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

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

= - \frac{4}{( x - 1 ) ^{3}}

\frac{d}{d x} (e^{3 x} sin (2 x))

y d x + x^{2} d y = 0

\int (\frac{8 x - 3}{12 x - 4 x ^{2} - 5}) d x

\int \frac{sin ( x )}{1 + 2 cos ( x )} d x

l a pl a ce t r an s f or m cos^{2} (3 t)