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The given function is:

\[
y = \frac{9x^2 + 1}{x^2 + 9}
\]

We need to find the derivative of this function with respect to \(x\). Since this is a rational function (a quotient of two functions), we will use the **quotient rule** for differentiation. The quotient rule states:

\[
\left( \frac{f(x)}{g(x)} \right)' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}
\]

Here:
- \( f(x) = 9x^2 + 1 \)
- \( g(x) = x^2 + 9 \)

Now, we differentiate \(f(x)\) and \(g(x)\):

1. \( f'(x) = \frac{d}{dx}(9x^2 + 1) = 18x \)
2. \( g'(x) = \frac{d}{dx}(x^2 + 9) = 2x \)

Using the quotient rule:

\[
y' = \frac{(18x)(x^2 + 9) - (9x^2 + 1)(2x)}{(x^2 + 9)^2}
\]

Now, let's simplify the numerator:

\[
y' = \frac{18x(x^2 + 9) - 2x(9x^2 + 1)}{(x^2 + 9)^2}
\]

Expanding both terms:

\[
18x(x^2 + 9) = 18x^3 + 162x
\]
\[
2x(9x^2 + 1) = 18x^3 + 2x
\]

Now subtract:

\[
(18x^3 + 162x) - (18x^3 + 2x) = 160x
\]

Thus, the derivative simplifies to:

\[
y' = \frac{160x}{(x^2 + 9)^2}
\]

So, the derivative of the function is:

\[
y' = \frac{160x}{(x^2 + 9)^2}
\]

Let me know if you'd like more details or have any questions!

Here are 5 related questions:
1. How do you apply the quotient rule to different types of functions?
2. What is the chain rule in differentiation, and how does it relate to the quotient rule?
3. Can you find the second derivative of this function?
4. How would the derivative change if the numerator or denominator were more complex?
5. How does the derivative behave at specific points like \(x = 0\) or as \(x \to \infty\)?

**Tip:** When simplifying derivatives, make sure to expand carefully and group similar terms to avoid calculation errors.

Find the derivative of the function y = (9x^2 + 1) / (x^2 + 9).

Learn how to find the derivative of the rational function y = (9x^2 + 1) / (x^2 + 9) using the quotient rule. This step-by-step guide covers differentiation techniques and simplifies the resulting expression. Suitable for undergraduate calculus students.

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