What is the derivative of xe^x-1 ?

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

f (x) = 5 x - 4 x^{2}, at x = - 1

\int_{0}^{3} (2^{x}) d x

y^{^{''}} + 2 y^{^{'}} + 4 y = 50 cos (x)

\int 2 e^{2 x} sin (e^{2 x}) d x

x y^{^{'}} + y = - 6 x y^{2}

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