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

Lösung

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

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

Schritte zur Lösung

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

Behandle

x

als Konstante

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial y} (ln (x y)) = \frac{1}{y}

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

Vereinfache

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

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

x \to 0 lim (\frac{7 ^{x} - 2 ^{x}}{x})

y^{^{'}} + (6 x + \frac{1}{x}) y = 1

x \frac{d y}{d x} + 4 y = 5

\int 6 x e^{3 x^{2}} d x

a y^{^{''}} + b y^{^{'}} = 0