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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

x^{'} = 1

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

= 1 \cdot e^{- x} + (- e^{- x}) x

Simplify

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

\int_{0}^{2 π} 20 + 16 t^{2} d t

\int sin (1) d x

x^{3} - 6 x^{2} + 8 x, [2, 4]

\int \frac{x ^{3}}{16 + x ^{2}} d x

x \to - 2 lim ((x + 1)^{4} (2 x^{2}))