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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (x^{2} x^{3} + a^{3})

\int \frac{3 x ^{3} + 3 x + 2}{x ^{3} + x} d x

f (x) = \frac{1}{x ^{4}}, at x = 3

\frac{\partial}{\partial z} (x e^{z})

\int x^{2} + 3 x - 1000000 d x

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