If $y=\sec \left( an ^{-1} x ight)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ is equal to

Question

If $y=\sec \left(	an ^{-1} xight)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & \begin{aligned} & y=\sec \left(	an ^{-1} xight) \ & 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\sec \left(	an ^{-1} xight) 	an \left(	an ^{-1} xight) \cdot \frac{1}{1+x^2} \ &=\sqrt{1+x^2} \cdot \frac{x}{1+x^2} \ & \ldots\left[\because 	an ^{-1} x=\sec ^{-1} \sqrt{1+x^2}ight] \ &=\frac{x}{\sqrt{1+x^2}} \ & 	herefore \quad\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}ight)_{x=1}=\frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}\end{aligned} .\end{aligned}$

If $y=\sec \left(\tan ^{-1} x\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ is equal to

Solution