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

Lösung

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

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

Schritte zur Lösung

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

Vereinfache

\frac{- y}{x ^{2} + y ^{2}} : - \frac{y}{x ^{2} + y ^{2}}

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

Behandle

x

als Konstante

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

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

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

\int_{- \infty}^{0} e^{- x} d x

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

t \to π lim (tan (\frac{5 t}{4}))

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = 2

\frac{\partial}{\partial x} (2 ln (x z))