What is the derivative of 2xe^x ?

The derivative of 2xe^x is 2(e^x+e^xx)

What is the first derivative of 2xe^x ?

The first derivative of 2xe^x is 2(e^x+e^xx)

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derivative of

2 x e^{x}

2 (e^{x} + e^{x} x)

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

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

f = x, g = e^{x}

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

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

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

= 2 (1 \cdot e^{x} + e^{x} x)

Simplify

= 2 (e^{x} + e^{x} x)

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

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

\frac{\partial}{\partial x} (4 e^{x^{7} y})

\int x^{2} - 4 x + 29 d x

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

What is the derivative of 2xe^x ?
The derivative of 2xe^x is 2(e^x+e^xx)
What is the first derivative of 2xe^x ?
The first derivative of 2xe^x is 2(e^x+e^xx)