derivative of (x^2+3/((x-1)^3+(x+1)^3))

解

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

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

解答ステップ

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

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

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

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

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