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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

d er i v a t i v e (arctan (8 x))^{2}

\int_{1}^{e^{2}} ln (x) d x

y^{^{''}} + 14 y^{^{'}} + 58 y = 0

d er i v a t i v e \frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

\int 2 (x - 2) \cdot ln (x - 2) d x