How do you differentiate #g(x) = (2sinx -e^x) ( 2x-x^2)# using the product rule?

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Answer 1 · 2018-01-04T21:17:34

#f'(x) = (2cosx - e^x)(2x-x^2)+(2sinx-e^x)(2-2x)#

The product rule tells us:

#f(x) = g(x)h(x)# then

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

We have the two products and there derivatives:

#g(x) = 2sinx - e^x#
#g'(x) = 2cosx-e^x#

and

#h(x)=2x-x^2#
#h'(x) = 2-2x#

So on substitution into the chain rule formula we get:

#f'(x) = (2cosx - e^x)(2x-x^2)+(2sinx-e^x)(2-2x)#