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Q. The derivative of $\cot \sqrt{x}$ is

Limits and Derivatives

Solution:

Let $f(x)=\cot \sqrt{x}$.
Then, $f(x+h)=\cot \sqrt{x+h}$
$\therefore \frac{d}{d x} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\Rightarrow \frac{d}{d x} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{\cot \sqrt{x+h}-\cot \sqrt{x}}{h}$
$\Rightarrow \frac{d}{d x}(f(x))=\displaystyle\lim _{h \rightarrow 0} \frac{-\sin (\sqrt{x+h}-\sqrt{x})}{h \sin \sqrt{x+h} \sin \sqrt{x}}$
$\Rightarrow \frac{d}{d x} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{-\sin (\sqrt{x+h}-\sqrt{x})}{[(x+h)-x] \sin \sqrt{x+h} \sin \sqrt{x}]}$
$\Rightarrow \frac{d}{d x} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{-\sin (\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})} x \sin \sqrt{x+h} \sin \sqrt{x}$
$\Rightarrow \frac{d}{d x} f(x)=\displaystyle\lim _{h \rightarrow 0} \frac{-\sin (\sqrt{x+h}-\sqrt{x})}{\sqrt{x+h}-\sqrt{x}}\times \displaystyle\lim _{h \rightarrow 0} \frac{1}{(\sqrt{x+h}+\sqrt{x}) \sin \sqrt{x+h} \sin \sqrt{x}}$
$\Rightarrow \frac{d}{d x} f(x)=\frac{-1}{2 \sqrt{x} \sin \sqrt{x} \sin \sqrt{x}}$
$\Rightarrow \frac{d}{d x} f(x)=\frac{-\operatorname{cosec}^2 \sqrt{x}}{2 \sqrt{x}}$