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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{4 x ( x - 1 ) ^{2} - 2 ( x - 1 ) ( 2 x ^{2} - 1 )}{( ( x - 1 ) ^{2} ) ^{2}} : \frac{2 ( - 2 x + 1 )}{( x - 1 ) ^{3}}

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

\frac{\partial}{\partial x} (x^{8} + 2^{y} + x^{y})

y^{^{'}} = \frac{x}{y ^{2}}

t \frac{d y}{d t} + 2 y = t^{2}

\int_{1}^{9} \frac{3 x - 2}{x} d x

\int sec^{3} (5 x) tan^{2} (5 x) d x