derivative of 2arctan(3x-ln(2+3x))

Lösung

\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 )}

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

(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}

Vereinfache

\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

d er i v a t i v e f (t) = ln (cos (t))

t an g e n t f (x) = e^{x} cos (x), (0, 1)

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

x y^{^{'}} + 2 y = - 2 e^{x}