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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

e^{x^{2}}

\frac{1}{s ^{2} + 16}

\frac{d y}{d x}, y = (sin (7 x))^{l n (x)}

\frac{s - 1}{2 s ^{2} + s + 3}

\frac{d}{d x} (5 x e^{x^{2}})

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