derivative ln(6x^3+3x^2+3x+3)

Lösung

ableitung von

ln (6 x^{3} + 3 x^{2} + 3 x + 3)

\frac{6 x ^{2} + 2 x + 1}{2 x ^{3} + x ^{2} + x + 1}

Schritte zur Lösung

\frac{d}{d x} (ln (6 x^{3} + 3 x^{2} + 3 x + 3))

Wende die Kettenregel an:

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

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

\frac{d}{d x} (6 x^{3} + 3 x^{2} + 3 x + 3) = 18 x^{2} + 6 x + 3

= \frac{1}{6 x ^{3} + 3 x ^{2} + 3 x + 3} (18 x^{2} + 6 x + 3)

Vereinfache

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

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

\int sin^{4} (3 x) cos^{4} (3 x) d x

t an g e n t f (x) = 9 + 5 x + 2 x e^{x}, at x = 0

\frac{\partial}{\partial y} (2 x y^{4} e^{y} + 2 x y^{3} + y)

t an g e n t f (x) = 19 - x^{2}, at x = 3

x y^{^{'}} - 2 y^{^{'}} - (x - 2) sin (x) + y = 0