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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = - 0.05 y

\frac{\partial}{\partial y} (2 x y^{4} - 4)

\int x^{3} (x^{2} + 1)^{8} d x

\frac{d}{d z} (x e^{\frac{y}{z}})

\int \frac{1}{( a x + b ) ^{4}} d x

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