What is (\partial)/(\partial v)(1/(u^2+v^2)) ?

The answer to (\partial)/(\partial v)(1/(u^2+v^2)) is -(2v)/((u^2+v^2)^2)

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(\partial)/(\partial v)(1/(u^2+v^2))

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

- \frac{2 v}{( u ^{2} + v ^{2} ) ^{2}}

Solution steps

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

Treat

u

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= - \frac{1}{( u ^{2} + v ^{2} ) ^{2}} \cdot 2 v

Simplify

- \frac{1}{( u ^{2} + v ^{2} ) ^{2}} \cdot 2 v : - \frac{2 v}{( u ^{2} + v ^{2} ) ^{2}}

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

\int \frac{7 + 6 x}{1 + x ^{2}} d x

x \to 3 lim (2 x - 5)

x \to \frac{π}{x} lim (x cos (x))

y = \frac{5 x}{x + 2} (3.3)

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