(\partial)/(\partial y)(ln(x^2+y^2+1))

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (3 - \frac{e ^{y}}{e ^{x + 1}})

d er i v a t i v e f (x) = 7 x^{6}

t an g e n t f (x) = 2 x^{2} + 3 x - 2, at x = 0

\int (3 x^{2} + 2 x - 1) d x

x \to 1 + lim (x^{3})