# Find the derivative of the given function $5 \sec x +4 \cos x$

Let $f(x) =5 \sec x+4 \cos x$. Accordingly, from the first principle,
$f'(x) = \lim\limits_{h \to 0}= \large\frac{f(x+h)-f(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{5 \sec (x+h) 4 \cos (x+h) - [5 \sec x + 4 \cos x]}{h}$
$=5 \lim\limits_{h \to 0} \large\frac{[ \sec (x+h)- \sec x]}{h} $$=4 \lim\limits_{h \to 0} \large\frac{[ \cos (x+h) - \cos x]}{h} =5 \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[ \large\frac{1}{\cos (x+h)}- \large\frac{1}{\cos x}\bigg]$$+4 \lim\limits_{h \to 0} \large\frac{1}{h}$$[ \cos (x+h)-\cos x]$
$=5 \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{\cos x- \cos(x+h)}{\cos x \cos(x+h)} \bigg]$$+4 \lim\limits_{h \to 0} \large\frac{1}{h}$$[\cos x \cos h - \sin x \sin h - \cos x] =\large\frac{5}{\cos x}$$ \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-2 \sin \Large\frac{x+x+h}{2} \sin \Large\frac{ x-x-h}{2}}{\cos (x+h)} \bigg]$$+4 \lim\limits_{h \to 0}\large\frac{1}{h}$$[ - \cos x(1- \cos h)- \sin x \sin h] =\large\frac{5}{\cos x}$$ \lim\limits_{h \to 0} \large\frac{1}{h}$$\bigg[ \large\frac{-2 \sin \bigg( \Large\frac{2x+h}{2} \bigg)\sin\bigg( -\Large\frac{ h}{2}\bigg)}{\cos (x+h)} \bigg]$$+4 \bigg[ - \cos x \lim\limits_{h \to 0} ( \large\frac{(1- \cos h)}{h} $$- \sin x \lim\limits_{h \to 0} \large\frac{\sin h}{h} \bigg] = \large\frac{5}{\cos x}$$. \lim\limits_{h \to 0}= \large\frac{ \bigg[\sin \bigg( \Large\frac{2x+h}{2} \bigg) \Large\frac{sin\bigg( \Large\frac{h}{2} \bigg)}{ \Large\frac{h}{2} } \bigg]}{\cos (x+h)}$$+4[(-\cos x).(0)-( \sin x).1] = \large\frac{5}{\cos x}$$ \bigg[ \lim\limits_{h \to 0}\large\frac{sin \bigg(\large\frac{ 2x+h}{2} \bigg)}{\cos (x+h)}$$. \lim\limits_{h \to 0} \Large\frac{sin\bigg( \Large\frac{h}{2} \bigg)}{ \Large\frac{h}{2} } \bigg]$$-4 \sin x$
$= \large\frac{5}{\cos x}$$. \large\frac{ \sin x}{\cos x}$$. 1 - 4 \sin x$
$= 5\sec x \tan x- 4 \sin x$