derivative of (2x^2-1/(xsqrt(1+x^2)))

解

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

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

解答ステップ

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

商の法則を適用:

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

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

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

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

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

簡素化

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

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