derivative of log_{5}(1+2x)

解答

\frac{d}{d x} (lo g_{5} (1 + 2 x))

\frac{2}{ln ( 5 ) ( 1 + 2 x )}

求解步骤

\frac{d}{d x} (lo g_{5} (1 + 2 x))

使用对数计算法则:

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

= \frac{d}{d x} (\frac{ln ( 1 + 2 x )}{ln ( 5 )})

将常数提出:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (1 + 2 x))

使用连锁律:

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

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

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

= \frac{1}{ln ( 5 )} \cdot \frac{1}{1 + 2 x} \cdot 2

化简

\frac{1}{ln ( 5 )} \cdot \frac{1}{1 + 2 x} \cdot 2 : \frac{2}{ln ( 5 ) ( 1 + 2 x )}

= \frac{2}{ln ( 5 ) ( 1 + 2 x )}

\frac{d}{d x} (\frac{x ^{\frac{1}{3}} - 4}{x ^{\frac{7}{2}}})

x \to 0 lim (x \cdot co tan (2 π x))

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

y^{^{''}} + 9 y = 0

t an g e n t y = 10 cot (\frac{π x}{20}), at x = 5