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

The derivative of e^{2x}+xe^{2x} is 3e^{2x}+2e^{2x}x

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

The first derivative of e^{2x}+xe^{2x} is 3e^{2x}+2e^{2x}x

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

3 e^{2 x} + 2 e^{2 x} x

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= 3 e^{2 x} + 2 e^{2 x} x

3 - x, 0

(2 x)^{^{'}}

\frac{d}{d x} (x^{2} - 25)

f (x) = x^{2} + 3 x, at x = 0

\frac{\partial}{\partial x} (x ln (t))

What is the derivative of e^{2x}+xe^{2x} ?
The derivative of e^{2x}+xe^{2x} is 3e^{2x}+2e^{2x}x
What is the first derivative of e^{2x}+xe^{2x} ?
The first derivative of e^{2x}+xe^{2x} is 3e^{2x}+2e^{2x}x