What is the derivative of 2e^x+xe^x ?

The derivative of 2e^x+xe^x is 3e^x+e^xx

What is the first derivative of 2e^x+xe^x ?

The first derivative of 2e^x+xe^x is 3e^x+e^xx

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\frac{d}{d x} (2 e^{x} + x e^{x})

3 e^{x} + e^{x} x

Solution steps

\frac{d}{d x} (2 e^{x} + x e^{x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (2 e^{x}) + \frac{d}{d x} (x e^{x})

\frac{d}{d x} (2 e^{x}) = 2 e^{x}

\frac{d}{d x} (x e^{x}) = e^{x} + e^{x} x

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

Simplify

= 3 e^{x} + e^{x} x

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What is the derivative of 2e^x+xe^x ?
The derivative of 2e^x+xe^x is 3e^x+e^xx
What is the first derivative of 2e^x+xe^x ?
The first derivative of 2e^x+xe^x is 3e^x+e^xx