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

Question

Isaiah Allen · Accepted Answer

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

For two functions, the product rule is

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

In this case, let #g(x)=e^x+x# and #h(x)=x^2-x#. Taking derivatives gives #g^'(x)=e^x+1# and #h^'(x)=2x-1#. Now plugging these expressions into the formula gives

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

Now divide and make things easier.

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

#f^'(x)=3x^2-2x+x^2e^x+xe^x-e^x#

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

Elijah Abram · Answer

undefined To differentiate $ f(x) = (e^x + x)(x^2 - x) $ using the product rule, apply the formula $ (uv)' = u'v + uv' $. Let $ u = e^x + x $ and $ v = x^2 - x $. Then, differentiate $ u $ with respect to $ x $ to get $ u' $, and differentiate $ v $ with respect to $ x $ to get $ v' $. Finally, substitute these values into the product rule formula and simplify to find the derivative of $ f(x) $.

$ u' = e^x + 1 $

$ v' = 2x - 1 $

Using the product rule formula:

$ f'(x) = (e^x + x)(2x - 1) + (x^2 - x)(e^x + 1) $

$ f'(x) = (2xe^x - e^x + 2x^2 - 2x) + (xe^x - xe^x + x^2 - x) $

$ f'(x) = 2xe^x - e^x + 2x^2 - 2x + x^2 - x $

$ f'(x) = 2xe^x - e^x + 3x^2 - 3x $