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

Lösung

ableitung von

f (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}}

\int \frac{3 - 18 x}{1 - 9 x ^{2}} d x

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

\frac{d}{d t} (\frac{1}{t ^{2} + 1})

t an g e n t f (x) = x + 1 - ln (x), at x = 0.5

(t^{2} + x^{2}) d t + (t x) d x = 0