How do you use the product Rule to find the derivative of #(10-2x)(6-2x)(x) + x^3#?

1 Answer
Aug 10, 2015

Answer:

#y^' = 15x^2 - 64x + 60#

Explanation:

Your function can be written as the sum of two functions, let's say #f(x)# and #g(x)#

#y = f(x) + g(x)#

where #f(x) = (10-2x)(6-2x) * x# and #g(x) = x^3#.

This means that the derivative of #y# will take the form

#d/dx(y) = d/dx(f(x)) + d/dx(g(x))#

Now, you can differentiate #f(x)#, which can be written as the product of three other functions, by using the product rule.

For a function #f(x)# that can be written as

#f(x) = h(x) * i(x) * k(x)#

you can find its derivative by using the formula

#color(blue)(d/dx(f(x)) = [d/dx(h(x))] * i(x) * k(x) + h(x) * [d/dx(i(x))] * k(x) + h(x) * i(x) * [d/dx(k(x))])#

In your case, you have

#d/dx(f(x)) = [d/dx(10-2x)] * (6-2x) * x + (10-2x) * [d/dx(6-2x)] * x + (10-2x) * (6-2x) * d/dx(x)#

#f^' = (-2) * (6-2x) * x + (10-2x) * (-2) * x + (10-2x) * (6-2x) * 1#

#f^' = -12x + 4x^2 -20x + 4x^2 + 60 - 32x + 4x^2#

#f^' = 12x^2 -64x + 60#

Your target derivative will thus be

#y^' = 12x^2 - 64x + 60 + d/dx(3x^2)#

#y^' = 12x^2 - 64x + 60 + 3x^2#

#y^' = color(green)(15x^2 - 64x + 60)#