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

Lösung

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

\frac{8}{( 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{8}{( x + 1 ) ^{3}}

= \frac{8}{( x + 1 ) ^{3}}

d er i v a t i v e ln (22 x)

a re a x + 1 = y^{2}, y^{2} + 2 x = 10

12 y^{^{''}} + 5 y^{^{'}} - 3 y = 0

y^{^{'}} + \frac{4}{x} y = x^{4}

\int 7 x^{3} e^{- x^{2}} d x