derivative f(x)=(x^3+1)/(x^3-1)

解答

导数

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

- \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

求解步骤

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

使用微分除法定则:

(\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) = 3 x^{2}

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

化简

\frac{3 x ^{2} ( x ^{3} - 1 ) - 3 x ^{2} ( x ^{3} + 1 )}{( x ^{3} - 1 ) ^{2}} : - \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}

= - \frac{6 x ^{2}}{( x ^{3} - 1 ) ^{2}}