Application of Derivatives
PYP_JEE_ADV_2025_P2
Grade None
Question:
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
$$f(x) = \log_e(x^2 + 2x + 4), \quad \text{and} \quad g(x) = \dfrac{4}{1 + e^{-2x}}.$$
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 the function $g$.
Then the value of the derivative of the composite function $f \circ g^{-1}$ at $x = 2$ is
Step-by-Step Solution
Key Concept: Using the derivative of inverse functions: $(g^{-1})'(x) = 1/g'(g^{-1}(x))$, and applying the chain rule to composition of functions.
Let $h(x) = f(g^{-1}(x))$. The derivative is:
$$h'(x) = f'(g^{-1}(x)) \cdot (g^{-1})'(x) = \dfrac{f'(g^{-1}(x))}{g'(g^{-1}(x))}$$
Let $y = g^{-1}(2) \implies g(y) = 2$:
$$\dfrac{4}{1 + e^{-2y}} = 2 \implies 1 + e^{-2y} = 2 \implies e^{-2y} = 1 \implies y = 0$$
Now, differentiate $f(x)$ and $g(x)$ at $x=0$:
- For $f'(x)$:
$$f'(x) = \dfrac{2x+2}{x^2+2x+4} \implies f'(0) = \dfrac{2}{4} = \dfrac{1}{2}$$
- For $g'(x)$:
$$g'(x) = \dfrac{8e^{-2x}}{(1+e^{-2x})^2} \implies g'(0) = \dfrac{8}{4} = 2$$
Substitute the evaluated derivatives:
$$h'(2) = \dfrac{f'(0)}{g'(0)} = \dfrac{1/2}{2} = 0.25$$
Correct Answer: