Suppose for a differentiable function h,h(0)=0,h(1)=1 and h^'(0)=h^'(1)=2.

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:

Accepted Answer

To determine $$g^{\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$.

The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by:

$(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$

Let's denote $$u(x) = h(\mathrm{e}^x)$$ and $$v(x) = \mathrm{e}^{h(x)}$$.

First, we need to find $u'(x)$ and $v'(x)$. Using the chain rule, we find:

$$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $$

Now, the derivative of $v(x)$ is:

$$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $$

Using the product rule, we get the derivative of $g(x)$:

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

Substituting $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the above expression, we get:

$$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $$

Next, we need to evaluate this at $x = 0$:

First, we know that:

$h(0) = 0$

$h(1) = 1$

$h'(0) = 2$

$h'(1) = 2$

Substituting $x = 0$ into the expressions, we get:

$$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $$

$$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $$

$$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $$

$$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $$

Therefore, evaluating $g'(0)$:

$$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $$

$$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $$

$g'(0) = 2 + 2 = 4$

Thus, the value of $$g^{\prime}(0)$$ is 4, which corresponds to Option A.

The correct answer is Option A: 4.

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