derivative of (50x/(0.01x^2+1))

解

\frac{d}{d x} (\frac{50 x}{0.01 x ^{2} + 1})

\frac{50 ( - 0.01 x ^{2} + 1 )}{( 0.01 x ^{2} + 1 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{50 x}{0.01 x ^{2} + 1})

定数を除去:

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

= 50 \frac{d}{d x} (\frac{x}{0.01 x ^{2} + 1})

商の法則を適用:

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

= 50 \cdot \frac{\frac{d x}{d x} ( 0.01 x ^{2} + 1 ) - \frac{d}{d x} ( 0.01 x ^{2} + 1 ) x}{( 0.01 x ^{2} + 1 ) ^{2}}

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

\frac{d}{d x} (0.01 x^{2} + 1) = 0.02 x

= 50 \cdot \frac{1 \cdot ( 0.01 x ^{2} + 1 ) - 0.02 xx}{( 0.01 x ^{2} + 1 ) ^{2}}

簡素化

50 \cdot \frac{1 \cdot ( 0.01 x ^{2} + 1 ) - 0.02 xx}{( 0.01 x ^{2} + 1 ) ^{2}} : \frac{50 ( - 0.01 x ^{2} + 1 )}{( 0.01 x ^{2} + 1 ) ^{2}}

= \frac{50 ( - 0.01 x ^{2} + 1 )}{( 0.01 x ^{2} + 1 ) ^{2}}