Question

How do you use the Product Rule to find the derivative of `3x(2x-1)^3`?

Answer 1

`y^' = 3(2x-1)^2 * (8x-1)`

Explanation:

To be able to use the product rule to differentiate this function, you need to write it as a product of two other functions, let's say `f(x)` and `g(x)`

`y = f(x) * g(x)`

This will allow you to use the formula

`color(blue)(d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x)))`

In your case, you have

`y = underbrace(3x)_(color(blue)(f(x))) * underbrace((2x-1)^3)_(color(blue)(g(x)))`

This means that you can write

`d/dx(y) = [d/dx(3x)] * (2x-1)^3 + 3x * d/dx(2x-1)^3`

To calculate `d/dx(2x-1)^3`, you can use the chain rule for `u^3`, with `u = 2x-1`.

`d/dx(u^3) = d/(du)(u^3) * d/dx(u)`

`d/dx(u^3) = 3u^2 * d/dx(2x-1)`

`d/dx(2x-1)^3 = 3(2x-1)^2 * 2`

Your target derivative will thus be

`y^' = 3 * (2x-1)^3 + 3x * 6(2x-1)^2`

`y^' = 3(2x-1)^3 + 18x(2x-1)^2`

This can be further simplified to

`y^' = 3(2x-1)^2 * (2x-1 + 6x)`

`y^' = color(green)(3(2x-1)^2 * (8x-1))`