(\partial)/(\partial y)(-8y^3+4xy+4x^2+9y^2)

Lösung

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 y^{2})

- 24 y^{2} + 18 y + 4 x

Schritte zur Lösung

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 y^{2})

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

= - \frac{\partial}{\partial y} (8 y^{3}) + \frac{\partial}{\partial y} (4 x y) + \frac{\partial}{\partial y} (4 x^{2}) + \frac{\partial}{\partial y} (9 y^{2})

\frac{\partial}{\partial y} (8 y^{3}) = 24 y^{2}

\frac{\partial}{\partial y} (4 x y) = 4 x

\frac{\partial}{\partial y} (4 x^{2}) = 0

\frac{\partial}{\partial y} (9 y^{2}) = 18 y

= - 24 y^{2} + 4 x + 0 + 18 y

Vereinfache

= - 24 y^{2} + 18 y + 4 x

\frac{\partial}{\partial x} (\frac{x}{10} cos (5 x))

\int_{- 2}^{3} - x^{2} + x + 6 d x

\int (x + y)^{- 2} (x^{2} + 2 x y - y^{2}) d x

\int - \frac{e ^{4 x}}{x} d x

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