derivative of (1-3x/(6+x))

Lösung

\frac{d}{d x} (\frac{1 - 3 x}{6 + x})

- \frac{19}{( 6 + x ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{1 - 3 x}{6 + x})

Wende die Quotientenregel an:

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

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

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

\frac{d}{d x} (6 + x) = 1

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

Vereinfache

\frac{( - 3 ) ( 6 + x ) - 1 \cdot ( 1 - 3 x )}{( 6 + x ) ^{2}} : - \frac{19}{( 6 + x ) ^{2}}

= - \frac{19}{( 6 + x ) ^{2}}

B^{^{'}}

t an g e n t y = 2 x^{3} - 2 x, (1, 0)

x \to 0 lim (arctan (3 x))

\frac{d x}{d t} = 3 x + e^{2 t} - 2 e^{3 t} + e^{4 t}

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