(\partial)/(\partial y)(x^3-3xy+y^3)

Lösung

\frac{\partial}{\partial y} (x^{3} - 3 x y + y^{3})

3 y^{2} - 3 x

Schritte zur Lösung

\frac{\partial}{\partial y} (x^{3} - 3 x y + y^{3})

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

= \frac{\partial}{\partial y} (x^{3}) - \frac{\partial}{\partial y} (3 x y) + \frac{\partial}{\partial y} (y^{3})

\frac{\partial}{\partial y} (x^{3}) = 0

\frac{\partial}{\partial y} (3 x y) = 3 x

\frac{\partial}{\partial y} (y^{3}) = 3 y^{2}

= 0 - 3 x + 3 y^{2}

Vereinfache

= 3 y^{2} - 3 x

d er i v a t i v e x^{2} sin^{3} (x) + x cos^{- 3} (x)

x \to \infty lim (x^{2} + 6)

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

\int 10 x - x^{2} d x

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