derivative of log_{2}(2x+4)

解答

\frac{d}{d x} (lo g_{2} (2 x + 4))

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

求解步骤

\frac{d}{d x} (lo g_{2} (2 x + 4))

使用对数计算法则:

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

= \frac{d}{d x} (\frac{ln ( 2 x + 4 )}{ln ( 2 )})

将常数提出:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (2 x + 4))

使用连锁律:

\frac{1}{2 x + 4} \frac{d}{d x} (2 x + 4)

= \frac{1}{2 x + 4} \frac{d}{d x} (2 x + 4)

\frac{d}{d x} (2 x + 4) = 2

= \frac{1}{ln ( 2 )} \cdot \frac{1}{2 x + 4} \cdot 2

化简

\frac{1}{ln ( 2 )} \cdot \frac{1}{2 x + 4} \cdot 2 : \frac{1}{ln ( 2 ) ( x + 2 )}

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

\int x^{7} a x^{8} - b d x

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

\frac{d}{d x} (tan^{2} (x))

x \to 0 lim (e^{4 x + 2})

d er i v a t i v e f (x) = (1 + tan (2 x))^{3}