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:

• A. 5
• B. 3
• C. 8
• D. 4

Answer 1

The Correct Option is D

We need to find the derivative of the function \( g(x) = h(e^x) \, e^{h(x)} \) at \( x = 0 \). To do this, we will use the product rule and the chain rule of differentiation.

First, let's apply the product rule to differentiate \( g(x) \), which is the product of two functions:

• \( u(x) = h(e^x) \)
• \( v(x) = e^{h(x)} \)

The derivative is:

\( g'(x) = u'(x) v(x) + u(x) v'(x) \)

Next, we need to find the derivatives \( u'(x) \) and \( v'(x) \).

The function \( u(x) = h(e^x) \), using the chain rule, gives us:

\( u'(x) = \frac{d}{dx}[h(e^x)] = h'(e^x) \frac{d}{dx}[e^x] = h'(e^x) e^x \)

Now, \( v(x) = e^{h(x)} \), using the chain rule, gives us:

\( v'(x) = \frac{d}{dx}[e^{h(x)}] = e^{h(x)} h'(x) \)

Now we substitute these into the expression for \( g'(x) \):

1. \( g'(x) = [h'(e^x) e^x] e^{h(x)} + h(e^x) [e^{h(x)} h'(x)] \)
2. \( = e^{h(x)} [h'(e^x) e^x + h(e^x) h'(x)] \)

Now we evaluate \( g'(x) \) at \( x = 0 \):

First, calculate each component at \( x = 0 \):

• \( e^0 = 1 \)
• \( e^x|_{x=0} = 1 \) implies \( e^0 = 1 \)
• \( h(e^0) = h(1) = 1 \)
• \( e^{h(0)} = e^0 = 1 \)
• \( h'(0) = 2 \) and \( h'(1) = 2 \)

Substitute these values into \( g'(0) \):

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

Thus, the value of \( g'(0) \) is 4. The correct answer is 4.

Answer 2

Differentiating with respect to \( x \):

\[ g(x) = h(e^x) \times e^{h(x)} \]

\[ g'(x) = h'(e^x) \times e^{h(x)} \times h'(x) + e^{h(x)} \times h'(e^x) \times e^x \]

\[ g'(0) = h(1)e^{h(0)}h'(0) + e^{h(0)}h'(1) \]

\[ = 2 + 2 = 4 \]