$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals:
A$\dfrac{2}{\sqrt{1-x^2}}$
B$\dfrac{-2}{1+x^2}$
C$\dfrac{1}{1+x^2}$
D$\dfrac{2}{1+x^2}$
Answer & Solution
Correct answer: D. $\dfrac{2}{1+x^2}$
Substitute $x = \tan\theta$. Then $(1-x^2)/(1+x^2) = \cos 2\theta$. So $\cos^{-1}\cos 2\theta = 2\theta = 2\tan^{-1}x$ (for $x > 0$, i.e. $0 < \theta < \pi/2$, so $0 < 2\theta < \pi$ — valid range of $\cos^{-1}$). Derivative = $2/(1+x^2)$.
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