derivative of (6x+x^2/((3+x)^2))

Lösung

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

\frac{18}{( 3 + x ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{( 6 + 2 x ) ( 3 + x ) ^{2} - 2 ( 3 + x ) ( 6 x + x ^{2} )}{( ( 3 + x ) ^{2} ) ^{2}} : \frac{18}{( 3 + x ) ^{3}}

= \frac{18}{( 3 + x ) ^{3}}

n \to \infty lim (0. 5^{n})

d er i v a t i v e 6 x e^{x}

in v erse l a pl a ce \frac{( 3 s + 2 )}{( s ^{2} + 4 s + 20 )}

x \to 0 + lim (\frac{x}{1 + ln ( x )})

y^{^{''}} + 7 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1