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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\int_{1}^{e^{3}} \frac{ln ^{5} ( x ^{2} )}{x} d x

d er i v a t i v e 35 cos (x)

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

x \to - 1 lim (\frac{x ^{2} - 2}{1 - x})

y^{^{'}} = 4 y^{2}