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

Lösung

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

2 x + 2 y

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

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

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

\frac{\partial}{\partial y} (2 x y) = 2 x

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

\frac{\partial}{\partial y} (8) = 0

= 0 + 2 x + 2 y - 0

Vereinfache

= 2 x + 2 y

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

f (x) = tan (5 x)

d er i v a t i v e ln (\frac{( x - 3 )}{x - 1})

\int t^{3} cos (t^{2}) d t

x \to 5 + lim (\frac{5 - x}{∣ 5 - x ∣})