Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
Consider the function $f(x) = \tan^{-1}\left(\frac{x^2-1}{x^2+1}\right), \forall x \geq 0$. If $g(x)$ is the inverse function of $f(x)$, then the value of $g'\left(\frac{\pi}{4}\right)$ is equal to
Step-by-Step Solution
Key Concept: Differentiation of inverse trigonometric functions and chain rule application
Given $f(x) = \tan^{-1}\left(\frac{x-1}{1+x}\right) = \tan^{-1}x - \tan^{-1}1$. Differentiating: $f'(x) = \frac{1}{1+x^2}$. Using the chain rule on $g(f(x))$ where $g'(f(x)) = x$, we get $g'(f(b)) \cdot f'(b) = b$. Putting $x = b$, we find $g'\left(\tan^{-1}(b)\right) = \frac{1}{1+b^2}$. Since $\tan^{-1}(1) = \frac{\pi}{4}$, we have $g'\left(\frac{\pi}{4}\right) = 26$.
Correct Answer: 26