What is the derivative of x+xe^x ?

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d x}{d x} = 1

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

= 1 + e^{x} + e^{x} x

\int \frac{x ^{2} + 6 x}{x + 3} d x

\int (0.1 x + 1) e^{0.03 x} d x

f (x) = (x - 6) (3 x + 5)

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

\frac{d}{d x} ((x^{3} - 3 x^{2} + 1)^{- 3})

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