derivative (x^2+2x-8)/(x-1)

Lösung

ableitung von

\frac{x ^{2} + 2 x - 8}{x - 1}

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

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 2 x - 8}{x - 1})

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 - 8 ) ( x - 1 ) - \frac{d}{d x} ( x - 1 ) ( x ^{2} + 2 x - 8 )}{( x - 1 ) ^{2}}

\frac{d}{d x} (x^{2} + 2 x - 8) = 2 x + 2

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

= \frac{( 2 x + 2 ) ( x - 1 ) - 1 \cdot ( x ^{2} + 2 x - 8 )}{( x - 1 ) ^{2}}

Vereinfache

\frac{( 2 x + 2 ) ( x - 1 ) - 1 \cdot ( x ^{2} + 2 x - 8 )}{( x - 1 ) ^{2}} : \frac{x ^{2} - 2 x + 6}{( x - 1 ) ^{2}}

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

\int \frac{a ^{l n (x)}}{x} d x

d er i v a t i v e \frac{1}{1 - 5 x ^{2}}

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

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

y^{^{''}} + \frac{14}{t} y^{^{'}} + \frac{42}{t ^{2}} y = 0