derivative of ln(3x^2+2x-1)

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

d er i v a t i v e y = sin^{5} (2 x + 1)

d er i v a t i v e f (x) = - csc (x) - sin (x)

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

\int_{1}^{e^{3}} \frac{ln ^{2} ( x ^{2} )}{x} d x

l a pl a ce t r an s f or m e^{t - 2}