derivative (24x^2+36x)/(2x+2)

Lösung

ableitung von

\frac{24 x ^{2} + 36 x}{2 x + 2}

\frac{6 ( 2 x ^{2} + 4 x + 3 )}{( x + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{24 x ^{2} + 36 x}{2 x + 2})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 24 x ^{2} + 36 x ) ( 2 x + 2 ) - \frac{d}{d x} ( 2 x + 2 ) ( 24 x ^{2} + 36 x )}{( 2 x + 2 ) ^{2}}

\frac{d}{d x} (24 x^{2} + 36 x) = 48 x + 36

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

= \frac{( 48 x + 36 ) ( 2 x + 2 ) - 2 ( 24 x ^{2} + 36 x )}{( 2 x + 2 ) ^{2}}

Vereinfache

\frac{( 48 x + 36 ) ( 2 x + 2 ) - 2 ( 24 x ^{2} + 36 x )}{( 2 x + 2 ) ^{2}} : \frac{6 ( 2 x ^{2} + 4 x + 3 )}{( x + 1 ) ^{2}}

= \frac{6 ( 2 x ^{2} + 4 x + 3 )}{( x + 1 ) ^{2}}

\frac{d}{d x} (7 sin (x) e^{- x} - 1)

\frac{\partial}{\partial x} (e^{2 x} - x cos (x y))

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

\int \frac{1}{x ^{3} - x} d x

2 y^{^{''}} + y^{^{'}} + 3 y = 3 cos (3 t) + 2 sin (3 t)