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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

\int cos (2 x) 4 - sin (2 x) d x

\int 3 e^{5 x + e^{5 x}} d x

\frac{d y}{d x} = \frac{3 x ^{2}}{1 - y}

x^{2} e^{- 5 x}

\int_{\frac{1}{3}}^{5} 6 x ln (3 x) d x

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