derivative of log_{6}(5x+1)

Lösung

\frac{d}{d x} (lo g_{6} (5 x + 1))

\frac{5}{ln ( 6 ) ( 5 x + 1 )}

Schritte zur Lösung

\frac{d}{d x} (lo g_{6} (5 x + 1))

Wende die log Regel an:

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

= \frac{d}{d x} (\frac{ln ( 5 x + 1 )}{ln ( 6 )})

Entferne die Konstante:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (5 x + 1))

Wende die Kettenregel an:

\frac{1}{5 x + 1} \frac{d}{d x} (5 x + 1)

= \frac{1}{5 x + 1} \frac{d}{d x} (5 x + 1)

\frac{d}{d x} (5 x + 1) = 5

= \frac{1}{ln ( 6 )} \cdot \frac{1}{5 x + 1} \cdot 5

Vereinfache

\frac{1}{ln ( 6 )} \cdot \frac{1}{5 x + 1} \cdot 5 : \frac{5}{ln ( 6 ) ( 5 x + 1 )}

= \frac{5}{ln ( 6 ) ( 5 x + 1 )}

\int_{0}^{1 - ∣ x - 1 ∣} 1 d y

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

\frac{\partial}{\partial x} (- cos^{2} (x))

\int \frac{x}{x ^{2} + 36} d x

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