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

The first (x^2e^x)^' is 2xe^x+e^xx^2

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(x^2e^x)^'

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

2 x e^{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}

y^{^{'}} + \frac{2 y}{x} = - \frac{4}{x}

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