derivative of log_{3}(4x+1)

Lösung

\frac{d}{d x} (lo g_{3} (4 x + 1))

\frac{4}{ln ( 3 ) ( 4 x + 1 )}

Schritte zur Lösung

\frac{d}{d x} (lo g_{3} (4 x + 1))

Wende die log Regel an:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 4 x + 1 )}{ln ( 3 )})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (4 x + 1))

Wende die Kettenregel an:

\frac{1}{4 x + 1} \frac{d}{d x} (4 x + 1)

= \frac{1}{4 x + 1} \frac{d}{d x} (4 x + 1)

\frac{d}{d x} (4 x + 1) = 4

= \frac{1}{ln ( 3 )} \cdot \frac{1}{4 x + 1} \cdot 4

Vereinfache

\frac{1}{ln ( 3 )} \cdot \frac{1}{4 x + 1} \cdot 4 : \frac{4}{ln ( 3 ) ( 4 x + 1 )}

= \frac{4}{ln ( 3 ) ( 4 x + 1 )}

\frac{\partial}{\partial y} (- 9 sin (3 x + yz))

x \to \frac{π}{8} lim (tan (24 x))

d er i v a t i v e tan (8 x)

\int_{0}^{1} 9 \frac{e ^{\frac{1}{x}}}{x ^{3}} d x

\frac{\partial}{\partial y} (8 e^{yz} cos (x yz))