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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

- \frac{1}{( x ^{2} + y ^{2} ) ^{2}} \cdot 2 x : - \frac{2 x}{( x ^{2} + y ^{2} ) ^{2}}

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

(x^{4} + y^{4}) d x - (x y^{3}) d y = 0

\int \frac{1}{( x ^{2} + 25 ) ^{2}} d x

f (x) = (\frac{9}{8}) x^{\frac{8}{9}}

\frac{\partial}{\partial y} (4 - x^{2} - 2 y^{2})

2 r (s^{2} + 1) d r + (r^{4} + 1) d s = 0

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