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."