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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int tan^{2} (x + c) + 1 d x

x^{2}, 1

y^{^{''''}} + y^{^{''}} = 0

\frac{\partial}{\partial x} (x z - e^{x} sin (y))

x \to - 2 lim (\frac{5 x ^{2} + 6}{x})

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