What is (\partial)/(\partial y)(1/(x^2)+1/(y^2)) ?

The answer to (\partial)/(\partial y)(1/(x^2)+1/(y^2)) is -2/(y^3)

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(\partial)/(\partial y)(1/(x^2)+1/(y^2))

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

- \frac{2}{y ^{3}}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (\frac{1}{x ^{2}}) = 0

\frac{\partial}{\partial y} (\frac{1}{y ^{2}}) = - \frac{2}{y ^{3}}

= 0 - \frac{2}{y ^{3}}

Simplify

= - \frac{2}{y ^{3}}

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{6}})

\frac{d}{d x} (\frac{2}{1 - 2 x})

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

\int x^{3} \cdot e^{12 x} d x

x^{3} + 3 x^{2}

What is (\partial)/(\partial y)(1/(x^2)+1/(y^2)) ?
The answer to (\partial)/(\partial y)(1/(x^2)+1/(y^2)) is -2/(y^3)