If $y = \tan^{-1}\left(\dfrac{\sqrt{1+x^2} - 1}{x}\right)$, then $\dfrac{dy}{dx}$ equals:
A$\dfrac{1}{1+x^2}$
B$\dfrac{1}{2(1+x^2)}$
C$1$
D$\dfrac{1}{\sqrt{1+x^2}}$
Answer & Solution
Correct answer: B. $\dfrac{1}{2(1+x^2)}$
Put $x = \tan\theta$. Then $\sqrt{1+x^2} = \sec\theta$ and the argument becomes $(\sec\theta - 1)/\tan\theta = (1 - \cos\theta)/\sin\theta = \tan(\theta/2)$. So $y = \tan^{-1}\tan(\theta/2) = \theta/2 = (1/2)\tan^{-1}x$. Hence $dy/dx = 1/(2(1+x^2))$.
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