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

The answer to (\partial)/(\partial z)(ln(sqrt(x^2+y^2+z^2))) is z/(x^2+y^2+z^2)

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(\partial)/(\partial z)(ln(sqrt(x^2+y^2+z^2)))

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

\frac{z}{x ^{2} + y ^{2} + z ^{2}}

Solution steps

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

Treat

x, y

as constants

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{z}{x ^{2} + y ^{2} + z ^{2}}

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

x \to 0 lim (e^{x + 1})

\int cos (t - 2 x) d x

\int (6 x^{2} - 3 x) d x

\frac{d}{d x} (lo g_{6} (2 x^{3} - 5 x^{2}))

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