What is the derivative of 2arctan(3x-ln(2+3x)) ?

The derivative of 2arctan(3x-ln(2+3x)) is (-27x^2+18x+9)/((3x+2)(9x^2+1))

What is the first derivative of 2arctan(3x-ln(2+3x)) ?

The first derivative of 2arctan(3x-ln(2+3x)) is (-27x^2+18x+9)/((3x+2)(9x^2+1))

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\frac{d}{d x} (2 arctan (3 x) - ln (2 + 3 x))

\frac{- 27 x ^{2} + 18 x + 9}{( 3 x + 2 ) ( 9 x ^{2} + 1 )}

Solution steps

\frac{d}{d x} (2 arctan (3 x) - ln (2 + 3 x))

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (2 arctan (3 x)) - \frac{d}{d x} (ln (2 + 3 x))

\frac{d}{d x} (2 arctan (3 x)) = \frac{6}{9 x ^{2} + 1}

\frac{d}{d x} (ln (2 + 3 x)) = \frac{3}{2 + 3 x}

= \frac{6}{9 x ^{2} + 1} - \frac{3}{2 + 3 x}

Simplify

\frac{6}{9 x ^{2} + 1} - \frac{3}{2 + 3 x} : \frac{- 27 x ^{2} + 18 x + 9}{( 3 x + 2 ) ( 9 x ^{2} + 1 )}

= \frac{- 27 x ^{2} + 18 x + 9}{( 3 x + 2 ) ( 9 x ^{2} + 1 )}

\int_{3}^{6} \frac{1}{36 - x ^{2}} d x

f (t) = ln (cos (t))

f (x) = e^{x} cos (x), (0, 1)

\frac{\partial}{\partial x} (ln (7 x^{2} + y^{2} + 6))

8 x^{3}

What is the derivative of 2arctan(3x-ln(2+3x)) ?
The derivative of 2arctan(3x-ln(2+3x)) is (-27x^2+18x+9)/((3x+2)(9x^2+1))
What is the first derivative of 2arctan(3x-ln(2+3x)) ?
The first derivative of 2arctan(3x-ln(2+3x)) is (-27x^2+18x+9)/((3x+2)(9x^2+1))