derivative ((3x^2+2x+7))/((x+1))

Lösung

ableitung von

\frac{( 3 x ^{2} + 2 x + 7 )}{( x + 1 )}

\frac{3 x ^{2} + 6 x - 5}{( x + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{3 x ^{2} + 2 x + 7}{x + 1})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 3 x ^{2} + 2 x + 7 ) ( x + 1 ) - \frac{d}{d x} ( x + 1 ) ( 3 x ^{2} + 2 x + 7 )}{( x + 1 ) ^{2}}

\frac{d}{d x} (3 x^{2} + 2 x + 7) = 6 x + 2

\frac{d}{d x} (x + 1) = 1

= \frac{( 6 x + 2 ) ( x + 1 ) - 1 \cdot ( 3 x ^{2} + 2 x + 7 )}{( x + 1 ) ^{2}}

Vereinfache

\frac{( 6 x + 2 ) ( x + 1 ) - 1 \cdot ( 3 x ^{2} + 2 x + 7 )}{( x + 1 ) ^{2}} : \frac{3 x ^{2} + 6 x - 5}{( x + 1 ) ^{2}}

= \frac{3 x ^{2} + 6 x - 5}{( x + 1 ) ^{2}}

\frac{d}{d x} ((x^{2} + 5 x - 3)^{3})

a re a f (x) = sin (x), 0, π

a re a \frac{1}{8 x ^{2}}, x, \frac{x}{20}

\int 6 x (x^{5} - \frac{1}{6}) d x

\frac{d Q}{d t} = (0.03 (1 - Q)^{\frac{2}{3}})