(\partial)/(\partial x)(3ln(sqrt(x^2+y^2)))

Lösung

\frac{\partial}{\partial x} (3 ln (x^{2} + y^{2}))

\frac{3 x}{x ^{2} + y ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial x} (3 ln (x^{2} + y^{2}))

Behandle

y

als Konstante

Entferne die Konstante:

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

= 3 \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}) = \frac{x}{x ^{2} + y ^{2}}

= 3 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot \frac{x}{x ^{2} + y ^{2}}

Vereinfache

3 \cdot \frac{1}{x ^{2} + y ^{2}} \cdot \frac{x}{x ^{2} + y ^{2}} : \frac{3 x}{x ^{2} + y ^{2}}

= \frac{3 x}{x ^{2} + y ^{2}}

\int \frac{x ^{3}}{x - 1} d x

(8 - 6 y + e^{- 3 x}) d x - 2 d y = 0

\frac{d}{d x} ((3 x - x^{3} + 1)^{4})

\int (40 - \frac{2}{sec ( θ )}) d θ

(5 + x^{8}) \frac{d y}{d x} = \frac{x ^{7}}{y}