(\partial)/(\partial y)(1/(2-x+y))

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

x \to - 1 lim (x^{3} - 3 x^{2} - 2 x + 1)

e x p an d x^{2} (1 - 2 x)

\frac{d}{d x} (\frac{1}{csc ( x )})

\int_{0}^{2} (2 x - x^{3}) d x

\frac{\partial}{\partial x} (x^{6} y^{5} + 2 x^{9} y)