derivative f(x)=(4/x)+(1/(5x^3+2x))

Lösung

ableitung von

f (x) = (\frac{4}{x}) + (\frac{1}{5 x ^{3} + 2 x})

\frac{- 100 x ^{4} - 95 x ^{2} - 18}{x ^{2} ( 5 x ^{2} + 2 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{4}{x} + \frac{1}{5 x ^{3} + 2 x})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{4}{x}) + \frac{d}{d x} (\frac{1}{5 x ^{3} + 2 x})

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

\frac{d}{d x} (\frac{1}{5 x ^{3} + 2 x}) = - \frac{15 x ^{2} + 2}{( 5 x ^{3} + 2 x ) ^{2}}

= - \frac{4}{x ^{2}} - \frac{15 x ^{2} + 2}{( 5 x ^{3} + 2 x ) ^{2}}

Vereinfache

- \frac{4}{x ^{2}} - \frac{15 x ^{2} + 2}{( 5 x ^{3} + 2 x ) ^{2}} : \frac{- 100 x ^{4} - 95 x ^{2} - 18}{x ^{2} ( 5 x ^{2} + 2 ) ^{2}}

= \frac{- 100 x ^{4} - 95 x ^{2} - 18}{x ^{2} ( 5 x ^{2} + 2 ) ^{2}}

\int_{0}^{\frac{π}{2}} sec^{2} (\frac{x}{2}) d x

x y^{^{'}} - 2 y = 0

y^{^{'}} + y = 4 x

\int 2 x + \frac{1}{x ^{2}} + 2 x - 3 d x

t a y l or \frac{1}{2 x}, 1