Question

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

• A. $xy/(1+x)$
• B. $xy\sqrt{1+x^2}$
• C. $y\sqrt{1+x^2}$
• D. $xy1+x^2$

Answer 1

The correct option is A $xy/(1+x)$

$y=sec\left(ta{n}^{-1}x\right)$

$\frac{dy}{dx}=sec\left(ta{n}^{-1}x\right).tan\left(ta{n}^{-1}x\right).\frac{1}{1+x^2}$

$y$ $x$

$\frac{dy}{dx}=\frac{yx}{1+x^2}$

Using

differentiation of $secx=secxtanx$

deff of $ta{n}^{-1}x=\frac{1}{1+x^2}$