(\partial)/(\partial x)(ln(x^2-4xy+y^2))

Lösung

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

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

Schritte zur Lösung

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

Behandle

y

als Konstante

Wende die Kettenregel an:

\frac{1}{x ^{2} - 4 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2})

= \frac{1}{x ^{2} - 4 x y + y ^{2}} \frac{\partial}{\partial x} (x^{2} - 4 x y + y^{2})

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

= \frac{1}{x ^{2} - 4 x y + y ^{2}} (2 x - 4 y)

Vereinfache

\frac{1}{x ^{2} - 4 x y + y ^{2}} (2 x - 4 y) : \frac{2 x - 4 y}{x ^{2} - 4 x y + y ^{2}}

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

x \to 0 lim (\frac{4}{17 + x})

\int \frac{1}{( x ^{2} + 9 )} d x

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

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{2}{x ^{4}}) (x + 8 x^{3}))

n = 1 \sum \infty (\frac{1}{2})^{\frac{1}{n}}