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

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

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

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

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

Solution steps

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

Treat

y, z

as constants

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to π lim (tan (\frac{x}{2}))

x \to - 4 lim (x^{2} - 5)

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

1^{^{'}}

x \to 2 + lim (\frac{3 x}{x - 2})

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