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

Lösung

ableitung von

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

- \frac{13}{( 3 + t ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

= - \frac{13}{( 3 + t ) ^{2}}

f (x) = \frac{sin ( x )}{x ^{2}}

\int x^{2} + 5 x + 5 d x

y y^{^{'}} - 2 e^{x} = 0

n \to \infty lim (sin (n))

\frac{\partial}{\partial x} (x^{2} y - 2 x y + y z^{2})