derivative of 1/(x^2cos(x))

Lösung

\frac{d}{d x} (\frac{1}{x ^{2} cos ( x )})

- \frac{2 cos ( x ) - x sin ( x )}{x ^{3} cos ^{2} ( x )}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{x ^{2} cos ( x )})

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

- \frac{1}{( x ^{2} cos ( x ) ) ^{2}} (2 x cos (x) - x^{2} sin (x)) : - \frac{2 cos ( x ) - x sin ( x )}{x ^{3} cos ^{2} ( x )}

= - \frac{2 cos ( x ) - x sin ( x )}{x ^{3} cos ^{2} ( x )}

y^{^{'}} = 135 y x^{4}

d er i v a t i v e f (x) = (5 x^{3} + 2) (3 x^{3} + 2)

n = 1 \sum \infty \frac{n - 1}{9 n - 1}

l a pl a ce t r an s f or m e^{- t} (t^{2} + 4 t + 5)

y^{^{'}} + 4 e^{x + y} = 0