Question

Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and h'(0) = h'(1) = 2. If g(x) = h(eˣ)eʰ⁽ˣ⁾, then g'(0) is equal to:

• A. 5
• B. 3
• C. 8
• D. 4

Answer 1

Answer: (D)

4

Answer 2

The given function is $g(x) = h(e^x)e^{h(x)}$. We need to find $g'(0)$.

First, find the derivative of $g(x)$ using the product rule $(uv)' = u'v + uv'$: Let $u = h(e^x)$ and $v = e^{h(x)}$. Then $u' = \frac{d}{dx}(h(e^x)) = h'(e^x) \cdot e^x$ (using chain rule). And $v' = \frac{d}{dx}(e^{h(x)}) = e^{h(x)} \cdot h'(x)$ (using chain rule).

So, $g'(x) = u'v + uv'$ $g'(x) = (h'(e^x) \cdot e^x) \cdot e^{h(x)} + h(e^x) \cdot (e^{h(x)} \cdot h'(x))$ $g'(x) = e^{h(x)} [e^x h'(e^x) + h(e^x) h'(x)]$

Now, substitute $x=0$: $g'(0) = e^{h(0)} [e^0 h'(e^0) + h(e^0) h'(0)]$ $g'(0) = e^{h(0)} [1 \cdot h'(1) + h(1) \cdot h'(0)]$

We are given the following values: $h(0) = 0$ $h(1) = 1$ $h'(0) = 2$ $h'(1) = 2$

Substitute these values into the expression for $g'(0)$: $g'(0) = e^0 [1 \cdot 2 + 1 \cdot 2]$ $g'(0) = 1 [2 + 2]$ $g'(0) = 1 \cdot 4$ $g'(0) = 4$