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

Question

1554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-06T11:23:24

Apply the product rule to the functions #f_1(x) = 2x^2 - 4x + 4# and #f_2(x) = e^x# to obtain #d/dxf_1(x)f_2(x) = 2x^2e^x#

The product rule states that for any functions #f(x)# and #g(x)#
#d/dxf(x)g(x) = f'(x)g(x) + f(x)g'(x)#

Let #f_1(x) = 2x^2 - 4x + 4# and #f_2(x) = e^x#

Then #f_1'(x) = 4x - 4# and #f_2'(x) = e^x#

Note that

#d/dx(2x^2 - 4x + 4)e^x = d/dxf_1(x)f_2(x)#

So, applying the product rule gives us

#d/dx(2x^2 - 4x + 4)e^x = f_1'(x)f_2(x) + f_1(x)f_2'(x)#
#=>d/dx(2x^2 - 4x + 4)e^x=(4x-4)e^x+(2x^2-4x+4)e^x#

Simplifying the result gives us the final answer of
#d/dx(2x^2 - 4x + 4)e^x = 2x^2e^x#