Differentiation Question?

This is a product, so, the product rule applies. But the part that reads (2x-5)^4 has an "inside" part whose derivative must be taken into account using the chain rule as well.

If you have two things multiplied, call them A and B, then the derivative is:

A'B + AB'

In this case,

A = 3x^2

B = (2x-5)^4

A' = (d/dx)3x^2 = 6x

B' = (d/dx)(2x-5)^4 = [4(2x-5)^3] * 2 = 8(2x-5)^3

Writing out the form: A'B + AB' the derivative is:

6x(2x-5)^4 + 3x^2[8(2x-5)^3]

Now, we can clean this up, if you wish...

6x(2x-5)^4 + 24x^2(2x-5)^3

= 6x(2x-5)^3[ (2x-5) + 4x ]

= 6x(2x-5)^3(6x-5)

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Now, for future reference, the reason you will want to factor out and simplify a derivative is that it will become essential that you know where the derivative is zero or undefined (the x-values where that occurs are called critical points). With critical points, you can find the original functions maximum and minimum values, and (doing the same with the second derivative) concavity and inflection points.

In this problem, we can see from the factored answer that the derivative is never undefined, but it is zero at 0, 5/6, and 5/2. These x values are critical points. If you graph the original function, it should have maximum or minimum values at 0, 5/6, and 5/2. And, in fact, this function, 3x² (2x-5)^4, has a minimum value at (0, 0), a maximum value at (5/6, 257.2), and a minimum at (5/2, 0).

Product Rule:

y=3x^2 (2x-5)^4

dy/dx = 3x^2 (4)(2x-5)^3 (2) + (2x-5)^4 (6x)

dy/dx = 24x^2(2x-5)^3 +6x(2x-5)^4