$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$…

Question

$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:

Accepted Answer

D. $\dfrac{2}{1+x^2}$

Substitute $x = 	an	heta$.
Then $(1-x^2)/(1+x^2) = \cos 2	heta$.
So $\cos^{-1}\cos 2	heta = 2	heta = 2	an^{-1}x$ (for $x > 0$, i.e. $0 < 	heta < \pi/2$, so $0 < 2	heta < \pi$ — valid range of $\cos^{-1}$).
Derivative = $2/(1+x^2)$.

Answer

$\dfrac{2}{\sqrt{1-x^2}}$

Answer

$\dfrac{-2}{1+x^2}$

Answer

$\dfrac{1}{1+x^2}$

$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:

Keep practising Unit 2: Differentiation and Fundamental Derivative Rules