What is the derivative of 2xe^{2x}+e^{2x} ?

The derivative of 2xe^{2x}+e^{2x} is 4e^{2x}x+4e^{2x}

What is the first derivative of 2xe^{2x}+e^{2x} ?

The first derivative of 2xe^{2x}+e^{2x} is 4e^{2x}x+4e^{2x}

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

4 e^{2 x} x + 4 e^{2 x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

2 (e^{2 x} + 2 e^{2 x} x) + e^{2 x} \cdot 2 : 4 e^{2 x} x + 4 e^{2 x}

= 4 e^{2 x} x + 4 e^{2 x}

What is the derivative of 2xe^{2x}+e^{2x} ?
The derivative of 2xe^{2x}+e^{2x} is 4e^{2x}x+4e^{2x}
What is the first derivative of 2xe^{2x}+e^{2x} ?
The first derivative of 2xe^{2x}+e^{2x} is 4e^{2x}x+4e^{2x}