derivative of sqrt((x^2-3x/(2x+1)))

解

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

\frac{2 x ^{2} + 2 x - 3}{2 x ^{2} - 3 x ( 2 x + 1 ) ^{\frac{3}{2}}}

解答ステップ

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

連鎖法則を適用:

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

= \frac{1}{2 \frac{x ^{2} - 3 x}{2 x + 1}} \frac{d}{d x} (\frac{x ^{2} - 3 x}{2 x + 1})

\frac{d}{d x} (\frac{x ^{2} - 3 x}{2 x + 1}) = \frac{2 x ^{2} + 2 x - 3}{( 2 x + 1 ) ^{2}}

= \frac{1}{2 \frac{x ^{2} - 3 x}{2 x + 1}} \cdot \frac{2 x ^{2} + 2 x - 3}{( 2 x + 1 ) ^{2}}

簡素化

\frac{1}{2 \frac{x ^{2} - 3 x}{2 x + 1}} \cdot \frac{2 x ^{2} + 2 x - 3}{( 2 x + 1 ) ^{2}} : \frac{2 x ^{2} + 2 x - 3}{2 x ^{2} - 3 x ( 2 x + 1 ) ^{\frac{3}{2}}}

= \frac{2 x ^{2} + 2 x - 3}{2 x ^{2} - 3 x ( 2 x + 1 ) ^{\frac{3}{2}}}