What is the derivative of x(e^x-1) ?

The derivative of x(e^x-1) is e^x-1+e^xx

What is the first derivative of x(e^x-1) ?

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

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = e^{x} - 1

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

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

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

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

Simplify

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

\frac{d}{d x} (- ln (x^{4}))

\int 1 + cos (π θ) d θ

\frac{x}{2}

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

f (x) = 4 x^{3} ln (x)

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