Aturan Rantai pada Turunan Pertama dan Turunan Kedua Fungsi Aljabar (Lanjutan Materi Turunan Fungsi Aljabar)

 Aturan Rantai

$\begin{array}{rl}& \\ \text{Untuk}:& \\ & \{\begin{array}{ll}a& \in \mathbb{R}\\ n& \in \mathbb{Q}\\ c& \text{konstanta}\\ U& =g(x)\\ V& =h(x)\end{array}\\ & \end{array}$.

$\begin{array}{|l|}\hline \mathbf{\text{Sifat-Sifat}}\\ \begin{array}{rl}y& =c\to y^{\prime }=0\\ y& =c.U\to y^{\prime }=c.U^{\prime }\\ y& =U\pm V\to y^{\prime }=U^{\prime }\pm V^{\prime }\\ y& =U.V\to y^{\prime }=U^{\prime }.V+U.V^{\prime }\\ y& ={\displaystyle \frac{U}{V}\to y^{\prime }={\displaystyle \frac{U^{\prime }.V-U.V^{\prime }}{V^2}}}\end{array}\\ \mathbf{\text{Fungsi Aljabar}}\\ \begin{array}{rl}y& =a.x^n\to y{}^{\prime }=n.a.x^{(n-1)}\\ y& =a.U^n\to y{}^{\prime }=n.a.U^{(n-1)}.U^{\prime }\end{array}\\ \mathbf{\text{Fungsi Trigonometri}}\\ \begin{array}{rlrl}y& =a\sin U\to & y^{\prime }& =(a\cos U).U^{\prime }\\ y& =a\cos U\to & y^{\prime }& =(-a\sin U).U^{\prime }\\ y& =a\tan U\to & y^{\prime }& =(a{\sec }^{2}U).U^{\prime }\end{array}\\ \mathbf{\text{Aturan rantai }}\\ \begin{array}{rl}& \text{pada turunan untuk}y=f(u),\text{jika}\\ & \text{untuk}u\text{merupakan fungsi}x,\text{maka}:\\ & y{}^{\prime }=f{}^{\prime }(x).u^{\prime }\text{atau}{\displaystyle \frac{dy}{dx}={\displaystyle \frac{dy}{du}.\frac{du}{dx}}}\\ & \end{array}\\ \hline\end{array}$.

Turunan Kedua Fungsi Aljabar

$\begin{array}{rl}& \\ \text{Suatu fungsi}& \\ & f:x\to y\\ & \end{array}$.

$\begin{array}{|cc|}\hline \text{Notasi}& \text{Proses}\\ \begin{array}{rl}& \\ & \{\begin{array}{ll}f{}^{\prime }(x)& ={\displaystyle \frac{dy}{dx}}\\ \\ f{}^{″}(x)& ={\displaystyle \frac{d^{2}y}{d{x}^2}}\end{array}\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \\ y& =a{x}^n\\ y^{\prime }& =n.a.x^{n-1}\\ y^{″}& =(n-1).n.a.x^{n-2}\\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai dari}{\displaystyle \frac{dy}{dx}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=\sqrt{3+\sqrt{3+\sqrt{x}}}\\ & \text{d}.y=\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{x}}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|cc|}\hline \begin{array}{rl}\text{a}.y& =\sqrt{x}\\ & =x^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}.x^{{}^{\frac{1}{2}-1}}}\\ & ={\displaystyle \frac{1}{2}{x}^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{2x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{2\sqrt{x}}}\\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{b}.y& =\sqrt{3+\sqrt{x}}\\ & ={(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}\\ y^{\prime }& ={\displaystyle \frac{1}{2}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}-1}}.{\displaystyle \frac{1}{2}{x}^{{}^{\frac{1}{2}-1}}}}\\ & ={\displaystyle \frac{1}{4}{(3+x^{{}^{\frac{1}{2}}})}^{{}^{-\frac{1}{2}}}.x^{{}^{-\frac{1}{2}}}}\\ & ={\displaystyle \frac{1}{4{(3+x^{{}^{\frac{1}{2}}})}^{{}^{\frac{1}{2}}}.x^{{}^{\frac{1}{2}}}}}\\ & ={\displaystyle \frac{1}{4\sqrt{3+\sqrt{x}}.\sqrt{x}}}\\ & \end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah turunannya}\\ & \begin{array}{l}\\ \text{a}.& f(x)=2\sin x\cos x& \text{k}.& f(x)=4\sin {(5x+\pi )}^3\\ \text{b}.& f(x)=x^2.\sin x& \text{l}.& f(x)={\cos }^3(x+5)& \\ \text{c}.& f(x)=3{\cos }^{2}x& \text{m}.& f(x)={\sin }^2(\pi -3x)& \\ \text{d}.& f(x)=3{\sin }^{2}x-x^3& \text{n}.& f(x)={\displaystyle \frac{{\sin }^{2}x}{\cos x}}& \\ \text{e}.& f(x)=5{\sin }^{4}x& \text{o}.& f(x)={\displaystyle \frac{\sin x^4}{x^2}}& \\ \text{f}.& f(x)=\sin \left({\displaystyle \frac{x+3}{x-2}}\right)\\ \text{g}.& f(x)=\cos (x^2-6x)\\ \text{h}.& f(x)={\displaystyle \frac{1}{\sin x}}\\ \text{i}.& f(x)={\displaystyle \frac{\sin x}{1+\cos x}}\\ \text{j}.& f(x)={\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}}\end{array}\end{array}$.

$\begin{array}{rl}& \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ 2.\text{a}& \text{Alternatif 1}\\ & f(x)=2\sin x\cos x\\ & \begin{array}{rl}f(x)& =\sin 2x\\ f{}^{\prime }(x)& =\cos 2x.(2)\\ & =2\cos 2x\end{array}\\ \\ & \text{Alternatif 2}\\ & f(x)=y=\sin 2x\\ & \{\begin{array}{ll}y=\sin u& \to {\displaystyle \frac{dy}{du}=\cos u}\\ \\ u=2x& \to {\displaystyle \frac{du}{dx}=2}\end{array}\\ & \text{sehingga}\\ & \begin{array}{rl}y& =\sin 2x\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dx}}\\ & =\cos u.2\\ & =2\cos u\\ & =2\cos 2x\end{array}\end{array}\end{array}$. 

$\begin{array}{l}\\ 2.\text{b}& f(x)=x^2\sin x\\ & \begin{array}{rl}f{}^{\prime }(x)& =2x^{2-1}.(\sin x)+x^2.(\cos x)\\ & =2x\sin x+x^2\cos x\end{array}\\ \text{c}& f(x)=3{\cos }^{2}x\\ & f^{\prime }(x)=2.3.{\cos }^{2-1}x.(-\sin x)\\ & =-6\sin x\cos x\\ & =-3\sin 2x\\ \text{d}& f(x)=3{\sin }^{2}x-x^3\\ & f^{\prime }(x)=2.3.{\sin }^{2-1}x.(\cos x)-3x^{3-1}\\ & =6\sin x\cos x-3x^2\\ & =3\sin 2x-3x^2\\ \text{e}& f(x)=5{\sin }^{4}x\\ & f^{\prime }(x)=4.5.{\sin }^{4-1}x.(\cos x)\\ & =20{\sin }^{3}x\cos x\\ & \text{atau boleh juga}\\ & =20{\sin }^{2}x\sin x\cos x\\ & =20{\sin }^{2}x\sin 2x=20\sin 2x{\sin }^{2}x\end{array}$ .

$\begin{array}{l}\\ 2.\text{m}& y=f(x)={\sin }^2(\pi -3x)\\ & \begin{array}{rl}f{}^{\prime }(x)& =2{\sin }^{2-1}(\pi -3x).\cos (\pi -3x).(-3)\\ & =-6\sin (\pi -3x)\cos (\pi -3x)\\ & =-3.2\sin (\pi -3x)\cos (\pi -3x)\\ & =-3\sin 2(\pi -3x)\\ & =-3\sin (2\pi -6x)\end{array}\\ \\ & \text{atau boleh juga}\\ & \begin{array}{rl}y=& f(x)={\sin }^2(\pi -3x)\\ & \{\begin{array}{ll}y=u^2& \to {\displaystyle \frac{dy}{du}=2u.}\\ \\ u=\sin w& \to {\displaystyle \frac{du}{dw}=\cos w}\\ \\ w=\pi -3x& \to {\displaystyle \frac{dw}{dx}=-3}\end{array}\\ {\displaystyle \frac{dy}{dx}}& ={\displaystyle \frac{dy}{du}.\frac{du}{dw}.\frac{dw}{dx}}\\ & =\left(2u\right).(\cos w).(-3)\\ & =-3.(2\sin w).(\cos w)\\ & =-3\sin 2w\\ & =-3\sin (2\pi -6x)\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukanlah nilai dari}{\displaystyle \frac{d^{2}y}{d{x}^2}}\\ & \text{a}.y=\sqrt{x}\\ & \text{b}.y=\sqrt{3+\sqrt{x}}\\ & \text{c}.y=15x^3-4x\\ & \text{d}.y={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Lihat pembahasan soal no.1, yaitu }\\ & y=\sqrt{x}\\ & y^{\prime }={\displaystyle \frac{1}{2\sqrt{x}}}\end{array}\\ & \text{maka}\\ & \begin{array}{rl}3.\text{a}.y& =\sqrt{x}\\ y^{\prime }& ={\displaystyle \frac{dy}{dx}={\displaystyle \frac{1}{2\sqrt{x}}={\displaystyle \frac{1}{2}{x}^{-\frac{1}{2}}}}}\\ y^{″}& ={\displaystyle \frac{d^{2}y}{d{x}^2}=(-{\displaystyle \frac{1}{2}}){\displaystyle \frac{1}{2}.x^{-\frac{1}{2}-1}}}\\ & =-{\displaystyle \frac{1}{4x^{\frac{1}{2}+1}}=-{\displaystyle \frac{1}{4x\sqrt{x}}}}\end{array}\end{array}$.

$.\begin{array}{|l|}\hline \begin{array}{rl}& 3.\text{d}\\ & \begin{array}{rl}y& ={\displaystyle \frac{x^4}{3}+\frac{x^2}{2}+x+\sqrt{x}+1+{\displaystyle \frac{1}{\sqrt{x}}+\frac{1}{x}+\frac{1}{x^2}+\frac{3}{x^4}}}\\ & \parallel \\ & \mathbf{\text{Turunan pertama}}\\ & \Downarrow \\ {\displaystyle \frac{dy}{dx}=y{}^{\prime }}& ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}+0-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & ={\displaystyle \frac{4x^3}{3}+x+1+{\displaystyle \frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}-\frac{1}{x^2}-\frac{2}{x^3}-\frac{12}{x^5}}}\\ & \parallel \\ & \mathbf{\text{Turunan kedua}}\\ & \Downarrow \\ {\displaystyle \frac{d^{2}y}{d{x}^2}=y{}^{″}}& =4x^2+1+0-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\\ & =4x^2+1-{\displaystyle \frac{1}{4x\sqrt{x}}+\frac{3}{4x^2\sqrt{x}}+\frac{2}{x^3}+\frac{6}{x^4}+\frac{60}{x^6}}\end{array}\\ & \end{array}\\ \hline\end{array}$.

$\text{LATIHAN SOAL}$.

Silahkan kerjakan soal yang belum diselesaikan atau dijawab

DAFTAR PUSTAKA

1. Kanginan, M., Terzalgi, Y. 2014. Matematika untuk SMA-MA/SMK Kelas XI (Wajib). Bandung: SEWU
2. Kartrini, Suprapto, Subandi, dan Setiyadi, U. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: INTAN PARIWARA.
3. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 1 Program IPA Standar Isi 2006. Jakarta: ESIS
4. Sunardi, Waluyo, S., Sutrisno, Subagya. 2004. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.