How do you find the derivative of #x^2*e^-x#?

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Answer 1 · 2015-05-31T15:53:01

#(uv)' = u'v+uv'#

#u = x^2#
#u' = 2x#

#v = e^(-x)#
We know that #(e^(ax))' = (ax)'*e^(ax) #.
#ax = -x#
#(-x)' = -1#
#v' = -e^(-x)#

Therefore :

#(uv)' = 2x ( e^(-x)) + x^2(-e^(-x))#

#(uv)' = ( 2x-x^2) (e^(-x))#

#(uv)' = (2-x) xe^(-x)#.