What is the derivative of xe^x+x ?

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

f (x) = x^{9} ln (4 x)

\frac{d}{d x} (1. 8^{0.045 x} - 1)

\int \frac{y ^{2} + 2 y + 1}{y} d y

\frac{4 x}{x + 3} (1.1)

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