derivative of ln(3x+2y)

Lösung

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

\frac{3 + 2 \frac{d y}{d x}}{3 x + 2 y}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

= \frac{3 + 2 \frac{d y}{d x}}{3 x + 2 y}

y^{^{''}} + 7 y^{^{'}} + 23 y = 0

x \to 2 lim (\frac{x ^{2} + 1}{x - 1})

\int \frac{2 z - 3}{7 z - 10} d z

\frac{\partial}{\partial x} (2 sin (x) cos (y))

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