Find the derivative of the following functions

Let $f(x)=\frac{\sin (x+a)}{\cos x}$

By quotient rule,

$f^{\prime}(x)=\frac{\cos x \frac{d}{d x}[\sin (x+a)]-\sin (x+a) \frac{d}{d x} \cos x}{\cos ^{2} x}$

$f^{\prime}(x)=\frac{\cos x \frac{d}{d x}[\sin (x+a)]-\sin (x+a)(-\sin x)}{\cos ^{2} x}$ $\ldots$ (i)

Let $g(x)=\sin (x+a)$. Accordingly, $g(x+h)=\sin (x+h+a)$

By first principle,

$g^{\prime}(x)=\lim _{h \rightarrow 0} \frac{g(x+h)-g(x)}{h}$

$=\lim _{h \rightarrow 0} \frac{1}{h}[\sin (x+h+a)-\sin (x+a)]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{x+h+a+x+a}{2}\right) \sin \left(\frac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{h \rightarrow 0} \frac{1}{h}\left[2 \cos \left(\frac{2 x+2 a+h}{2}\right) \sin \left(\frac{h}{2}\right)\right]$

$=\lim _{h \rightarrow 0}\left[\cos \left(\frac{2 x+2 a+h}{2}\right)\left\{\frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right\}\right]$

$=\lim _{h \rightarrow 0} \cos \left(\frac{2 x+2 a+h}{2}\right), \lim _{\frac{h}{2} \rightarrow 0}\left\{\frac{\sin \left(\frac{h}{2}\right)}{\left(\frac{h}{2}\right)}\right\} \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \frac{h}{2} \rightarrow 0\right]$

$=\left(\cos \frac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{h \rightarrow 0} \frac{\sin h}{h}=1\right]$

$=\cos (x+a)$ (ii)

From (i) and (ii), we obtain

$f^{\prime}(x)=\frac{\cos x \cdot \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x}$

$=\frac{\cos (x+a-x)}{\cos ^{2} x}$

$=\frac{\cos a}{\cos ^{2} x}$