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

Lösung

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

\frac{4 x}{x ^{2} + y ^{2}}

Schritte zur Lösung

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

Behandle

y

als Konstante

Entferne die Konstante:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

2 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot 2 x : \frac{4 x}{x ^{2} + y ^{2}}

= \frac{4 x}{x ^{2} + y ^{2}}

\frac{\partial}{\partial x} (e^{x + t})

\frac{\partial}{\partial x} (6 e^{x y})

\frac{d}{d x} (e^{x} - x - 1)

y^{^{'}} + y - x^{2} - 2 x = 0

(\frac{2 y}{x} + \frac{2 x ^{4}}{3}) d x - d y = 0