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  4. If y= tan -1( sec x), then (d y/d x) at x= operatornamecosec-1 √2 is equal to

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Q. If $y=\tan ^{-1}(\sec x)$, then $\frac{d y}{d x}$ at $x=\operatorname{cosec}^{-1} \sqrt{2}$ is equal to

Continuity and Differentiability

Solution:

$y=\tan ^{-1}(\sec x)$
So, $\left.\frac{ dy }{ dx }\right]_{ x =\frac{\pi}{4}}=\frac{1}{1+\sec ^2 x }(\sec x \cdot \tan x )=\frac{\sqrt{2}}{3}$