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

 $\text{C. Sifat-Sifat Turunan Fungsi Trigonometri}$

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

$\begin{array}{|cll|}\hline \text{No}& \text{Fungsi}& \text{Turunan Pertama}\\ 1.& y=k.u& y^{\prime }=k.u^{\prime }\\ 2.& y=u\pm v& y^{\prime }=u^{\prime }\pm v^{\prime }\\ 3.& y=u.v& y^{\prime }=v.u^{\prime }+u.v^{\prime }\\ 4.& y=k.u^n& =n.k.u^{(n-1)}.u^{\prime }\\ 5.& y={\displaystyle \frac{u}{v}}& y^{\prime }={\displaystyle \frac{u^{\prime }.v-u.v^{\prime }}{v^2}}\\ \hline\end{array}$

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

$y^{\prime }={\displaystyle \frac{dy}{dx}=f^{\prime }(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}}$

dan untuk turunan keduanya dari fungsi di atas adalah:

$y^{″}={\displaystyle \frac{d^{2}y}{d{x}^2}=f^{″}(x)\text{atau kadang dituliskan}{\displaystyle \frac{d{f}^{\prime }(x)}{dx}=\frac{d^{2}f}{d{x}^2}}}$

Selanjutnya perhatikanlah tabel berikut

$\begin{array}{|ll|}\hline \text{Turunan}& \text{Notasi}\\ \text{Pertama}& y^{\prime }=f^{\prime }(x)={\displaystyle \frac{dy}{dx}=\frac{df}{dx}}\\ & \\ \text{Kedua}& y^{″}=f^{″}(x)={\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{d^{2}f}{d{x}^2}}\\ & \\ \text{Ketiga}& y^{‴}=f^{‴}(x)={\displaystyle \frac{d^{3}y}{d{x}^3}=\frac{d^{3}f}{d{x}^3}}\\ & \\ \cdots & \cdots \cdots \cdots \cdots \\ \text{Ke-n}& y^n=f^n(x)={\displaystyle \frac{d^{n}y}{d{x}^n}=\frac{d^{n}f}{d{x}^n}}\\ \hline\end{array}$

$\boxed{\text{CONTOH SOAL}}$

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

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

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

$.\begin{array}{rl}2.(\text{a})\text{dike}& \text{tahui}f(x)={\displaystyle \frac{1+\sin x}{\cos x}}\\ \text{gun}& \text{akan formula}y={\displaystyle \frac{u}{v}\Rightarrow y^{\prime }={\displaystyle \frac{u^{\prime }v-u{v}^{\prime }}{v^2}}}\\ f^{\prime }(x)& ={\displaystyle \frac{(\cos x)(\cos x)-(1+\sin x)(-\sin x)}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+\sin x+{\sin }^{2}x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{{\cos }^{2}x+{\sin }^{2}x+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1+\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{\sin x}{{\cos }^{2}x}}\\ & ={\displaystyle \frac{1}{{\cos }^{2}x}+\frac{1}{\cos x}.\frac{\sin x}{\cos x}}\\ & ={\sec }^{2}x+\sec x\tan x\end{array}$

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

$.\begin{array}{rl}2.(\text{c})\text{dike}& \text{tahui}h(x)=\sin x\tan x\\ \text{gun}& \text{akan formula}y=u.v\Rightarrow y^{\prime }=u^{\prime }.v+u.v^{\prime }\\ h^{\prime }(x)& =\cos x.(\tan x)+\sin x.({\sec }^{2}x)\\ h\left({45}^{\circ }\right)& =\cos \left({45}^{\circ }\right)\tan \left({45}^{\circ }\right)+\sin \left({45}^{\circ }\right){\sec }^2\left({45}^{\circ }\right)\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right).1+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right){\left(\sqrt{2}\right)}^2\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+\sqrt{2}}\\ & ={\displaystyle \frac{3}{2}\sqrt{2}}\end{array}$

$.\begin{array}{rl}2.(\text{d})\text{dike}& \text{tahui}k(x)=\sin x+n\cos x\\ k^{\prime }(x)& =\cos x-n\sin x,\text{dengan}{k}^{\prime }\left({\displaystyle \frac{\pi }{3}}\right)=0\\ k^{\prime }\left({\displaystyle \frac{\pi }{6}}\right)& =\cos \left({\displaystyle \frac{\pi }{3}}\right)-n\sin \left({\displaystyle \frac{\pi }{3}}\right)\\ 0& =\cos {60}^{\circ }-n\sin {60}^{\circ }={\displaystyle \frac{1}{2}-n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}\\ \iff & n\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)={\displaystyle \frac{1}{2}}\\ \iff & n={\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ \iff & n={\displaystyle \frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}}\\ \iff & n={\displaystyle \frac{1}{3}\sqrt{3}}\end{array}$

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

DAFTAR PUSTAKA

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