derivative of (((-2x-1)/((-4x+2)))^5)

Lösung

\frac{d}{d x} ((\frac{( - 2 x - 1 )}{( - 4 x + 2 )})^{5})

- \frac{5 ( - 2 x - 1 ) ^{4}}{8 ( 2 x - 1 ) ^{6}}

Schritte zur Lösung

\frac{d}{d x} ((\frac{( - 2 x - 1 )}{( - 4 x + 2 )})^{5})

Wende die Kettenregel an:

5 (\frac{- 2 x - 1}{- 4 x + 2})^{4} \frac{d}{d x} (\frac{- 2 x - 1}{- 4 x + 2})

= 5 (\frac{- 2 x - 1}{- 4 x + 2})^{4} \frac{d}{d x} (\frac{- 2 x - 1}{- 4 x + 2})

\frac{d}{d x} (\frac{- 2 x - 1}{- 4 x + 2}) = - \frac{2}{( 2 x - 1 ) ^{2}}

= 5 (\frac{- 2 x - 1}{- 4 x + 2})^{4} (- \frac{2}{( 2 x - 1 ) ^{2}})

Vereinfache

5 (\frac{- 2 x - 1}{- 4 x + 2})^{4} (- \frac{2}{( 2 x - 1 ) ^{2}}) : - \frac{5 ( - 2 x - 1 ) ^{4}}{8 ( 2 x - 1 ) ^{6}}

= - \frac{5 ( - 2 x - 1 ) ^{4}}{8 ( 2 x - 1 ) ^{6}}

d er i v a t i v e \frac{1}{x ^{\frac{2}{3}}}

in v erse l a pl a ce \frac{6}{s ( s + 2 )}

d er i v a t i v e 3 x^{2} - 8 x - x

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

\frac{\partial}{\partial y} (x e^{x y} + y^{2} sin (x y))