The image you uploaded presents a mathematical problem involving twice-differentiable functions. Here is a breakdown of the content:

We have two functions, $f(x)$ and $g(x)$, that are twice differentiable. The relations are given as:

• $g(x) = e^{f(x)}$
• $g''(x) = h(x) e^{f(x)}$

We need to find the function $h(x)$.

Let's proceed step by step. I will differentiate $g(x)$ and its second derivative to express $h(x)$.

Step 1: First derivative of $g(x)$

Given $g(x) = e^{f(x)}$, applying the chain rule:

$g'(x) = e^{f(x)} \cdot f'(x)$

Step 2: Second derivative of $g(x)$

Now, differentiating $g'(x)$:

$g''(x) = \frac{d}{dx} \left( e^{f(x)} \cdot f'(x) \right)$

We apply the product rule here:

$g''(x) = e^{f(x)} \cdot (f''(x)) + e^{f(x)} \cdot (f'(x))^2$

Thus, we have:

$g''(x) = e^{f(x)} \cdot (f''(x) + (f'(x))^2)$

Step 3: Match with the given expression for $g''(x)$

The given relation for $g''(x)$ is $g''(x) = h(x) e^{f(x)}$. Comparing this with the expression derived:

$e^{f(x)} \cdot (f''(x) + (f'(x))^2) = h(x) e^{f(x)}$

Canceling $e^{f(x)}$ from both sides (assuming $e^{f(x)} \neq 0$):

$f''(x) + (f'(x))^2 = h(x)$

Final Answer:

Thus, the function $h(x)$ is:

$h(x) = f''(x) + (f'(x))^2$

Let me know if you need any more details or clarifications on the steps.

Would you like me to explain further or explore related questions?

Here are 5 questions to help expand on this:

1. How does the chain rule apply to the differentiation of $g(x) = e^{f(x)}$?
2. Can you explain the product rule in more detail when applying it to $g'(x)$?
3. What happens if we use a different form of $g(x)$ in this problem?
4. How would this method apply if we had more functions involved in $g(x)$?
5. What are the implications of the function $h(x)$ in the context of differential equations?

Tip: Make sure to keep track of the chain and product rules in differential calculus—they come in handy for many problems involving compositions of functions.