How do you differentiate $y = x^{2} e^{- x}$ $y=x^2e^-x$ ?

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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.

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

How do you differentiate y=x^2e^-xy=x2e−xy=x^2e^-x?