What is the derivative of 2xe^{4x} ?

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

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

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

derivative of

2 x e^{4 x}

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

Solution steps

\frac{d}{d x} (2 x e^{4 x})

Take the constant out:

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

= 2 \frac{d}{d x} (x e^{4 x})

Apply the Product Rule:

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

f = x, g = e^{4 x}

= 2 (\frac{d x}{d x} e^{4 x} + \frac{d}{d x} (e^{4 x}) x)

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

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

= 2 (1 \cdot e^{4 x} + e^{4 x} \cdot 4 x)

Simplify

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

\int \frac{2}{x ^{3} x ^{2} - 3} d x

d er i v a t i v e 2 x + 3 x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x}

x \to \infty lim (\frac{1}{x} + 5)

\int sin (9 x) cos (7 x) d x

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