How do you differentiate #f(x)=(x+4x^2)(x^2-3x)# using the product rule?

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Answer 1 · 2016-01-09T16:03:56

#f'(x)=16x^3-33x^2-6x#

The product rule states that for a function #f(x)=g(x)h(x)#,

#f'(x)=g'(x)h(x)+h'(x)g(x)#

Thus,

#f'(x)=(x^2-3x)d/dx(x+4x^2)+(x+4x^2)d/dx(x^2-3x)#

These derivatives are simple to find through the power rule:

#d/dx(x+4x^2)=1+8x#

#d/dx(x^2-3x)=2x-3#

Plug these back in.

#f'(x)=(x^2-3x)(1+8x)+(x+4x^2)(2x-3)#

Now, distribute and simplify.

#f'(x)=8x^3-23x^2-3x+8x^3-10x^2-3x#

#f'(x)=16x^3-33x^2-6x#