JEE Advanced 2025 Paper 2 · Q14 · Differentiation Rules
Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to (0, 4)$ be functions defined by $$f(x) = \log_e(x^2 + 2x + 4),\ \text{and}\ g(x) = \frac{4}{1 + e^{-2x}}.$$
Define the composite function $f \circ g^{-1}$ by $(f \circ g^{-1})(x) = f(g^{-1}(x))$, where $g^{-1}$ is the inverse of $g$.
Then the value of the derivative of $f \circ g^{-1}$ at $x = 2$ is __________.
Reveal answer + step-by-step solution
Correct answer:0.25
Solution
Chain rule: $(f \circ g^{-1})'(x) = \frac{f'(g^{-1}(x))}{g'(g^{-1}(x))}$. Solve $g(t) = 2$: $\frac{4}{1 + e^{-2t}} = 2$ $\to$ $1 + e^{-2t} = 2$ $\to$ $e^{-2t} = 1$ $\to$ $t = 0$. So $g^{-1}(2) = 0$. $f'(t) = \frac{2t + 2}{t^2 + 2t + 4}$; $f'(0) = \frac{2}{4} = \frac{1}{2}$. $g'(x) = \frac{8 e^{-2x}}{(1 + e^{-2x})^2}$; $g'(0) = \frac{8 \cdot 1}{2^2} = 2$. Derivative $= \frac{1/2}{2} = \frac{1}{4} = 0.25$.
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