# Find the derivative of the given function $2\tan x - 7 \sec x$

Let $f(x) = 2\tan x - 7 \sec x$. Accordingly, from the first principle,
$f'(x) = \lim\limits_{h \to 0}= \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$[2 \tan (x+h)-7 \sec(x+h)-2\tan x+7 \sec x] = \lim\limits_{h \to 0} \large\frac{1}{h}$$ [2 \{\tan (x+h)-tan x\}-7 \{\sec(x+h) - \sec x\}]$
$=2 \lim\limits_{h \to 0} \large\frac{1}{h}$$[ \tan(x+h)-\tan x]-7 \lim\limits_{h \to 0} \large\frac{1}{h}$$[ \sec(x+h)-\sec x]$
$=2 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin(x+h)}{\cos (x+h)}- \large\frac{\sin x}{\cos x} \bigg]$$ -7 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{1}{\cos (x+h)}- \large\frac{1}{\cos x} \bigg] =2 \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{\sin (x+h)\cos x - \sin x \cos (x+h) }{\cos x \cos(x+h)} \bigg] $$-7 \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{\cos x - \cos(x+h)}{\cos x\cos(x+h)} \bigg]$
$=2 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\sin(x+h-x)}{\cos x \cos (x+h)}\bigg]$$ -7 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-2 \sin \bigg(\Large\frac{x+x+h}{2}\bigg) \sin \bigg(\Large\frac{ x-x-h}{2}\bigg)}{\cos x \cos (x+h)} \bigg] 2 \lim\limits_{h \to 0} \bigg[ \bigg( \large\frac{\sin h}{h} \bigg) \large\frac{1}{\cos x \cos(x+h)} \bigg]$$-7\lim\limits_{h \to 0}\large\frac{1}{h}$$\bigg[ \large\frac{-2 \sin \bigg( \Large\frac{2x+h}{2} \bigg) \sin \bigg( - \Large\frac{\sin h}{h} \bigg) }{\cos x \cos(x+h)}\bigg] = 2 \bigg( \lim\limits_{h \to 0} \large\frac{\sin h}{h} \bigg)\bigg( \lim\limits_{h \to 0} \large\frac{1}{\cos x \cos(x+h)} \bigg)$$-7 \bigg( \lim\limits_{\large\frac{h}{2} \to 0} \large\frac{sin \Large\frac{h}{2}}{ \Large\frac{h}{2}}\bigg)$ $\bigg[ \lim\limits_{h \to 0} \Large\frac{\sin \bigg(\Large\frac{2x+h}{2}\bigg)}{\cos x \cos(x+h)}\bigg]$
$= 2.1. \large\frac{1}{\cos x \cos x}$$-7.1 \bigg( \large\frac{\sin x}{\cos x \cos x} \bigg)$
$= 2\sec^2x-7\sec x \tan x$
edited Apr 10, 2014