Question 28.7.4: Differentiate e^√cot x....

Let y=e^{\sqrt{\cot x}} . Put \cot x=t and \sqrt{\cot x}=\sqrt{t}=u , so that

y=e^{u}, u=\sqrt{t} \text { and } t=\cot x \text {. }\\  \\\therefore \quad \frac{d y}{d u}=e^{u}, \frac{d u}{d t}=\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} \text { and } \frac{d t}{d x}=-\operatorname{cosec}^{2} x \text {. }

So, \frac{d y}{d x}=\left(\frac{d y}{d u} \times \frac{d u}{d t} \times \frac{d t}{d x}\right)=-\frac{1}{2} \cdot \frac{\operatorname{cosec}^{2} x}{\sqrt{t}} e^{u} \\  \\\begin{array}{l}=\frac{-\operatorname{cosec}^{2} x}{2 \sqrt{t}} \cdot e^{\sqrt{t}} \quad[\because   u=\sqrt{t}]\\  \\=\frac{-\operatorname{cosec}^{2} x}{2 \sqrt{\cot x}} \cdot e^{\sqrt{\cot x}} \quad[\because   t=\cot x] .\end{array}