Question-38 The derivative of \tan^{-1}(\sqrt{1+x^2}-1) is
(A) \frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{x+1}+1)}
(B) \frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{1+x^2}+3)}
(C) \frac{x}{\sqrt{1+x^2}(x^2-2\sqrt{x^2+1}+2)}
(D) \frac{x}{\sqrt{1+x^2}(x^2+2\sqrt{1+x^2}-3)}
Answer: (B)
Explanation:
Let y=\tan^{-1}(\sqrt{1+x^2}-1)
Differentiating w.r.t. x, we get
\frac{dy}{dx}=\frac{1}{1+(\sqrt{1+x^2}-1)^2}\times\frac{d}{dx}(\sqrt{1+x^2}-1) =\frac{1}{1+2+x^2-2\sqrt{1+x^2}}\times\frac{1}{2\sqrt{1+x^2}}\times2x =\frac{x}{(\sqrt{1+x^2})(x^2-2\sqrt{1+x^2}+3)}