derivative y=((2x-3)/((x^2+4)^2))

Lösung

ableitung von

y = (\frac{2 x - 3}{( x ^{2} + 4 ) ^{2}})

\frac{2 ( - 3 x ^{2} + 6 x + 4 )}{( x ^{2} + 4 ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 2 x - 3 ) ( x ^{2} + 4 ) ^{2} - \frac{d}{d x} ( ( x ^{2} + 4 ) ^{2} ) ( 2 x - 3 )}{( ( x ^{2} + 4 ) ^{2} ) ^{2}}

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

\frac{d}{d x} ((x^{2} + 4)^{2}) = 4 x (x^{2} + 4)

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

Vereinfache

\frac{2 ( x ^{2} + 4 ) ^{2} - 4 x ( x ^{2} + 4 ) ( 2 x - 3 )}{( ( x ^{2} + 4 ) ^{2} ) ^{2}} : \frac{2 ( - 3 x ^{2} + 6 x + 4 )}{( x ^{2} + 4 ) ^{3}}

= \frac{2 ( - 3 x ^{2} + 6 x + 4 )}{( x ^{2} + 4 ) ^{3}}

s l o p e in t erce pt (1, 4), (4, 1)

d er i v a t i v e y = (2 x + e^{- x^{2}})^{3}

\frac{d P}{d t} = P (a - b P)

a re a y = 2 x, y = 2 x^{2} - 4

\frac{d}{d x} (∣ 2 x + 1 ∣)