((x-2ln(x)-1)/(x-1))^'

Lösung

(\frac{x - 2 ln ( x ) - 1}{x - 1})^{^{'}}

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

Schritte zur Lösung

(\frac{x - 2 ln ( x ) - 1}{x - 1})^{'}

Wende die Quotientenregel an:

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

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

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

(x - 1)^{'} = 1

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

Vereinfache

\frac{( - \frac{2}{x} + 1 ) ( x - 1 ) - 1 \cdot ( x - 2 ln ( x ) - 1 )}{( x - 1 ) ^{2}} : \frac{2 + 2 x ln ( x ) - 2 x}{x ( x - 1 ) ^{2}}

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

\frac{d}{d x} (x^{7} e^{4 x})

\int x^{5} sin (x^{6}) d x

\frac{d}{d x} \frac{3}{( ( 56 ) ( 2 x - 1 ) ^{7} )} - \frac{1}{( ( 24 ) ( 2 x - 1 ) ^{6} )} - \frac{1}{( ( 40 ( 2 x - 1 ) ^{5} ) )}

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

\frac{d}{d x} (\frac{1}{2 x + 1})