The derivative of

yequals

StartFraction 3 x squared minus 5 Over 7 x cubed plus 9 EndFraction
 
is

enter your response here

.

The derivative of the function \( y = \frac{3x^2 - 5}{7x^3 + 9} \) can be found using the **quotient rule**, which is:

\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

Here, \( u = 3x^2 - 5 \) and \( v = 7x^3 + 9 \).

### Step 1: Find \( u' \) and \( v' \)
\[
u' = \frac{d}{dx}(3x^2 - 5) = 6x
\]
\[
v' = \frac{d}{dx}(7x^3 + 9) = 21x^2
\]

### Step 2: Apply the quotient rule
\[
\frac{dy}{dx} = \frac{(7x^3 + 9)(6x) - (3x^2 - 5)(21x^2)}{(7x^3 + 9)^2}
\]

### Step 3: Simplify the expression
Expanding both terms in the numerator:
\[
(7x^3 + 9)(6x) = 42x^4 + 54x
\]
\[
(3x^2 - 5)(21x^2) = 63x^4 - 105x^2
\]

Thus, the derivative becomes:
\[
\frac{dy}{dx} = \frac{42x^4 + 54x - (63x^4 - 105x^2)}{(7x^3 + 9)^2}
\]

Simplifying further:
\[
\frac{dy}{dx} = \frac{42x^4 + 54x - 63x^4 + 105x^2}{(7x^3 + 9)^2}
\]
\[
\frac{dy}{dx} = \frac{-21x^4 + 105x^2 + 54x}{(7x^3 + 9)^2}
\]

So the derivative is:
\[
\boxed{\frac{-21x^4 + 105x^2 + 54x}{(7x^3 + 9)^2}}
\]

Would you like further details or clarifications on any step?

Here are some related questions:
1. How does the quotient rule compare to the product rule for derivatives?
2. What is the derivative of a quotient when the numerator is a constant?
3. How would the second derivative of this function be computed?
4. How does this derivative behave as \( x \to \infty \)?
5. How do we interpret the critical points of this function?

**Tip:** Always check if simplification of terms is possible after applying derivative rules for a cleaner final expression!

The derivative of y equals (3x^2 - 5) / (7x^3 + 9) is enter your response here.

Learn how to find the derivative of the function y = (3x^2 - 5) / (7x^3 + 9) by applying the quotient rule. This solution covers key calculus concepts and includes a step-by-step simplification process. Suitable for students in Grades 11-12 or introductory college courses.

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