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

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

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

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

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

Solution steps

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

Simplify

\frac{- y}{2 x ^{2} + y ^{2}} : - \frac{y}{2 x ^{2} + y ^{2}}

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(x^{2} - 2 x) (2 x - x^{2})

\frac{\partial}{\partial t} (t e^{s t})

\frac{d}{d x} ((x - 1) tan (x) + x sin (π x))

(2 x y) d x + (x^{2} - 1) d y = 0

\frac{8}{x ^{6}}

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