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

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

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

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

Vereinfache

\frac{1}{2 x ^{2} + y ^{2}} \cdot 2 y : \frac{2 y}{2 x ^{2} + y ^{2}}

= \frac{2 y}{2 x ^{2} + y ^{2}}

x \to 4 lim (\frac{x - 4}{x ^{2} + 3 x - 28})

x \to \infty lim (\frac{x}{12})

\frac{d}{d x} (\frac{x - 7}{x + 5})

\frac{\partial}{\partial x} (x^{2} yz - x y z^{5})

y^{^{'''}} - y^{^{''}} + 7 y^{^{'}} - 7 y = 0