What is the derivative of-2xsin(2x+cos(2x)) ?

The derivative of-2xsin(2x+cos(2x)) is -4xcos(2x)-4sin(2x)

What is the first derivative of-2xsin(2x+cos(2x)) ?

The first derivative of-2xsin(2x+cos(2x)) is -4xcos(2x)-4sin(2x)

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

- 4 x cos (2 x) - 4 sin (2 x)

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= - 2 (sin (2 x) + 2 x cos (2 x)) - sin (2 x) \cdot 2

Simplify

- 2 (sin (2 x) + 2 x cos (2 x)) - sin (2 x) \cdot 2 : - 4 x cos (2 x) - 4 sin (2 x)

= - 4 x cos (2 x) - 4 sin (2 x)

\frac{( s + 1 )}{s ( s ^{2} + s + 1 )}

\frac{d y}{d x} + 1 = 2 y

\int e^{r} d r

\int (15 x^{- 1} - 4 sinh (x)) d x

\frac{d}{d x} ((x^{6} + 1)^{\frac{5}{2}})

What is the derivative of-2xsin(2x+cos(2x)) ?
The derivative of-2xsin(2x+cos(2x)) is -4xcos(2x)-4sin(2x)
What is the first derivative of-2xsin(2x+cos(2x)) ?
The first derivative of-2xsin(2x+cos(2x)) is -4xcos(2x)-4sin(2x)