Question:
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:
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:
Updated On: Mar 20, 2025
- 5
- 3
- 8
- 4
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The Correct Option is D
Solution and Explanation
Differentiating with respect to \( x \):
\[ g(x) = h(e^x) \times e^{h(x)} \]
\[ g'(x) = h'(e^x) \times e^{h(x)} \times h'(x) + e^{h(x)} \times h'(e^x) \times e^x \]
\[ g'(0) = h(1)e^{h(0)}h'(0) + e^{h(0)}h'(1) \]
\[ = 2 + 2 = 4 \]
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