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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

= 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

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

Simplify

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

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

6 x y^{^{'}} = 2 x e^{x} - 6 y + 6 x^{2}

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

\int \frac{1}{2 k ^{2}} \cdot e^{- \frac{x}{2 k ^{2}}} d x

\int \frac{1 - u}{1 + u ^{2}} d u

\int x^{2} (x^{3} + 5)^{8} cos ((x^{3} + 5)^{9}) d x

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