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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

5 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- \frac{1}{4}})

x \to 0 lim (\frac{x - 2}{x})

\frac{\partial}{\partial y} (- 8 y^{3} + 4 x y + 4 x^{2} + 9 y^{2})

\int_{1}^{n} \frac{1}{x ^{1.1}} d x

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