To find the derivative of the function $h(x) = \frac{x^2}{x + 5}$, we can apply the Quotient Rule. The Quotient Rule states that if we have a function of the form $f(x) = \frac{g(x)}{u(x)}$, then the derivative is given by:

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

For $h(x) = \frac{x^2}{x + 5}$, we identify:

• $g(x) = x^2$
• $u(x) = x + 5$

Step 1: Compute the derivatives of $g(x)$ and $u(x)$

• $g'(x) = 2x$ (derivative of $x^2$)
• $u'(x) = 1$ (derivative of $x + 5$)

Step 2: Apply the Quotient Rule

Using the formula:

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

Substitute the values:

$h'(x) = \frac{(2x)(x + 5) - (x^2)(1)}{(x + 5)^2}$

Step 3: Simplify the numerator

$h'(x) = \frac{2x(x + 5) - x^2}{(x + 5)^2}$ $h'(x) = \frac{2x^2 + 10x - x^2}{(x + 5)^2}$ $h'(x) = \frac{x^2 + 10x}{(x + 5)^2}$

Thus, the derivative of $h(x)$ is:

$h'(x) = \frac{x^2 + 10x}{(x + 5)^2}$

---

Would you like more details about the process? Or have any other questions?

Related questions:

1. What happens to the function when $x = -5$?
2. How can we use the product rule in a similar situation?
3. What is the second derivative of the function?
4. How do we analyze the limits of the function as $x \to \infty$?
5. Can the quotient rule be used for more complex functions involving trigonometric or exponential terms?

Tip: Always double-check the signs when simplifying the numerator in the quotient rule—careful handling of terms will save you from errors.