How do you differentiate #y=x^2e^-x#?

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Answer 1 · 2016-12-14T16:59:36

#d/dx(uv) = u'v+uv'#

#d/dx(e^u) = e^u (du)/dx#

#y = x^2e^-x#

#dy/dx = (2x)(e^-x)+(x^2)(e^-x (-1))#

# = 2xe^-x - x^2e^-x#

If you like, remove the common factor.

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