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

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

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

d er i v a t i v e \frac{16}{( 1 - x ) ^{3}}

\int \frac{( 9 - x ^{2} )}{x ^{2}} d x

\int \frac{x}{( x ^{2} + 8 ) ^{2}} d x

d er i v a t i v e f (x) = x^{- 12}