How do you find the 1st and 2nd derivative of #x^2e^x#?

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Answer 1 · 2016-11-14T14:23:02

#dy/dx = x(x + 2)e^x#
#(d^2y)/(dx^2) = (x^2 +4x + 2)e^x#

#dy/dx = 2x xx e^x + e^x xx x^2#

#dy/dx = 2xe^x + x^2e^x#

#dy/dx = x(x + 2)e^x#

#(d^2y)/(dx^2) = 2 xx e^x + 2x xx e^x + 2x xx e^x + x^2xxe^x#

#(d^2y)/(dx^2) = 2e^x + 2xe^x + 2xe^x + x^2e^x#

#(d^2y)/(dx^2) = 2e^x + 4xe^x + x^2e^x#

#(d^2y)/(dx^2) = (x^2 + 4x +2)e^x#

Hopefully this helps!