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

解答

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

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

求解步骤

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

使用对数计算法则:

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

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

将常数提出:

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

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

使用连锁律:

\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

化简

\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 )}

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