derivative (2t^2+1)/t

Lösung

ableitung von

\frac{2 t ^{2} + 1}{t}

\frac{2 t ^{2} - 1}{t ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

\frac{d t}{d t} = 1

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

Vereinfache

\frac{4 tt - 1 \cdot ( 2 t ^{2} + 1 )}{t ^{2}} : \frac{2 t ^{2} - 1}{t ^{2}}

= \frac{2 t ^{2} - 1}{t ^{2}}

\frac{d}{d x} (x e^{- x})

t a y l or f (x) = cos (4 x)

t an g e n t f (x) = 2 x, at x = 4

y^{^{'}} = \frac{3 x ^{2} + 2 x - 4}{2 y - 4}, y (2) = 0

in v erse l a pl a ce \frac{2 s}{( s ^{2} + 1 ) ^{2} ( s + 1 )}