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

The (xe^{x^2})^' is e^{x^2}+2e^{x^2}x^2

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

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

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

e^{x^{2}} + 2 e^{x^{2}} x^{2}

Solution steps

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

Apply the Product Rule:

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

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

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

x^{'} = 1

(e^{x^{2}})^{'} = e^{x^{2}} \cdot 2 x

= 1 \cdot e^{x^{2}} + e^{x^{2}} \cdot 2 xx

Simplify

1 \cdot e^{x^{2}} + e^{x^{2}} \cdot 2 xx : e^{x^{2}} + 2 e^{x^{2}} x^{2}

= e^{x^{2}} + 2 e^{x^{2}} x^{2}

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