Method of Differentiation
Inverse Trigonometric Functions
JEE Main 2018
Grade 12
Question:
If $f(x) = \tan^{-1}\left(\frac{\sqrt{1 + x^2} - 1}{x}\right)$, then $f'(x)$ equals:
(1) $\frac{1}{1 + x^2}$
(2) $\frac{1}{2(1 + x^2)}$
(3) $\frac{2}{1 + x^2}$
(4) $-\frac{1}{2(1 + x^2)}$
Step-by-Step Solution
Key Concept: Substitute $x = \tan \theta$; the expression inside the inverse tangent collapses to $\tan(\theta/2)$, so $f(x)$ reduces to $\frac{1}{2} \tan^{-1} x$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (2)