(\partial)/(\partial y)(ln(x+2y+3z))

Lösung

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

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

Schritte zur Lösung

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

Behandle

x, z

als Konstanten

Wende die Kettenregel an:

\frac{1}{x + 2 y + 3 z} \frac{\partial}{\partial y} (x + 2 y + 3 z)

= \frac{1}{x + 2 y + 3 z} \frac{\partial}{\partial y} (x + 2 y + 3 z)

\frac{\partial}{\partial y} (x + 2 y + 3 z) = 2

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

Vereinfache

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

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

\int \frac{n}{n ^{2} + 6 n + 10}

\int \frac{4 x - 3}{4 x ^{2} - 6 x} d x

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

\int tanh (\frac{x}{6}) d x

x \to 0 lim (2 cos (\frac{1}{x}))