\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}\\ g'(x)&=\displaystyle \frac{1}{2}\left ( \cos ^{2}x \right )^{.^{-\frac{1}{2}}}.\left ( 2\cos x \right ).\left ( -\sin x \right )\\ &=\displaystyle \frac{-2\sin x\cos x}{2\left ( \cos ^{2}x \right )^{.^{\frac{1}{2}}}}\\ &=-\displaystyle \frac{\sin 2x}{2\sqrt{\cos ^{2}x}}\\ &=\color{red}-\displaystyle \frac{\sin 2x}{2\left | \cos x \right |} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 45.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\sin x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\cos x+\sin x}{x^{2}}\\ \color{red}\textrm{b}.&\displaystyle \frac{x\cos x-\sin x}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{-x\cos x-\sin x}{x^{2}}\\ \textrm{d}.&\displaystyle \frac{\cos x-x\sin x}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{\cos x+x\sin x}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x}{x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x\Rightarrow u'=\cos x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\cos x.(x)-\sin x.1}{x^{2}}\\ &=\color{red}\displaystyle \frac{x\cos x-\sin x}{x^{2}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 46.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{1-\cos x}{x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{x\sin x+\cos x+1}{x^{2}}\\ \textrm{b}.&\displaystyle \frac{x\cos x+\sin x-1}{x^{2}}\\ \textrm{c}.&\displaystyle \frac{x\sin x-\cos x+1}{x^{2}}\\ \color{red}\textrm{d}.&\displaystyle \frac{x\sin x+\cos x-1}{x^{2}}\\ \textrm{e}.&\displaystyle \frac{x\cos x-\sin x+1}{x^{2}} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{1-\cos x}{x}\\ \textrm{Guna}&\textrm{kan formula}\\ &\color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=1-\cos x\Rightarrow u'=\sin x\\ v&=x\Rightarrow v'=1\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sin x.(x)-(1-\cos x).1}{x^{2}}\\ &=\color{red}\displaystyle \frac{x\sin x+\cos x-1}{x^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Turunan pertama dari}\: \: f(x)=\displaystyle \frac{\tan x}{\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1+\cos ^{2}x}{\cos ^{3}x}\\ \textrm{b}.&\displaystyle \frac{1-\cos x}{\cos ^{3}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x}\\ \textrm{d}.&\displaystyle \frac{1+\sin x}{\cos ^{3}x}\\ \textrm{e}.&\displaystyle \frac{1-\sin ^{2}x}{\cos ^{3}x} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\tan x}{\cos x}\\ \textrm{Guna}&\textrm{kan formula}\\ &\color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\tan x\Rightarrow u'=\sec ^{2}x\\ v&=\cos x\Rightarrow v'=-\sin x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{\sec ^{2}x.(\cos x)-(\tan x).(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\sec ^{2}x.\cos x+\tan x\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\left ( \displaystyle \frac{1}{\cos ^{2}x} \right )\cos x+\left ( \displaystyle \frac{\sin x}{\cos x} \right )\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\displaystyle \frac{1}{\cos x}+\displaystyle \frac{\sin ^{2}x}{\cos x}}{\cos ^{2}x}\\ &=\color{red}\displaystyle \frac{1+\sin ^{2}x}{\cos ^{3}x} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Turunan pertama dari}\: \: g(t)=\displaystyle \frac{\cos t+2t}{\sin t} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{2\sin t+2t\cos t-1}{\sin^{2} t}\\ \textrm{b}.&\displaystyle \frac{2\sin t-2t\cos t+1}{\sin^{2} t}\\ \textrm{c}.&\displaystyle \frac{2\sin t+2t\cos t+1}{\sin^{2} t}\\ \color{red}\textrm{d}.&\displaystyle \frac{2\sin t-2t\cos t-1}{\sin^{2} t}\\ \textrm{e}.&\displaystyle \frac{-2\sin t+2t\cos t-1}{\sin^{2} t} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ g(t)&=\displaystyle \frac{\cos t+2t}{\sin t}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\cos t+2t\Rightarrow u'=-\sin t+2\\ v&=\sin t\Rightarrow v'=\cos t\\ \color{red}\textrm{maka}&\\ g'(t)&=\displaystyle \frac{(-\sin t+2)(\sin t)-(\cos t+2t)(\cos t)}{\sin ^{2}t}\\ &=\displaystyle \frac{-\sin ^{2}t+2\sin t-\cos ^{2}t-2t\cos t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\sin ^{2}t-\cos ^{2}t}{\sin ^{2}t}\\ &=\displaystyle \frac{t+2\sin t-2t\cos t-\left (\sin ^{2}t+\cos ^{2}t \right )}{\sin ^{2}t}\\ &=\color{red}\displaystyle \frac{2\sin t-2t\cos t-1}{\sin ^{2}t} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&\textrm{Turunan pertama dari}\: \: h(x)=\displaystyle \frac{\sin x}{\sin x+\cos x} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{\cos ^{2}x-\sin ^{2}x}\\ \textrm{b}.&\displaystyle \frac{1}{\sin ^{2}x-\cos ^{2}x}\\ \color{red}\textrm{c}.&\displaystyle \frac{1}{(\sin x+\cos x)^{2}}\\ \textrm{d}.&\displaystyle \sin ^{2}x-\cos ^{2}x\\ \textrm{e}.&\displaystyle 1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ h(x)&=\displaystyle \frac{\sin x}{\sin x+\cos x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x \Rightarrow u'=\cos x\\ v&=\sin x+\cos x\Rightarrow v'=\cos x-\sin x\\ \color{red}\textrm{maka}&\\ h'(x)&=\displaystyle \frac{\cos x.(\sin x+\cos x)-\sin x.(\cos x-\sin x)}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\cos x\sin x+\cos ^{2}x-\sin x\cos x+\sin ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\displaystyle \frac{\sin ^{2}x+\cos ^{2}x}{(\sin x+\cos x)^{2}}\\ &=\color{red}\displaystyle \frac{1}{(\sin x+\cos x)^{2}} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{\sin x-\cos x}{\tan x}. \: \: \textrm{Nilai}\\ &\textrm{turunan pertama fungsi}\: \: f\: \: \textrm{saat}\: \: x=45^{\circ}\\ &\textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{1}{2}\sqrt{2}\\ \textrm{b}.&\displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{c}.&\displaystyle 1\\ \color{red}\textrm{d}.&\displaystyle \sqrt{2}\\ \textrm{e}.&\displaystyle \sqrt{3} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\textrm{Diket}&\textrm{ahui}\\ f(x)&=\displaystyle \frac{\sin x-\cos x}{\tan x}\\ \textrm{Guna}&\textrm{kan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-u.v'}{v^{2}}\\ u&=\sin x-\cos x \Rightarrow u'=\cos x+\sin x\\ v&=\tan x\Rightarrow v'=\sec ^{2}x\\ \color{red}\textrm{maka}&\\ f'(x)&=\displaystyle \frac{(\cos x+\sin x).\tan x-(\sin x-\cos x).\sec ^{2}x}{\tan ^{2}x}\\ f'\left ( 45^{\circ} \right )&=\displaystyle \frac{\left ( \displaystyle \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{2} \right ).1-\left ( \displaystyle \frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{2} \right ).\left ( \sqrt{2} \right )^{2}}{1^{2}}\\ &=\displaystyle \frac{\sqrt{2}-0}{1}\\ &=\color{red}\sqrt{2} \end{aligned} \end{array}$