derivative f(x)=(x^2+3x-2)/(x^2-5x+6)

Lösung

ableitung von

f (x) = \frac{x ^{2} + 3 x - 2}{x ^{2} - 5 x + 6}

\frac{- 8 x ^{2} + 16 x + 8}{( x ^{2} - 5 x + 6 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} + 3 x - 2}{x ^{2} - 5 x + 6})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x ^{2} + 3 x - 2 ) ( x ^{2} - 5 x + 6 ) - \frac{d}{d x} ( x ^{2} - 5 x + 6 ) ( x ^{2} + 3 x - 2 )}{( x ^{2} - 5 x + 6 ) ^{2}}

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

\frac{d}{d x} (x^{2} - 5 x + 6) = 2 x - 5

= \frac{( 2 x + 3 ) ( x ^{2} - 5 x + 6 ) - ( 2 x - 5 ) ( x ^{2} + 3 x - 2 )}{( x ^{2} - 5 x + 6 ) ^{2}}

Multipliziere aus

(2 x + 3) (x^{2} - 5 x + 6) - (2 x - 5) (x^{2} + 3 x - 2) : - 8 x^{2} + 16 x + 8

= \frac{- 8 x ^{2} + 16 x + 8}{( x ^{2} - 5 x + 6 ) ^{2}}

d er i v a t i v e f (x) = e^{x} - x^{3}

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (1) = 0, y^{^{'}} (1) = - 1

y^{^{'}} = \frac{( x - y )}{( x + 2 y )}, y (0) = 1

\frac{d}{d x} (lo g_{e} (cos (4 x)))

\int x ln (x^{5}) d x