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

1 Answer
Aug 9, 2015

Answer:

#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))#