derivative 6/(x^3)-6/(x^2)-1/x+350

Lösung

ableitung von

\frac{6}{x ^{3}} - \frac{6}{x ^{2}} - \frac{1}{x} + 350

\frac{12}{x ^{3}} + \frac{1}{x ^{2}} - \frac{18}{x ^{4}}

Schritte zur Lösung

\frac{d}{d x} (\frac{6}{x ^{3}} - \frac{6}{x ^{2}} - \frac{1}{x} + 350)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (\frac{6}{x ^{3}}) - \frac{d}{d x} (\frac{6}{x ^{2}}) - \frac{d}{d x} (\frac{1}{x}) + \frac{d}{d x} (350)

\frac{d}{d x} (\frac{6}{x ^{3}}) = - \frac{18}{x ^{4}}

\frac{d}{d x} (\frac{6}{x ^{2}}) = - \frac{12}{x ^{3}}

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

\frac{d}{d x} (350) = 0

= - \frac{18}{x ^{4}} - (- \frac{12}{x ^{3}}) - (- \frac{1}{x ^{2}}) + 0

Vereinfache

= \frac{12}{x ^{3}} + \frac{1}{x ^{2}} - \frac{18}{x ^{4}}

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

\frac{d}{d x} (2 x + 2 x ln (x))

t an g e n t (4 x + 4)^{\frac{1}{3}}, at x = 2

t an g e n t y = \frac{- 6 x}{x ^{2} + 1}, (- 1, 3)

y^{^{''}} + 4 y^{^{'}} - 32 y = 0