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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{\partial}{\partial x} (x) = 1

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

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

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

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

\int \frac{- 2 x - 1}{( x ^{2} + x + 1 ) ^{6}} d x

\frac{d y}{d x} = \frac{y ^{2}}{x ^{2}}

\frac{\partial}{\partial x} (12 cos (6 x) cos (4 y))

\int_{0}^{3} \frac{1}{x + 1} d x

y = \frac{7}{x + 5}

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