derivative of 1/(x^4*(5-x))

Lösung

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

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

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (x^{4} (5 - x)) = 20 x^{3} - 5 x^{4}

= - \frac{1}{( x ^{4} ( 5 - x ) ) ^{2}} (20 x^{3} - 5 x^{4})

Vereinfache

- \frac{1}{( x ^{4} ( 5 - x ) ) ^{2}} (20 x^{3} - 5 x^{4}) : - \frac{5 ( - x + 4 )}{x ^{5} ( - x + 5 ) ^{2}}

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

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

\frac{\partial}{\partial x} (x^{4} - 3 x^{2} y^{2} + y^{4})

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = 0, y (\frac{π}{2}) = 1

d er i v a t i v e f (x) = 4 x^{2} - 1

t an g e n t y = (1 + 5 x)^{12}, (0, 1)