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

Lösung

ableitung von

(\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}}

\frac{\partial}{\partial y} (x^{2} + x y^{4})

\int_{1}^{x} \frac{1}{t} d t

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

d er i v a t i v e y = \frac{x - x}{x ^{2}}

\frac{d}{d t} (5 t + 4 t^{5})