QJEEM24Maf348: JEE Main 2024 Mathematics - Class 12 JEEM24Maf348 Solution: Suppose for a differentiable function and . If , then is equal…

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.