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

The answer to (\partial)/(\partial x)((2xy)/((x^2-y^2)^2)) is (2y(-3x^2-y^2))/((x^2-y^2)^3)

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

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

\frac{2 y ( - 3 x ^{2} - y ^{2} )}{( x ^{2} - y ^{2} ) ^{3}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

2 y \frac{1 \cdot ( x ^{2} - y ^{2} ) ^{2} - 4 x ( x ^{2} - y ^{2} ) x}{( ( x ^{2} - y ^{2} ) ^{2} ) ^{2}} : \frac{2 y ( - 3 x ^{2} - y ^{2} )}{( x ^{2} - y ^{2} ) ^{3}}

= \frac{2 y ( - 3 x ^{2} - y ^{2} )}{( x ^{2} - y ^{2} ) ^{3}}

What is (\partial)/(\partial x)((2xy)/((x^2-y^2)^2)) ?
The answer to (\partial)/(\partial x)((2xy)/((x^2-y^2)^2)) is (2y(-3x^2-y^2))/((x^2-y^2)^3)