derivative of (x^4+2x-1/(5x^2-2x+1))

解

\frac{d}{d x} (\frac{x ^{4} + 2 x - 1}{5 x ^{2} - 2 x + 1})

\frac{10 x ^{5} - 6 x ^{4} + 4 x ^{3} - 10 x ^{2} + 10 x}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

解答ステップ

\frac{d}{d x} (\frac{x ^{4} + 2 x - 1}{5 x ^{2} - 2 x + 1})

商の法則を適用:

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

= \frac{\frac{d}{d x} ( x ^{4} + 2 x - 1 ) ( 5 x ^{2} - 2 x + 1 ) - \frac{d}{d x} ( 5 x ^{2} - 2 x + 1 ) ( x ^{4} + 2 x - 1 )}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

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

\frac{d}{d x} (5 x^{2} - 2 x + 1) = 10 x - 2

= \frac{( 4 x ^{3} + 2 ) ( 5 x ^{2} - 2 x + 1 ) - ( 10 x - 2 ) ( x ^{4} + 2 x - 1 )}{( 5 x ^{2} - 2 x + 1 ) ^{2}}

拡張

(4 x^{3} + 2) (5 x^{2} - 2 x + 1) - (10 x - 2) (x^{4} + 2 x - 1) : 10 x^{5} - 6 x^{4} + 4 x^{3} - 10 x^{2} + 10 x

= \frac{10 x ^{5} - 6 x ^{4} + 4 x ^{3} - 10 x ^{2} + 10 x}{( 5 x ^{2} - 2 x + 1 ) ^{2}}