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

Lösung

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

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

Schritte zur Lösung

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

Behandle

y

als Konstante

Wende die Kettenregel an:

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

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

\frac{\partial}{\partial x} (2 x^{2} + y^{2}) = 4 x

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

Vereinfache

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

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

n = 1 \sum \infty \frac{3}{n ^{2}}

(x y^{2} + x) d x + (y - x^{2} y) d y = 0

x \to 0 lim (x sin (\frac{4 π}{x}))

\int_{1}^{2} 27 x^{2} 4^{x^{3}} d x

\frac{d}{d x} (cos (x^{2} + x y))