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

1 Answer
Aug 9, 2015

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))