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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

cos (x sin (x))

\frac{d}{d x} (\frac{- 6 x}{1 - x})

\frac{s + 4}{s ^{2} + 6 s + 18}

\int a x^{5} d x

y = x^{2}, y = 8 - x^{2}, y = 4 x + 11

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