What is the derivative of 1+xe^x ?

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

x \to 4 + lim (ln (x - 4))

\frac{3 s + 2}{( s + 1 ) ^{3}}

\int_{- 1}^{3} (x^{3} + 1) d x

\frac{d}{d x} (2 x^{3} - 42 x^{2} + 78 x + k)

t \to 1 lim (t + 1)

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