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

Lösung

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

\frac{6 y}{2 x ^{3} + 3 y ^{2}}

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

\frac{\partial}{\partial y} (2 x^{3} + 3 y^{2}) = 6 y

= \frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y

Vereinfache

\frac{1}{2 x ^{3} + 3 y ^{2}} \cdot 6 y : \frac{6 y}{2 x ^{3} + 3 y ^{2}}

= \frac{6 y}{2 x ^{3} + 3 y ^{2}}

\int_{0}^{2} (x^{2})^{2} d x

x \to 0 lim (\frac{1}{1 - e ^{x}} - \frac{1}{x})

\int e^{x^{2}} x d x

x \to - 5 + lim (\frac{∣ x - 5 ∣}{x - 5})

\frac{d}{d x} (9 cos (x) - cos (9 x))