Differentiate each of the following functions from the first principal :

We have to find the derivative of $\mathrm{e}^{\sqrt{\cot x}}$ with the first principle method, so,

$f(x)=e^{\sqrt{\cot x}}$

by using the first principle formula, we get,

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

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot (x+h)}}-e^{\sqrt{\cot x}}}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right)}{h}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}\left(e^{\sqrt{\cot (x+h)}-\sqrt{\cot x}}-1\right)}{h} \times \frac{(\sqrt{\cot (x+h)}-\sqrt{\cot x})}{\sqrt{\cot (x+h)}-\sqrt{\cot x}}$

[By using $\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1$ ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}}{h} \times(\sqrt{\cot (x+h)}-\sqrt{\cot x}) \times \frac{(\sqrt{\cot (x+h)}+\sqrt{\cot x})}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$

[Rationalizing]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}}{h} \times(\cot (x+h)-\cot x) \times \frac{1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}}{h} \times \frac{\cos (x+h) \sin x-\sin (x+h) \cos x}{\sin x \sin (x+h)} \times \frac{1}{\sqrt{\cot (x+h)+\sqrt{\cot x}}}$

$[\sin A \cos B-\cos A \sin B=\sin (A-B)]$

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}}{h} \times \frac{\sin (x-x-h)}{\sin x \sin (x+h)} \times \frac{1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$

[By using $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$ ]

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{e^{\sqrt{\cot x}}}{\sin x \sin (x+h)} \times \frac{-1}{\sqrt{\cot (x+h)}+\sqrt{\cot x}}$

$f^{\prime}(x)=\frac{-\operatorname{cosec}^{2} x e^{\sqrt{\cot x}}}{2 \sqrt{\cot x}}$