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

解

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

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

解答ステップ

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

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

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

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

拡張

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

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