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

解

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

\frac{3 x ^{5} - 9 x ^{2}}{2 ( x ^{3} - 1 ) x ^{3} - 1}

解答ステップ

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

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

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

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

簡素化

\frac{3 x ^{2} x ^{3} - 1 - \frac{3 x ^{2}}{2 x ^{3} - 1} ( x ^{3} + 1 )}{( x ^{3} - 1 ) ^{2}} : \frac{3 x ^{5} - 9 x ^{2}}{2 ( x ^{3} - 1 ) x ^{3} - 1}

= \frac{3 x ^{5} - 9 x ^{2}}{2 ( x ^{3} - 1 ) x ^{3} - 1}