How do you differentiate #f(x)=(2x^2+1)(3x-3)# using the product rule?

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Answer 1 · 2015-12-31T07:33:02

#f'(x)=18x^2-12x+3#

The product rule states that for some function

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

Here, we have

#{(g(x)=2x^2+1),(h(x)=3x-3):}#

Differentiate each function.

#{(g'(x)=4x),(h'(x)=3):}#

Relate these to find #f'(x)#.

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

Distribute and simplify.

#f'(x)=18x^2-12x+3#