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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= - \frac{2}{x ^{3}} + 0

Simplify

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

\frac{d}{d x} (co tan (3 - 2 x))

\frac{\partial}{\partial x} (8 z^{\frac{y}{x}})

\frac{d y}{d x} = x + 6 y, y (0) = 7

5 x y \frac{d y}{d x} = 2

\int \frac{1}{( x + 4 ) ^{5}} d x

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