derivative of (\sqrt[3]{x}/(x^2+x))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

t an g e n t f (x) = \frac{3}{x}, (1, 3)

y^{^{''}} + 3 y^{^{'}} - 10 y = x (e^{x} + 1)

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

f^{^{'}} (x) = 3 x^{2} (8 - x^{- 3})

\int_{0}^{1} 5^{2 x} d x