Lanjutan Materi (4) Turunan fungsi Trigonometri (Matematika Peminatan Kelas XII)

 $\color{blue}\textrm{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

Sebelumnya silahkan ingat kembali pada dalil-dalil yang berlaku pada materi turunan fungsi aljabar di kelas XI, maka turunan fungsi trigonometri pun serupa, yaitu:

$\begin{array}{|c|l|l|}\hline \textrm{No}&\color{red}\textrm{Fungsi}&\color{blue}\textrm{Turunan Pertama}\\\hline 1.&y=k.u&y'=k.u'\\\hline 2.&y=u\pm v&y'=u'\pm v'\\\hline 3.&y=u.v&y'=v.u'+u.v'\\\hline 4.&y=k.u^{n}&=n.k.u^{(n-1)}.u'\\\hline 5.&y=\displaystyle \frac{u}{v}&y'=\displaystyle \frac{u'.v-u.v'}{v^{2}}\\\hline \end{array}$

Selanjutnya untuk turunan pertama fungsi di atas semisal fungsi  $y=f(x)$ diturunkan terhadap  $x$, maka turunan pertamnya dapat dituliskan dengan

$\color{blue}y'=\displaystyle \frac{dy}{dx}=f'(x)=\underset{h\rightarrow 0}{\textrm{lim}}\: \frac{f(x+h)-f(x)}{h}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$\color{blue}y''=\displaystyle \frac{d^{2}y}{dx^{2}}=f''(x)\: \: \color{black}\textrm{atau kadang dituliskan}\: \: \displaystyle \color{purple}\frac{df'(x)}{dx}=\frac{d^{2}f}{dx^{2}}$

Selanjutnya perhatikanlah tabel berikut

$\color{blue}\begin{array}{|l|l|}\hline \textrm{Turunan}&\qquad\quad\quad\textrm{Notasi}\\\hline \textrm{Pertama}&y'=f'(x)=\displaystyle \frac{dy}{dx}=\frac{df}{dx}\\ &\\ \textrm{Kedua}&y''=f''(x)=\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{d^{2}f}{dx^{2}}\\ &\\ \textrm{Ketiga}&y'''=f'''(x)=\displaystyle \frac{d^{3}y}{dx^{3}}=\frac{d^{3}f}{dx^{3}}\\ &\\ \cdots &\cdots \qquad\cdots \qquad\cdots \qquad\cdots \\ \textrm{Ke-n}&y^{n}=f^{n}(x)=\displaystyle \frac{d^{n}y}{dx^{n}}=\frac{d^{n}f}{dx^{n}}\\\hline \end{array}$

$\LARGE\color{magenta}\fbox{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Tentukanlah turunan pertama dari}\\ &\begin{array}{ll}\\ \textrm{a}.&y=\sin 2x\\ \textrm{b}.&y=\cos 4x\\ \textrm{c}.&y=\sin^{2} x\\ \textrm{d}.&y=\cos^{4} x\\ \textrm{e}.&y=-\sin x\\ \textrm{f}.&y=-5\cos 2x\\ \textrm{g}.&y=-7\tan x\\ \textrm{h}.&y=3\sin^{3} x\\ \textrm{i}.&y=-10\cos^{5} x\\ \textrm{j}.&y=-4\tan^{2} x\\ \textrm{k}.&y=\sqrt{\cos x}\\ \textrm{l}.&y=2\sin x+5x\\ \textrm{m}.&y=3\cos^{2} x+2x^{2}\\ \textrm{n}.&y=\csc x-2\tan ^{2}x+4x\\ \end{array} \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}\textbf{Jawab}&\\ \textrm{Turun}&\textrm{an pertamanya masing}\\ \textrm{fungsi}&\: \textrm{di atas adalah berikut}:\\ (\textrm{a}).\: \: y&=\sin 2x\\ y'&=2\cos 2x\\ (\textrm{b}).\: \: y&=\cos 4x\\ y'&=-4\sin 4x\\ (\textrm{c}).\: \: y&=\sin^{2} x\\ y'&=2\sin x\cos x,\: \: \color{red}\textrm{atau boleh juga}\\ &=\sin 2x\\ (\textrm{d}).\: \: y&=\cos^{4} x\\ y'&=4\cos ^{3}(-\sin x)=-4\cos ^{3}x.\sin x\\ (\textrm{e}).\: \: y&=-2\sin x\\ y'&=-2\cos x\\ (\textrm{f}).\: \: y&=-5\cos 2x\\ y'&=-5(-\sin 2x.(2))=10x\sin 2x\\ (\textrm{g}).\: \: y&=-7\tan x\\ y'&=-7\sec ^{2}x\\ (\textrm{h}).\: \: y&=3\sin^{3} x\\ y'&=3.\left ( 3\sin ^{2}x \right ).(\cos x)=9\sin ^{2}x\cos x\\ (\textrm{i}).\: \: y&=-10\cos^{5} x\\ y'&=5\left ( -10\cos ^{4}x \right ).(-\sin x)\\ &=50\cos ^{4}x\sin x\\ (\textrm{j}).\: \: y&=-4\tan^{2} x\\ &=2\left ( -4\tan x \right ).\left ( \sec ^{2}x \right )\\ &=-8\tan x\sec ^{2}x\\ (\textrm{k}).\: \: y&=\sqrt{\cos x}=\cos ^{.^{\frac{1}{2}}}x\\ y'&=\displaystyle \frac{1}{2}\left ( \cos ^{.^{-\frac{1}{2}}}x \right ).\left ( -\sin x \right )\\ &=-\displaystyle \frac{1}{2}\cos ^{.^{-\frac{1}{2}}}x\sin x\\ &=-\displaystyle \frac{\sin x}{2\sqrt{\cos x}}\\ (\textrm{l}).\: \: y&=2\sin x+5x\\ y'&=2\cos x+5\\ (\textrm{m}).\: \: y&=3\cos^{2} x+2x^{2}\\ y'&=2\left ( 3\cos x \right ).(-\sin x)+4x\\ &=-6\cos x\sin x+4x,\: \: \color{red}\textrm{atau}\\ &=-3\sin 2x+4x=4x-3\sin 2x\\ (\textrm{n}).\: \: y&=\csc x-2\tan ^{2}x+4x\\ &=-\csc x\cot x-2\left ( 2\tan x \right ).\left ( \sec ^{2}x \right )+4\\ &=-\csc x\cot x-4\tan x\sec ^{2}x+4 \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Jika diketahui}\\ &\begin{array}{ll}\\ \textrm{a}.&f(x)=\displaystyle \frac{1+\sin x}{\cos x}.\: \: \textrm{Tentukanlah}\: \: f'(x)\\ \textrm{b}.&g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=\displaystyle \frac{\pi }{6}\\ \textrm{c}.&h(x)=\sin x\tan x.\: \: \textrm{Tentukanlah nilai}\\ &\textrm{saat}\: \: x=45^{\circ}\\ \textrm{d}.&k(x)=\sin x+n\cos x\: \: \textrm{dan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0.\\ &\textrm{Tentukanlah nilai}\: \: n \end{array}\\\\ &\color{blue}\textbf{Jawab}: \end{array}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{a})\: \: \textrm{dike}&\textrm{tahui}\: \: f(x)=\displaystyle \frac{1+\sin x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ f'(x)&=\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x+\sin x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1+\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{\sin x}{\cos ^{2}x}\\ &=\displaystyle \frac{1}{\cos ^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}\\ &=\sec ^{2}x+\sec x\tan x \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{b})\: \: \textrm{dike}&\textrm{tahui}\: \: g(x)=\displaystyle \frac{\sin x+\cos x}{\cos x}\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=\displaystyle \frac{u}{v}\Rightarrow y'=\displaystyle \frac{u'v-uv'}{v^{2}}\\ g'(x)&=\displaystyle \frac{(\cos x-\sin x)(\cos x)-(\sin x+\cos x)(-\sin x)}{\cos ^{2}x}\\ &=\displaystyle \frac{\cos ^{2}x-\sin x\cos x+\sin x+\sin ^{2}x+\sin x\cos x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}\\ &=\displaystyle \frac{\color{red}1}{\cos ^{2}x}\\ g\left ( \displaystyle \frac{\pi }{6} \right )&=\displaystyle \frac{1}{\cos ^{2}\left ( \displaystyle \frac{\pi }{6} \right )}\\ &=\left (\displaystyle \frac{1}{\cos \left ( \displaystyle \frac{\pi }{6} \right )} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\cos 30^{\circ}} \right )^{2}\\ &=\left (\displaystyle \frac{1}{\displaystyle \frac{1}{2}\sqrt{3}} \right )^{2}\\ &=\left ( \displaystyle \frac{2}{\sqrt{3}} \right )^{2}\\ &=\frac{4}{3}\: \: \color{red}\textrm{Jika Anda tidak terganggu dengan nilai}\\ &\color{red}\textrm{perbandingan trigonometri, Anda bisa langsung saja}\\ &\color{red}\textrm{ke jawabannya, yaitu}\: \: \color{blue}\displaystyle \frac{4}{3} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{c})\: \: \textrm{dike}&\textrm{tahui}\: \: h(x)=\sin x\tan x\\ \textrm{gun}&\textrm{akan formula}\: \: \color{red}y=u.v\Rightarrow y'=u'.v+u.v'\\ h'(x)&=\cos x.(\tan x)+\sin x.\left ( \sec ^{2}x \right )\\ h\left ( 45^{\circ} \right )&=\cos \left ( 45^{\circ} \right )\tan \left ( 45^{\circ} \right )+\sin \left ( 45^{\circ} \right )\sec ^{2}\left ( 45^{\circ} \right )\\ &=\left ( \displaystyle \frac{1}{2}\sqrt{2} \right ).1+\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \sqrt{2} \right )^{2}\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}\\ &=\displaystyle \frac{3}{2}\sqrt{2} \end{aligned}$

$.\: \: \quad\color{blue}\begin{aligned}2.(\textrm{d})\: \: \textrm{dike}&\textrm{tahui}\: \: k(x)=\sin x+n\cos x\\ k'(x)&=\cos x-n\sin x,\: \: \color{red}\textrm{dengan}\: \: k'\left ( \displaystyle \frac{\pi }{3} \right ) =0\\ k'\left ( \displaystyle \frac{\pi }{6} \right )&=\cos \left ( \displaystyle \frac{\pi }{3} \right )-n\sin \left ( \displaystyle \frac{\pi }{3} \right )\\ 0&=\cos 60^{\circ}-n\sin 60^{\circ}=\displaystyle \frac{1}{2}-n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ \Leftrightarrow \quad&n\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )=\displaystyle \frac{1}{2}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{\sqrt{3}}\times \color{black}\frac{\sqrt{3}}{\sqrt{3}}\\ \Leftrightarrow \quad&n\: \: \, \,\quad\quad\quad =\displaystyle \frac{1}{3}\sqrt{3} \end{aligned}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui fungsi}\: \: y=\displaystyle \frac{1}{2}\sin ^{2}x.\: \: \textrm{Tentukanlah}\\ &\textrm{Turunan pertama, kedua, ketiga, keempat},\\ &\textrm{dan kelima dari fungsi tersebut di atas}\\\\ &\textrm{Jawab}:\\ &\color{blue}\begin{aligned}\qquad y&=\displaystyle \frac{1}{2}\sin ^{2}x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan pertama}\\ \displaystyle \frac{dy}{dx}&=2\left ( \displaystyle \frac{1}{2}\sin ^{1}x \right )\times \cos x\\ &=\sin x\cos x=\displaystyle \frac{1}{2}\left ( 2\sin x\cos x \right )=\frac{1}{2}\sin 2x\\ \color{black} \textrm{Turu}&\color{black}\textrm{nan keduanya}\\ \displaystyle \frac{d^{2}y}{dx^{2}}&= \displaystyle \frac{1}{2}(\cos 2x).(2)=\color{red}\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan ketiganya}\\ \displaystyle \frac{d^{3}y}{dx^{3}}&=-\sin 2x.(2)=\color{red}-2\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keempatnya}\\ \displaystyle \frac{d^{4}y}{dx^{4}}&=-2\cos 2x.(2)=\color{red}-4\cos 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan kelimanya}\\ \displaystyle \frac{d^{5}y}{dx^{5}}&=-4(-\sin 2x),(2)=\color{red}8\sin 2x\\ \color{black}\textrm{Turu}&\color{black}\textrm{nan keenamnya}\\ \displaystyle \frac{d^{6}y}{dx^{6}}&=8\cos 2x.(2)=\color{red}16\cos 2x \end{aligned} \end{array}$

DAFTAR PUSTAKA

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