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

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

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

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

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

Solution steps

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

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

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

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

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

(ln (2 x))^{^{'}}

\int_{b}^{1} 8 ln (x) d x

\int \frac{y}{( y ^{2} + 1 ) ^{\frac{1}{2}}} d y

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

\frac{\partial}{\partial x} (e^{8 x} cos (- 9 y))

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