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

The answer to (\partial)/(\partial y)(aln(x^2+y^2)) is (2ay)/(x^2+y^2)

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

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

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

Solution steps

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

Treat

a, x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 3 - lim (\frac{1}{6 - 2 x})

\frac{1}{s ^{2} - 2 s + 5}

\frac{8 + x f ( x )}{x}

x^{2} - 2 x + 1

x \to 0 + lim (\frac{5}{x ^{2}})

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