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

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

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

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

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

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (z) = 0

= \frac{2 y}{x ^{2} + y ^{2}} - 0

Simplify

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

(2 + y) d x - (3 - x) d y = 0

\frac{d y}{d x} = \frac{2 x y}{x ^{2} - y ^{2}}

\int \frac{7}{x ^{5}} + 8 e^{2 x} - 4 x d x

\frac{d}{d x} (\frac{x ^{2} - 4}{x + 2})

\frac{d}{d x} (3 ln ∣ x - e^{x} ∣)

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