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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

= \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

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

Simplify

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

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

\int_{0}^{1} \frac{5 x}{( 4 + x ^{2} ) ^{2}} d x

\int_{1}^{\infty} x^{- 1} d x

f (x) = \frac{1}{x - 2}, (1, - 1)

\int sin (7 x) sin (4 x) d x

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