To differentiate the function $f(x) = \frac{x^2 e^x}{x^2 + e^x}$, we can apply the quotient rule. The quotient rule states that if $f(x) = \frac{g(x)}{h(x)}$, then

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$

In this case:

• $g(x) = x^2 e^x$
• $h(x) = x^2 + e^x$

Step 1: Differentiate $g(x) = x^2 e^x$

Using the product rule for $g(x) = x^2 e^x$: $g'(x) = \frac{d}{dx}(x^2) \cdot e^x + x^2 \cdot \frac{d}{dx}(e^x)$ $g'(x) = 2x e^x + x^2 e^x = e^x (2x + x^2)$

Step 2: Differentiate $h(x) = x^2 + e^x$

$h'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(e^x)$ $h'(x) = 2x + e^x$

Step 3: Apply the Quotient Rule

Now we substitute $g(x)$, $g'(x)$, $h(x)$, and $h'(x)$ into the quotient rule: $f'(x) = \frac{g'(x) h(x) - g(x) h'(x)}{[h(x)]^2}$ Substituting the values: $f'(x) = \frac{\left(e^x (2x + x^2)\right)(x^2 + e^x) - (x^2 e^x)(2x + e^x)}{(x^2 + e^x)^2}$

Expanding and simplifying this expression would give the final derivative. Would you like me to continue with the full simplification?

Additional Questions to Explore

1. What is the simplified form of the derivative $f'(x)$ for this function?
2. How does the product rule apply within the quotient rule in this example?
3. How do we find the critical points of $f(x)$ after finding $f'(x)$?
4. What is the behavior of $f(x)$ as $x \to \infty$?
5. How can the second derivative $f''(x)$ be used to determine concavity?

Tip:

When working with derivatives involving exponentials and polynomials, break down each term carefully, especially with rules like the product and quotient rules.