JEE Main 2024 — Differentiation Question with Solution

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:

$$(u(x)v(x){)}^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$

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

First, we need to find $$u^{\prime }(x)$$ and $$v^{\prime }(x)$$. Using the chain rule, we find:

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

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

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

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

$$g^{\prime }(x)=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$

Substituting $$u(x)$$, $$v(x)$$, $$u^{\prime }(x)$$, and $$v^{\prime }(x)$$ into the above expression, we get:

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

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

First, we know that:

$$h(0)=0$$

$$h(1)=1$$

$$h^{\prime }(0)=2$$

$$h^{\prime }(1)=2$$

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

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

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

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

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

Therefore, evaluating $$g^{\prime }(0)$$:

$$g^{\prime }(0)=\left(u^{\prime }(0)v(0)\right)+\left(u(0)v^{\prime }(0)\right)$$

$$g^{\prime }(0)=\left(2\cdot 1\right)+\left(1\cdot 2\right)$$

$$g^{\prime }(0)=2+2=4$$

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

The correct answer is Option A: 4.