Question

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

• A. $\frac{1}{2}$
• B. $1$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

$y=sec\left(ta{n}^{-1}x\right)\therefore sec\left(ta{n}^{-1}x\right)tan\left(ta{n}^{-1}x\right)\cdot \left(\frac{1}{1+x^2}\right)putx=1then\left(\frac{dy}{dx}\right)=sec\left(ta{n}^{-1}1\right)tan\left(ta{n}^{-1}1\right)\cdot \left(\frac{1}{1+1^2}\right)=sec\left(\frac{\pi }{4}\right)\cdot tan\left(\frac{\pi }{4}\right)\cdot \left(\frac{1}{2}\right)=\sqrt{2}\cdot 1\cdot \left(\frac{1}{2}\right)=\left(\frac{1}{\sqrt{2}}\right)$