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

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{d}{d x} (\frac{A}{x ^{9}} + B e^{x})

(4 + x) y^{^{'}} = 8 y

\int (x^{2} - 7 x + 2) d x

t an g e n t y = e^{5 x}, at x = \frac{1}{5} ln (5)

\int 5 sin^{3} (x co) s^{7} x d x