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

Answer

We have, $\text{y}=\log\sqrt{\tan\text{x}}$
$\Rightarrow\text{y}=\log(\tan\text{x})^\frac{1}{2}$
$\Rightarrow\text{y}=\frac{1}{2}\log(\tan\text{x})\big[\because\log\text{a}^\text{b}=\text{b}\log\text{a}\big]$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}\frac{\text{d}}{\text{dx}}(\tan\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\times\frac{1}{\tan\text{x}}(\sec^2\text{x})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\frac{\sin\text{x}}{\cos\text{x}}}\times\cos^2\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sin\text{x}\cos\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin2\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{cosec }2\text{x}$

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