Animated Solution for Mathematics - Differentiation: Suppose for a differentiable function $h,h(0)=0,h(1)=1$ and $h^{\prime }(0)=h^{\prime }(1)=2$. If $g(x)=h(e^x)e^{h(x)}$, then $g^{\prime }(0)$ is equal to:

Unraveling the Composite Function

Welcome, future engineer. Today, we stand at the precipice of a classic JEE Advanced calculus challenge. You might look at the function $g(x)=h(e^x)\cdot e^{h(x)}$ and feel a momentary shiver.

It looks complex, perhaps even intimidating, with its nested functions and exponential layers. But I want you to take a deep breath. In the world of JEE, complexity is often just a mask for elegance. Our job is to peel back that mask, layer by layer, until we find the simple, beautiful truth underneath.

The Anatomy of the Problem

First, let us observe the structure. We are given a differentiable function $h(x)$ with specific boundary conditions: $h(0)=0$, $h(1)=1$, $h^{\prime }(0)=2$, and $h^{\prime }(1)=2$. We are asked to find $g^{\prime }(0)$ for the function $g(x)=h(e^x)\cdot e^{h(x)}$.

Notice the structure immediately. This is not a single function; it is a product of two distinct entities. Let us define them clearly to keep our minds organized. Let $u(x)=h(e^x)$ and $v(x)=e^{h(x)}$.

Now, our function is simply $g(x)=u(x)\cdot v(x)$. The Product Rule is our guiding star here. It tells us that the derivative of a product is:

This is our roadmap. We do not need to solve the whole thing at once; we just need to find $u^{\prime }(x)$ and $v^{\prime }(x)$ separately.

The Chain Rule Dance

Now, let us tackle $u(x)=h(e^x)$. This is a composite function. The outer function is $h$, and the inner function is $e^x$.

The Chain Rule is the key that unlocks this door. We differentiate the outer function while keeping the inner function intact, and then we multiply by the derivative of the inner function:

Next, we move to $v(x)=e^{h(x)}$. Again, we see a composition. The outer function is the exponential $e^u$, and the inner function is $h(x)$.

Applying the Chain Rule again, we differentiate the exponential, which remains $e^{h(x)}$, and multiply by the derivative of the exponent, $h^{\prime }(x)$:

The Assembly

Now, we bring it all together. We substitute our findings back into the Product Rule formula:

This looks like a long, intimidating expression, but do not let the length deceive you. It is just a collection of terms waiting to be evaluated at $x=0$.

The Final Reveal

This is the moment of truth. We substitute $x=0$ into our expression. Let us do this carefully, step by step:

Recall that $e^0=1$. This simplifies our expression beautifully:

We are given $h(0)=0$, so $e^{h(0)}=e^0=1$. We are also given $h(1)=1$. Substituting these values, the expression becomes:

Finally, we use the given derivative values: $h^{\prime }(1)=2$ and $h^{\prime }(0)=2$. Plugging these in, we get:

There it is. The final answer is 4. You have successfully navigated the composition, the product rule, and the substitution. Keep this clarity, and you will conquer any problem the JEE throws your way.