derivative of (1-3x/(2+x))

Lösung

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

- \frac{7}{( 2 + x ) ^{2}}

Schritte zur Lösung

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

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

\frac{d}{d x} (2 + x) = 1

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

Vereinfache

\frac{( - 3 ) ( 2 + x ) - 1 \cdot ( 1 - 3 x )}{( 2 + x ) ^{2}} : - \frac{7}{( 2 + x ) ^{2}}

= - \frac{7}{( 2 + x ) ^{2}}

\int sec^{5} (x) d x

t an g e n t f (x) = x x, at x = 0

\int \frac{1}{x ^{2} + 8 x + 32} d x

x \to 7 - lim (\frac{x}{x ^{2} - 49})

a re a x + 1, 9 - x^{2}, - 1, 2