derivative v(t)=(4t^2+2)/(5t+5)

Lösung

ableitung von

v = \frac{4 t ^{2} + 2}{5 t + 5}

\frac{2 ( 2 t ^{2} + 4 t - 1 )}{5 ( t + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (\frac{4 t ^{2} + 2}{5 t + 5})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d t} ( 4 t ^{2} + 2 ) ( 5 t + 5 ) - \frac{d}{d t} ( 5 t + 5 ) ( 4 t ^{2} + 2 )}{( 5 t + 5 ) ^{2}}

\frac{d}{d t} (4 t^{2} + 2) = 8 t

\frac{d}{d t} (5 t + 5) = 5

= \frac{8 t ( 5 t + 5 ) - 5 ( 4 t ^{2} + 2 )}{( 5 t + 5 ) ^{2}}

Vereinfache

\frac{8 t ( 5 t + 5 ) - 5 ( 4 t ^{2} + 2 )}{( 5 t + 5 ) ^{2}} : \frac{2 ( 2 t ^{2} + 4 t - 1 )}{5 ( t + 1 ) ^{2}}

= \frac{2 ( 2 t ^{2} + 4 t - 1 )}{5 ( t + 1 ) ^{2}}

\int (x + 2) 2 - x d x

y^{^{'}} = k y (1 - y)

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

t an g e n t 5 x^{2} - x^{3}

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