What is (\partial)/(\partial u)(ln(u^2+v^2+w^2)) ?

The answer to (\partial)/(\partial u)(ln(u^2+v^2+w^2)) is (2u)/(u^2+v^2+w^2)

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(\partial)/(\partial u)(ln(u^2+v^2+w^2))

\frac{\partial}{\partial u} (ln (u^{2} + v^{2} + w^{2}))

\frac{2 u}{u ^{2} + v ^{2} + w ^{2}}

Solution steps

\frac{\partial}{\partial u} (ln (u^{2} + v^{2} + w^{2}))

Treat

v, w

as constants

Apply the chain rule:

\frac{1}{u ^{2} + v ^{2} + w ^{2}} \frac{\partial}{\partial u} (u^{2} + v^{2} + w^{2})

= \frac{1}{u ^{2} + v ^{2} + w ^{2}} \frac{\partial}{\partial u} (u^{2} + v^{2} + w^{2})

\frac{\partial}{\partial u} (u^{2} + v^{2} + w^{2}) = 2 u

= \frac{1}{u ^{2} + v ^{2} + w ^{2}} \cdot 2 u

Simplify

\frac{1}{u ^{2} + v ^{2} + w ^{2}} \cdot 2 u : \frac{2 u}{u ^{2} + v ^{2} + w ^{2}}

= \frac{2 u}{u ^{2} + v ^{2} + w ^{2}}

What is (\partial)/(\partial u)(ln(u^2+v^2+w^2)) ?
The answer to (\partial)/(\partial u)(ln(u^2+v^2+w^2)) is (2u)/(u^2+v^2+w^2)