# Find the derivative of the following functions ( it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers ) $\large\frac{\sin (x+a)}{ \cos x}$

Let $f(x) = \large\frac{\sin (x+a)}{ \cos x}$
By quotient rule,
$f'(x) = \large\frac{\cos x \large\frac{d}{dx}[ \sin (x+a)]- \sin (x+a)\large\frac{d}{dx} \cos x}{\cos^2x}$
$f'(x) = \large\frac{\cos x \large\frac{d}{dx}[ \sin (x+a)]- \sin (x+a)- \sin x}{\cos^2x}$----(1)
Let $g(x) = \sin(x+a)$. Accordingly, $g(x+h)=\sin (x+h+a)$
By first principle,
$g'(x) = \lim\limits_{ h \to 0} \large\frac{g(x+h)-g(x)}{h}$
$= \lim\limits_{h \to 0} \large\frac{1}{h}$$[\sin (x+h+a)- \sin (x+a)] = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[2 \cos \bigg( \large\frac{x+h+a+x+a}{2}\bigg)$$\sin \bigg( \large\frac{x+h+a-x-a}{2} \bigg) \bigg] = \lim\limits_{h \to 0} \large\frac{1}{h}$$ \bigg[2 \cos \bigg( \large\frac{2x+2a+h}{2}\bigg)$$\sin \bigg( \large\frac{h}{2} \bigg) \bigg] = \lim\limits_{h \to 0} \bigg[ cos \bigg( \large\frac{2x+2a+h}{2} \bigg)$$\bigg \{ \large\frac{ \sin\bigg(\large\frac{h}{2}\bigg)}{\bigg(\large\frac{h}{2} \bigg) } \bigg\} \bigg]$
$= \lim\limits_{h \to 0} cos \bigg( \large\frac{2x+2a+h}{2} \bigg)$$\lim\limits_{\large\frac{ h}{2} \to 0}\bigg \{ \large\frac{ \sin\bigg(\large\frac{h}{2}\bigg)}{\bigg(\large\frac{h}{2} \bigg) } \bigg\}$$\qquad \bigg[ As \: h \to 0 \Rightarrow \large\frac{h}{2} \to 0\bigg]$
$= \bigg(\cos \large\frac{2x+2a}{2} \bigg) $$\times 1 \qquad \bigg[ \lim\limits_{h \to 0} \large\frac{\sin h}{h}$$ = 1 \bigg]$
$= \cos(x+a)$-------------(2)
From (1) and (2), we obtain,
$f'(x) = \large\frac{\cos x.\cos(x+a)+ \sin x \sin (x+a)}{\cos^2x}$
$= \large\frac{\cos (x+a-x)}{\cos^2x}$
$= \large\frac{\cos a}{\cos^2x}$