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

Lösung

ableitung von

\frac{1 - 2 x}{3 + x}

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

= - \frac{7}{( 3 + x ) ^{2}}

p a r i t y x^{2} tan (x)

x \to + 0 lim (\frac{tan ( 6 x )}{sin ( 4 x )})

x \to 0 lim (\frac{cos ( a x ) - cos ( b x )}{x ^{2}})

y^{^{'}} = 1 - sin (y), y (0) = 0

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