derivative G(t)=(1-3t)/(4+t)

Lösung

ableitung von

G (t) = \frac{1 - 3 t}{4 + t}

- \frac{13}{( 4 + t ) ^{2}}

Schritte zur Lösung

\frac{d}{d t} (\frac{1 - 3 t}{4 + t})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d t} ( 1 - 3 t ) ( 4 + t ) - \frac{d}{d t} ( 4 + t ) ( 1 - 3 t )}{( 4 + t ) ^{2}}

\frac{d}{d t} (1 - 3 t) = - 3

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

= \frac{( - 3 ) ( 4 + t ) - 1 \cdot ( 1 - 3 t )}{( 4 + t ) ^{2}}

Vereinfache

\frac{( - 3 ) ( 4 + t ) - 1 \cdot ( 1 - 3 t )}{( 4 + t ) ^{2}} : - \frac{13}{( 4 + t ) ^{2}}

= - \frac{13}{( 4 + t ) ^{2}}

\int \frac{x ^{2}}{5 x ^{3} + 8} d x

t an g e n t f (x) = 3 e^{x} + x, (0, 3)

in v erse l a pl a ce \frac{1}{( s ^{2} + 5 )}

d er i v a t i v e f (x) = \frac{1 - 4 x}{1 + 4 x}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 2)