What is the derivative of e^x+xe^x-1 ?

The derivative of e^x+xe^x-1 is e^xx+2e^x

What is the first derivative of e^x+xe^x-1 ?

The first derivative of e^x+xe^x-1 is e^xx+2e^x

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\frac{d}{d x} (e^{x} + x e^{x} - 1)

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

Solution steps

\frac{d}{d x} (e^{x} + x e^{x} - 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (e^{x}) + \frac{d}{d x} (x e^{x}) - \frac{d}{d x} (1)

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

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

\frac{d}{d x} (1) = 0

= e^{x} + e^{x} + e^{x} x - 0

Simplify

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

What is the derivative of e^x+xe^x-1 ?
The derivative of e^x+xe^x-1 is e^xx+2e^x
What is the first derivative of e^x+xe^x-1 ?
The first derivative of e^x+xe^x-1 is e^xx+2e^x