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

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

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

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

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

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

Solution steps

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

Apply the Product Rule:

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

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

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

x^{'} = 1

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

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

Simplify

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

\frac{3}{5} x^{\frac{5}{3}} - \frac{3}{4} x^{\frac{1}{3}} + 8

\frac{d y}{d x} y = e^{x}

x \to 1 lim (\frac{5}{x ^{2} - 1})

\frac{d}{d x} (\frac{x ^{3}}{e ^{\frac{1}{x}}})

f (x) 6 - 7 x + 5 x^{2}