Question

Differentiate with respect to x :

sec-1 ( x2 + 1 / x2 - 1 )

The answer: -2 / ( 1 + x2 )

Answer 1

Answer:

Step-by-step explanation:

Concept:

$sec^{-1}x=\pi-sec^{-1}x\\\\cos2A= \frac{1-tan^2\theta}{1+tan^2\theta}$

$y=sec^{-1}[\frac{x^2+1}{x^2-1}]\\\\y=sec^{-1}[-(\frac{1+x^2}{1-x^2})]\\\\y=\pi-sec^{-1}[(\frac{1+x^2}{1-x^2})]$

take

$x=tan\theta\\\theta=tan^{-1}x\\\\y=\pi-sec^{-1}[(\frac{1+tan^2\theta}{1-tan^2\theta})]\\\\y=\pi-sec^{-1}[\frac{1}{\frac{1-tan^2\theta}{1+tan^2\theta}}]\\\\y=\pi-sec^{-1}[\frac{1}{cos2\theta}]\\\\y=\pi-sec^{-1}[sec2\theta]\\\\y=\pi-2\theta\\\\y=\pi-2tan^{-1}x$

differentiate with respect to x

$\frac{dy}{dx}=-2(\frac{1}{1+x^2})\\\\\frac{dy}{dx}=\frac{-2}{1+x^2}$

Answer 2

Answer:

your answer is this