Solve this


If $\mathrm{y}=\log \sqrt{\tan \mathrm{x}}$, write $\frac{\mathrm{dy}}{\mathrm{dx}}$


This particular problem is a perfect way to demonstrate how simple but powerful the Chain Rule of Differentiation is.

It is important to identify and break the problem into the individual functions with respect to which successive differentiation shall be done.

In this case, this is the way to break down the problem -

$\frac{d y}{d x}=\frac{d y}{d(\sqrt{\tan x})} \cdot \frac{d(\sqrt{\tan x})}{d(\tan x)} \cdot \frac{d(\tan x)}{d x}$

i.e., $\frac{d y}{d x}=\frac{d(\log \sqrt{\tan x})}{d(\sqrt{\tan x})} \cdot \frac{d(\sqrt{\tan x})}{d(\tan x)} \cdot \frac{d(\tan x)}{d x}$

$=\frac{1}{\sqrt{\tan x}} \cdot \frac{1}{2 \sqrt{\tan x}} \cdot \sec ^{2} x$

$=\frac{\sec ^{2} x}{2 \tan x}$

$=\frac{1+\tan ^{2} x}{2 \tan x}$

$=\frac{1}{2}(\tan x+\cot x)$ (Ans)

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