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

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

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Find the Derivative - d/dx 2cos(2x)

\int - \frac{d}{d x} (2 cos (2 x)) d x

- x \frac{d}{d x} (2 cos (2 x)) - 4 x sin (2 x) - 2 cos (2 x) + C

Solution steps

\int - \frac{d}{d x} (2 cos (2 x)) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integration By Parts

: x \frac{d}{d x} (2 cos (2 x)) - \int - 8 x cos (2 x) d x

= - (x \frac{d}{d x} (2 cos (2 x)) - \int - 8 x cos (2 x) d x)

\int - 8 x cos (2 x) d x = - 2 (2 x sin (2 x) + cos (2 x))

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

Simplify

- (x \frac{d}{d x} (2 cos (2 x)) - (- 2 (2 x sin (2 x) + cos (2 x)))) : - x \frac{d}{d x} (2 cos (2 x)) - 4 x sin (2 x) - 2 cos (2 x)

= - x \frac{d}{d x} (2 cos (2 x)) - 4 x sin (2 x) - 2 cos (2 x)

Add a constant to the solution

= - x \frac{d}{d x} (2 cos (2 x)) - 4 x sin (2 x) - 2 cos (2 x) + C

- \frac{d}{d x} (\frac{1}{( e ^{x} )})

\frac{d y}{d x} y = x sin (x)

\frac{1}{( y ^{2} )}

- \frac{d}{d x} \frac{1}{( ( 1 - x ) ^{2} )}

f (x) = cos (x)

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