(\partial)/(\partial x)(({f}(x,y))(x^2-2xy))

Lösung

\frac{\partial}{\partial x} ((f (x, y)) (x^{2} - 2 x y))

\frac{\partial}{\partial x} (f (x, y)) (x^{2} - 2 x y) + (2 x - 2 y) f (x, y)

Schritte zur Lösung

\frac{\partial}{\partial x} ((f (x, y)) (x^{2} - 2 x y))

Behandle

y

als Konstante

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = f (x, y), g = x^{2} - 2 x y

= \frac{\partial}{\partial x} (f (x, y)) (x^{2} - 2 x y) + \frac{\partial}{\partial x} (x^{2} - 2 x y) f (x, y)

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

= \frac{\partial}{\partial x} (f (x, y)) (x^{2} - 2 x y) + (2 x - 2 y) f (x, y)

e x p an d \frac{e ^{- x} \cdot ( x ^{2} + 2 )}{x ^{2} + 3 \cdot x + 2}

e x p an d (3 y^{\frac{1}{3}} = \frac{1}{2} x^{2} + C) \frac{1}{3}

e x p an d \frac{4 ( r + 1 ) ^{3}}{3}

e x p an d (a x^{2} y^{4} e^{x^{3} y^{4}} + \frac{3}{x ^{2} + 1}) d x

e x p an d (4 - e^{x}) e^{x}