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

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

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

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

2 y ln (x^{2} + y^{2}) + 2 y

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + y^{2}, g = ln (x^{2} + y^{2})

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

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

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

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

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

= 2 y ln (x^{2} + y^{2}) + 2 y

5 t^{- \frac{3}{5}}

\frac{5}{3 - 2 x}

\int sec^{4} (4 x) d x

x \to - 2 lim (x^{2} + 1)

\frac{6 x}{x ^{2} + 1}

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