derivative of (x^3+2x+6(4+1/(x^2)))

Lösung

\frac{d}{d x} ((x^{3} + 2 x + 6) (4 + \frac{1}{x ^{2}}))

\frac{12 x ^{5} - 2 x ^{3} - 2 x - 12}{x ^{3}} + 11

Schritte zur Lösung

\frac{d}{d x} ((x^{3} + 2 x + 6) (4 + \frac{1}{x ^{2}}))

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3} + 2 x + 6, g = 4 + \frac{1}{x ^{2}}

= \frac{d}{d x} (x^{3} + 2 x + 6) (4 + \frac{1}{x ^{2}}) + \frac{d}{d x} (4 + \frac{1}{x ^{2}}) (x^{3} + 2 x + 6)

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

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

= (3 x^{2} + 2) (4 + \frac{1}{x ^{2}}) + (- \frac{2}{x ^{3}}) (x^{3} + 2 x + 6)

Vereinfache

(3 x^{2} + 2) (4 + \frac{1}{x ^{2}}) + (- \frac{2}{x ^{3}}) (x^{3} + 2 x + 6) : \frac{12 x ^{5} - 2 x ^{3} - 2 x - 12}{x ^{3}} + 11

= \frac{12 x ^{5} - 2 x ^{3} - 2 x - 12}{x ^{3}} + 11

\int_{0}^{\infty} e^{- 5 x} d x

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

d er i v a t i v e (1 + e^{- t})^{- 1} + e^{- 1}

\int 5 v (v^{2} + 2)^{2} d v

\frac{d y}{d x} y = \frac{sec ( x )}{1 + sec ( x )}