derivative of (5x^4+2x/(2x^9))

Lösung

\frac{d}{d x} (\frac{5 x ^{4} + 2 x}{2 x ^{9}})

\frac{- 25 x ^{3} - 16}{2 x ^{9}}

Schritte zur Lösung

\frac{d}{d x} (\frac{5 x ^{4} + 2 x}{2 x ^{9}})

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{d}{d x} (\frac{5 x ^{4} + 2 x}{x ^{9}})

Wende die Quotientenregel an:

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

= \frac{1}{2} \cdot \frac{\frac{d}{d x} ( 5 x ^{4} + 2 x ) x ^{9} - \frac{d}{d x} ( x ^{9} ) ( 5 x ^{4} + 2 x )}{( x ^{9} ) ^{2}}

\frac{d}{d x} (5 x^{4} + 2 x) = 20 x^{3} + 2

\frac{d}{d x} (x^{9}) = 9 x^{8}

= \frac{1}{2} \cdot \frac{( 20 x ^{3} + 2 ) x ^{9} - 9 x ^{8} ( 5 x ^{4} + 2 x )}{( x ^{9} ) ^{2}}

Vereinfache

\frac{1}{2} \cdot \frac{( 20 x ^{3} + 2 ) x ^{9} - 9 x ^{8} ( 5 x ^{4} + 2 x )}{( x ^{9} ) ^{2}} : \frac{- 25 x ^{3} - 16}{2 x ^{9}}

= \frac{- 25 x ^{3} - 16}{2 x ^{9}}

\int \frac{5 ^{x}}{5 ^{x} - 1} d x

x \to 0 lim (\frac{( 4 + x ) - 16}{x})

d er i v a t i v e f (x) = \frac{5}{6 x ^{5}}

\frac{d}{d x} (9 k \cdot ln (\frac{x + 9}{x - 9}))

y^{^{''}} + b y = 0