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

The derivative of xe^{4x}+cos(2x) is e^{4x}+4e^{4x}x-sin(2x)*2

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (cos (2 x)) = - sin (2 x) \cdot 2

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

\frac{d y}{d x} = y^{2} - 16

x \to 0 lim (f (x) - g (x))

\int \frac{1}{x 6 x + 25} d x

\frac{d}{d x} (x - 8 3 x)

\int arctan (8 x) d x

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