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Q. If $y=tan^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)$, then $y'(1)$ is equal to

Continuity and Differentiability

Solution:

$y=tan^{-1}\left(\frac{\sqrt{x}-x}{1+x^{3/2}}\right)$
$= tan^{-1}\left(\frac{\sqrt{x}-x}{1+\sqrt{x}\cdot x}\right)$
$=tan^{-1}\left(\sqrt{x}\right)-tan^{-1}\left(x\right)$
Differentiating $w$.$r$.$t$. $x$, we get
$y'=\frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}-\frac{1}{1+x^{2}}$
$\Rightarrow y'\left(1\right)=\frac{1}{2}\cdot\frac{1}{2}-\frac{1}{2}$
$=-\frac{1}{4}$