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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

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

(1 + x^{2})^{2}

\int (7 x^{5} + 6 x^{4} - 3 x^{3} - 9 x + 5) d x

\int \frac{x ^{2} - 64}{x ( x - 1 ) ^{2}} d x

\int \frac{1}{y 1 + ( ln ( y ) ) ^{2}} d y

y = \frac{( x ^{2} + 1 )}{( x ^{3} - 1 )}

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