Suppose for a differentiable function h , h ( 0 ) = 0 , h ( 1 ) = 1 and h ' ( 0 ) = h ' ( 1 ) = 2 . If g ( x ) = h ( e x ) e h ( x ) , then g ' ( 0 ) is equal to [2024]

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Solution

Q.
Suppose for a differentiable function $h, h (0) = 0,$ $h (1) = 1$ and $h' (0) = h' (1) = 2 .$ If $g (x) = h (e^{x}) e^{h (x)},$ then $g' (0)$ is equal to [2024]

1 8

2 5

3 3

4 4

Ans.
(4)

$We have, g (x) = h (e^{x}) e^{h (x)}$

$g' (x) = h (e^{x}) e^{h (x)} h' (x) + e^{h (x)} h' (e^{x}) \cdot e^{x}$

$g' (0) = h (1) e^{h (0)} h' (0) + e^{h (0)} h' (1) = 1 \times 2 + 2 = 4$

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