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

The answer to (\partial)/(\partial x)(-(2y)/(x^2+y^2+1)) is (4yx)/((x^2+y^2+1)^2)

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

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

\frac{4 y x}{( x ^{2} + y ^{2} + 1 ) ^{2}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

- 2 y (- \frac{1}{( x ^{2} + y ^{2} + 1 ) ^{2}} \cdot 2 x) : \frac{4 y x}{( x ^{2} + y ^{2} + 1 ) ^{2}}

= \frac{4 y x}{( x ^{2} + y ^{2} + 1 ) ^{2}}

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What is (\partial)/(\partial x)(-(2y)/(x^2+y^2+1)) ?
The answer to (\partial)/(\partial x)(-(2y)/(x^2+y^2+1)) is (4yx)/((x^2+y^2+1)^2)