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

The -derivative of tan(2x) is -x derivative of tan(2x)+2xtan^2(2x)+2x-tan(2x)+C

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

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

- x \frac{d}{d x} (tan (2 x)) + 2 x tan^{2} (2 x) + 2 x - tan (2 x) + C

Solution steps

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

Apply Integration By Parts

: \frac{d}{d x} (tan (2 x)) x - \int 8 sec^{2} (2 x) tan (2 x) x d x

= - (\frac{d}{d x} (tan (2 x)) x - \int 8 sec^{2} (2 x) tan (2 x) x d x)

\int 8 sec^{2} (2 x) tan (2 x) x d x = 2 x tan^{2} (2 x) + 2 x - tan (2 x)

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

Simplify

- (\frac{d}{d x} (tan (2 x)) x - (2 x tan^{2} (2 x) + 2 x - tan (2 x))) : - x \frac{d}{d x} (tan (2 x)) + 2 x tan^{2} (2 x) + 2 x - tan (2 x)

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

Add a constant to the solution

= - x \frac{d}{d x} (tan (2 x)) + 2 x tan^{2} (2 x) + 2 x - tan (2 x) + C

- \frac{d}{d x} (1) - sin (x)

f (x) = cos (2 x)

\frac{- d}{d θ sin ( θ ) cos ( θ )}

(4, 5), (9, 11)

- \frac{d}{d x} (\frac{c}{x})

What is the -derivative of tan(2x) ?
The -derivative of tan(2x) is -x derivative of tan(2x)+2xtan^2(2x)+2x-tan(2x)+C