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You've provided an image with the function \( y = \frac{4x^2}{2x-5} \) and asked to find the derivative.

To find the derivative \( y' \) of the function \( y = \frac{4x^2}{2x-5} \), we use the quotient rule which states:
\[
y' = \frac{v(u') - u(v')}{v^2}
\]
where \( u = 4x^2 \) and \( v = 2x - 5 \).

First, compute the derivatives \( u' \) and \( v' \):
\[
u' = 8x \quad \text{and} \quad v' = 2
\]

Now substitute these into the quotient rule formula:
\[
y' = \frac{(2x - 5)(8x) - (4x^2)(2)}{(2x - 5)^2}
\]

Simplify the expression:
\[
y' = \frac{16x^2 - 40x - 8x^2}{(2x - 5)^2} = \frac{8x^2 - 40x}{(2x - 5)^2}
\]

The derivative \( y' \) of the function \( y = \frac{4x^2}{2x-5} \) is:
\[
y' = \frac{8x^2 - 40x}{(2x - 5)^2}
\]

Would you like to go through any of these steps in more detail or do you have any other questions? Here are some follow-up questions you might consider:

1. What happens to the derivative as \( x \) approaches infinity?
2. How does the derivative behave near the point where the denominator equals zero?
3. Can you find the critical points of the function \( y = \frac{4x^2}{2x-5} \)?
4. Would you like to know how the graph of the derivative looks compared to the original function?
5. What is the physical interpretation of this derivative in a real-world context?

**Tip**: Remember that the quotient rule is particularly useful when dealing with ratios of polynomials and can be memorized as "low d high minus high d low, square the bottom and away we go."

Find the derivative of y = (4x^2) / (2x - 5).

Learn how to find the derivative of the function y = (4x^2) / (2x - 5) using the quotient rule. This step-by-step solution covers key calculus concepts like derivatives, quotient rule applications, and polynomial differentiation.

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