Applications of Derivatives
Chain Rule / Product Rule
nta_pyq_2024_apr
Grade 12
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:
Step-by-Step Solution
Key Concept: $g'(x)=h'(e^x)\cdot e^x\cdot e^{h(x)}+h(e^x)\cdot e^{h(x)}\cdot h'(x)$. At $x=0$: $e^0=1$.
$g'(0)=h'(1)\cdot e^0+h(1)\cdot e^0\cdot h'(0)=2+2=4$.
Correct Answer: 2