derivative of (cos(2x+x^3)/(tan(2x)-3x))

Lösung

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

\frac{- 6 x ^{3} - 2 x ^{3} sec ^{2} ( 2 x ) + 3 x ^{2} tan ( 2 x ) + 6 x sin ( 2 x ) - 2 sin ( 2 x ) tan ( 2 x ) - 2 sec ^{2} ( 2 x ) cos ( 2 x ) + 3 cos ( 2 x )}{( tan ( 2 x ) - 3 x ) ^{2}}

Schritte zur Lösung

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

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

\frac{d}{d x} (tan (2 x) - 3 x) = sec^{2} (2 x) \cdot 2 - 3

= \frac{( - sin ( 2 x ) \cdot 2 + 3 x ^{2} ) ( tan ( 2 x ) - 3 x ) - ( sec ^{2} ( 2 x ) \cdot 2 - 3 ) ( cos ( 2 x ) + x ^{3} )}{( tan ( 2 x ) - 3 x ) ^{2}}

Multipliziere aus

(- sin (2 x) \cdot 2 + 3 x^{2}) (tan (2 x) - 3 x) - (sec^{2} (2 x) \cdot 2 - 3) (cos (2 x) + x^{3}) : - 6 x^{3} - 2 x^{3} sec^{2} (2 x) + 3 x^{2} tan (2 x) + 6 x sin (2 x) - 2 sin (2 x) tan (2 x) - 2 sec^{2} (2 x) cos (2 x) + 3 cos (2 x)

= \frac{- 6 x ^{3} - 2 x ^{3} sec ^{2} ( 2 x ) + 3 x ^{2} tan ( 2 x ) + 6 x sin ( 2 x ) - 2 sin ( 2 x ) tan ( 2 x ) - 2 sec ^{2} ( 2 x ) cos ( 2 x ) + 3 cos ( 2 x )}{( tan ( 2 x ) - 3 x ) ^{2}}

x \to 2 - lim (\frac{x + 5}{x - 2})

in v erse l a pl a ce - \frac{e ^{- π s}}{s ^{2} + 1}

s l o p e (1950, 38300), (2000, 82500)

f^{^{'}} (x) = x^{2} - 5

\int \frac{1}{3 x ^{2} + 6 x + 78} d x