What is (\partial)/(\partial x)((x^{-2}+y^{-2})^3) ?

The answer to (\partial)/(\partial x)((x^{-2}+y^{-2})^3) is -(6(y^2+x^2)^2)/(x^7y^4)

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(\partial)/(\partial x)((x^{-2}+y^{-2})^3)

\frac{\partial}{\partial x} ((x^{- 2} + y^{- 2})^{3})

- \frac{6 ( y ^{2} + x ^{2} ) ^{2}}{x ^{7} y ^{4}}

Solution steps

\frac{\partial}{\partial x} ((x^{- 2} + y^{- 2})^{3})

Treat

y

as a constant

Apply the chain rule:

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

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

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

= 3 (x^{- 2} + y^{- 2})^{2} (- \frac{2}{x ^{3}})

Simplify

3 (x^{- 2} + y^{- 2})^{2} (- \frac{2}{x ^{3}}) : - \frac{6 ( y ^{2} + x ^{2} ) ^{2}}{x ^{7} y ^{4}}

= - \frac{6 ( y ^{2} + x ^{2} ) ^{2}}{x ^{7} y ^{4}}

- 3 4 2 - 9 x

(\frac{2}{x ^{2}})^{^{'}}

\int x^{3} - 3 x^{- 4} - 4 d x

\int \frac{1 - 4 x + 3 x ^{2}}{x ^{2}} d x

f (x) = \frac{3}{x}, at x = 5

What is (\partial)/(\partial x)((x^{-2}+y^{-2})^3) ?
The answer to (\partial)/(\partial x)((x^{-2}+y^{-2})^3) is -(6(y^2+x^2)^2)/(x^7y^4)