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

The answer to (\partial)/(\partial y)(ln(x^3+y^2)-ln(xy)) is (y^2-x^3)/(y(x^3+y^2))

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(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}

Simplify

\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

What is (\partial)/(\partial y)(ln(x^3+y^2)-ln(xy)) ?
The answer to (\partial)/(\partial y)(ln(x^3+y^2)-ln(xy)) is (y^2-x^3)/(y(x^3+y^2))