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

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

n = 1 \sum \infty \frac{x ^{n}}{1 ^{n}}

\int (1 + x)^{3} d x

d er i v a t i v e f (x) = \frac{x ^{2} - 3 x + 4}{x + 2}

a re a 1 + x - x^{2} - x^{3}, 1 - x

\frac{d}{d x} (\frac{6}{2 x + 1})