What is the (x^2e^{-x})^' ?

The (x^2e^{-x})^' is 2e^{-x}x-e^{-x}x^2

What is the first (x^2e^{-x})^' ?

The first (x^2e^{-x})^' is 2e^{-x}x-e^{-x}x^2

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(x^{2} e^{- x})^{^{'}}

2 e^{- x} x - e^{- x} x^{2}

Solution steps

(x^{2} e^{- x})^{'}

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = e^{- x}

= (x^{2})^{^{'}} e^{- x} + (e^{- x})^{^{'}} x^{2}

(x^{2})^{'} = 2 x

(e^{- x})^{'} = - e^{- x}

= 2 x e^{- x} + (- e^{- x}) x^{2}

Simplify

= 2 e^{- x} x - e^{- x} x^{2}