derivative y=(2(2x^5-9))/(x^4)

Lösung

ableitung von

y = \frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

\frac{4 ( x ^{5} + 18 )}{x ^{5}}

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (2 x^{5} - 9) = 10 x^{4}

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

= 2 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 9 )}{( x ^{4} ) ^{2}}

Vereinfache

2 \cdot \frac{10 x ^{4} x ^{4} - 4 x ^{3} ( 2 x ^{5} - 9 )}{( x ^{4} ) ^{2}} : \frac{4 ( x ^{5} + 18 )}{x ^{5}}

= \frac{4 ( x ^{5} + 18 )}{x ^{5}}

\frac{d y}{d t} = y^{3}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 cos (3 t) - 20 sin (3 t)

t an g e n t f (x) = - x^{2} + 25, at x = 3

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

\int_{0}^{\infty} \frac{1}{x ^{2} + 4} d x