What is the derivative of f(x)=2xe^x ?

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

What is the first derivative of f(x)=2xe^x ?

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

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derivative of

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

\int (x^{2} - 3 x + 5) e^{x} d x

[x + (x + sin (2) (x))^{7}]^{3}

7 x - x^{2}

f (x) = \frac{3}{4} x + 7

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