derivative f(x)=(2500x)/(ln(8x+8))

解

導関数

f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

\frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}

解答ステップ

\frac{d}{d x} (\frac{2500 x}{ln ( 8 x + 8 )})

定数を除去:

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

= 2500 \frac{d}{d x} (\frac{x}{ln ( 8 x + 8 )})

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 2500 \cdot \frac{\frac{d x}{d x} ln ( 8 x + 8 ) - \frac{d}{d x} ( ln ( 8 x + 8 ) ) x}{( ln ( 8 x + 8 ) ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} (ln (8 x + 8)) = \frac{1}{x + 1}

= 2500 \cdot \frac{1 \cdot ln ( 8 x + 8 ) - \frac{1}{x + 1} x}{( ln ( 8 x + 8 ) ) ^{2}}

簡素化

2500 \cdot \frac{1 \cdot ln ( 8 x + 8 ) - \frac{1}{x + 1} x}{ln ^{2} ( 8 x + 8 )} : \frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}

= \frac{2500 ( ln ( 8 x + 8 ) ( x + 1 ) - x )}{ln ^{2} ( 8 x + 8 ) ( x + 1 )}