(\partial)/(\partial y)(ye^{xy}+2x-1)

Lösung

\frac{\partial}{\partial y} (y e^{x y} + 2 x - 1)

e^{y x} + e^{y x} y x

Schritte zur Lösung

\frac{\partial}{\partial y} (y e^{x y} + 2 x - 1)

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

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

= \frac{\partial}{\partial y} (y e^{x y}) + \frac{\partial}{\partial y} (2 x) - \frac{\partial}{\partial y} (1)

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

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

\frac{\partial}{\partial y} (1) = 0

= e^{x y} + x e^{x y} y + 0 - 0

Vereinfache

= e^{y x} + e^{y x} y x

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

(3 x)^{^{'}}

y^{^{''}} - 6 y^{^{'}} + 34 y = 0

\frac{\partial}{\partial x} (\frac{1}{3} x^{3})

\int_{- 1}^{2} (7 - 3 t) d t