A. BENTUK DASAR
| \(f(x)\) | \(f'(x)\) |
|---|---|
| \(\sin x\) | \(\cos x\) |
| \(\cos x\) | \(- \sin x\) |
| \(\tan x\) | \(\sec^2 x\) |
| \(\cot x\) | \(- \csc^2 x\) |
| \(\sec x\) | \(\sec x \tan x\) |
| \(\csc x\) | \(- \csc x \cot x\) |
ATURAN BERANTAI
\(\dfrac {dy}{dx} = \dfrac {dy}{du} \:.\: \dfrac {du}{dx}\)
Contoh 01
Tentukan turunan pertama dari \(y = \sin \: (2x + 5)\)
Cara pemisalan
\begin{equation*} \begin{split} & u =2x + 5 \\\\ & \frac {du}{dx} = 2 \end{split} \end{equation*}
\begin{equation*} \begin{split} & y = \sin u \\\\ & \frac {dy}{dx} = \frac {dy}{du} \:.\: \frac {du}{dx} \\\\ & \frac {dy}{dx} = \cos u \:.\: 2 \\\\ & \frac {dy}{dx} = 2 \:.\: \cos u \\\\ & \frac {dy}{dx} = \bbox[5px, border: 2px solid magenta] {2 \:.\: \cos \: (2x + 5)} \end{split} \end{equation*}
Cara langsung
\begin{equation*} \begin{split} & y = \sin \: ({\color {red} 2x + 5}) \\\\ & {\color {blue} \text{turunan dari } \sin \: (2x + 5) \text{ adalah } \cos \: (2x + 5)} \\\\ & {\color {blue} \text{turunan dari } 2x + 5 \text{ adalah } 2} \\\\ & \frac {dy}{dx} = \bbox[5px, border: 2px solid magenta] {2 \:.\: \cos \: (2x + 5)} \end{split} \end{equation*}
Contoh 02
Tentukan turunan pertama dari \(y = \cos^3 x\)
Cara pemisalan
\begin{equation*} \begin{split} & u =\cos x \\\\ & \frac {du}{dx} = - \sin x \end{split} \end{equation*}
\begin{equation*} \begin{split} & y = u^3 \\\\ & \frac {dy}{dx} = \frac {dy}{du} \:.\: \frac {du}{dx} \\\\ & \frac {dy}{dx} = 3 \:.\: u^2 \:.\: -\sin x \\\\ & \frac {dy}{dx} = -3 \:.\: \sin x \:.\: u^2 \\\\ & \frac {dy}{dx} = \bbox[5px, border: 2px solid magenta] {-3 \:.\: \sin x \:.\: \cos^2 x} \end{split} \end{equation*}
Cara langsung
\begin{equation*} \begin{split} & y = \cos^3 x \\\\ & {\color {blue} \text{turunan dari } \cos^3 x \text{ adalah } 3 \cos^2 x} \\\\ & {\color {blue} \text{turunan dari } \cos x \text{ adalah } - \sin x} \\\\ & \frac {dy}{dx} = - \sin x \:.\: 3 \:.\: \cos^2 x \\\\ & \frac {dy}{dx} = \bbox[5px, border: 2px solid magenta] {-3 \:.\: \sin x \:.\: \cos^2 x} \end{split} \end{equation*}
B. PERKALIAN DAN PEMBAGIAN
\(y = u \:.\: v \rightarrow y' = u' \:.\: v + u \:.\: v'\)
\(y = u \:.\: v \:.\: w \rightarrow y' = u' \:.\: v \:.\: w + u \:.\: v' \:.\: w + u \:.\: v \:.\: w'\)
\(y = \dfrac uv \rightarrow y' = \dfrac {u' \:.\: v - u \:.\: v'}{v^2}\)
Contoh 01
\(y = 5x \:.\: \sin x\)
\begin{equation*} \begin{split} & u = 5x \\\\ & u' = 5 \end{split} \end{equation*}
\begin{equation*} \begin{split} & v = \sin x \\\\ & v' = \cos x \end{split} \end{equation*}
\begin{equation*} \begin{split} & y' = u' \:.\: v + u \:.\: v' \\\\ & y' = \bbox[5px, border: 2px solid magenta] {5 \:.\: \sin x + 5x \:.\:\cos x} \end{split} \end{equation*}
Contoh 02
\(y = \dfrac {x}{\tan 3x}\)
\begin{equation*} \begin{split} & u = x \\\\ & u' = 1 \end{split} \end{equation*}
\begin{equation*} \begin{split} & v = \tan 3x \\\\ & v' = 3 \:.\: \sec^2 x \end{split} \end{equation*}
\begin{equation*} \begin{split} & y' = \frac {u' \:.\: v - u \:.\: v'}{v^2} \\\\ & y' = \frac {1 \:.\: \tan 3x - x \:.\: 3 \:.\: \sec^2 3x}{(\tan 3x)^2} \\\\ & y' = \bbox[5px, border: 2px solid magenta] {\frac {\tan 3x - 3x \:.\: \sec^2 3x}{\tan^2 3x}} \end{split} \end{equation*}
C. BENTUK IMPLISIT
Bentuk implisit dapat diturunkan dengan cara melakukan penurunan secara keseluruhan.
Contoh 01
\begin{equation*} \begin{split} & \sin x + \cos y = 1 \\\\ & \cos x + y' \:.\: - \sin y = 0 \\\\ & -y \:.\: \sin y = - \cos x \\\\ & y' = \frac {\cos x}{\sin x} \end{split} \end{equation*}
Catatan:
- turunan dari \(y\) adalah \(y'\), sehingga turunan dari \(\cos y\) adalah \(y' \:.\: - \sin y\)
Contoh 02
Diketahui \(x^2 + y^2 = \tan xy\)
Tentukan nilai dari \(y'\) untuk \(x = 0\) dan \(y = 1\)
\begin{equation*} \begin{split} & x^2 + y^2 = \tan xy \\\\ & 2x + y' \:.\: 2y = (1 \:.\: y + x \:.\: y') \:.\: \sec^2 xy \\\\ & 2(0) + y' \:.\: 2 (1) = (1 \:.\: 1 + 0 \:.\: y') \:.\: \sec^2 \: (0 \:.\: 1) \\\\ & 2y' = \sec^2 \: 0 \\\\ & 2y' = \frac {1}{\cos^2 \: 0} \\\\ & 2y' = \frac {1}{1} \\\\ & y' = \frac 12 \end{split} \end{equation*}
D. INVERS TRIGONOMETRI
| F(x) | F'(x) |
| \(\arcsin x\) | \(\dfrac {1}{\sqrt{1 - x^2}}\) |
| \(\arccos x\) | \(- \dfrac {1}{\sqrt{1 - x^2}}\) |
| \(\arctan x\) | \(\dfrac {1}{1 + x^2}\) |