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

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

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

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

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

Solution steps

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

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

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

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

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

\int tan^{3} (x) sec^{6} (x) d x

\int \frac{1}{9} x^{- \frac{2}{9}} d x

\frac{\partial}{\partial x} (y^{2} \cdot x)

\int_{1}^{9} \frac{1}{2 x ^{2}} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = - 1, y^{^{'}} (1) = 10

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