MATEMATIKA INDONESIA: Turunan 5

Apa yang diajarkan di bagian ini adalah lanjutan dari turunan 1

Contoh 8

Tentukan turunan pertama dari fungsi f(x) = sec x

Jawab :

$f'(x)=\begin{matrix} lim\\ h\rightarrow 0 \end{matrix} \frac{\sec(x+h)-\sec x}{h}$

$f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\frac{1}{cos(x+h)}-\frac{1}{cosx}}{h} \end{matrix}$

$f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\frac{\cos x-cos(x+h)}{cos(x+h)\cos x}}{h} \end{matrix}$

$f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\cos x-cos(x+h)}{hcos(x+h)\cos x} \end{matrix}$

$f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{-2sin\frac{1}{2}(2x+h)sin\frac{1}{2}(-h)}{hcos(x+h)\cos x} \end{matrix}$

$f'(x)=\begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{2\sin \frac{1}{2}h}{h}.\frac{sin\frac{1}{2}(2x+h)}{\cos (x+h)\cos x} \end{matrix}$

$f'(x)= 2.\frac{1}{2}.\frac{sin\frac{1}{2}(2x+0)}{\cos x.\cos x}=\frac{1}{\cos x}.\frac{\sin x}{\cos x}=\sec x.\tan x$

Contoh 9 :
Tentukan turunan pertama dari f(x) = cot x

Jawab :

$f'(x)= \begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\cot (x+h)-\cot x}{h} \end{matrix}$

$f'(x)= \begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\frac{1}{tan(x+h)}-\frac{1}{tanx}}{h} \end{matrix}$

$f'(x)= \begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\frac{\tan x -\tan (x+h)}{tan(x+h).tanx}}{h} \end{matrix}$

$f'(x)= \begin{matrix} \begin{matrix} lim\\h \to \0 \end{matrix} & \frac{\tan x -\tan (x+h)}{h.tan(x+h).tanx} \end{matrix}$

Dengan mengingat tan A - tan B = tan (A - B)(1 + tan A tan B) maka diperoleh

$f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}$ $\frac{\tan \left ( x-(x+h) \right )(1+\tan x.\tan(x+h))}{h.tan(x+h).\tan x}$

$f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}$ $\frac{\tan \left ( -h)(1+\tan x.\tan(x+h))}{h.tan(x+h).\tan x}$

$f'(x)=\begin{matrix} lim\\h \to \0 \end{matrix}$ $-\frac{\tan h}{h}.\frac{\left (1+\tan x.\tan(x+h))}{tan(x+h).\tan x}$

$f'(x)=-1.\frac{\left (1+\tan x.\tan(x+0))}{tan(x+0).\tan x}$

$f'(x)=-\frac{ 1+\tan^{2} x}{\tan^{2} x}=-\frac{\sec^{2}x}{\tan^{2} x}=-\sec^{2}x.\frac{1}{\tan^{2} x}$

$f'(x)=-\sec^{2}x.\frac{1}{\tan^{2} x}=-\frac{1}{\cos^{2}x}.\frac{\cos^{2}x}{\sin^{2}x}=-\frac{1}{\sin^{2} x}=-\csc^{2}x$

materi sebelumnya (turunan 4)
materi berikutnya (turunan trigonometri)