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

The answer to (\partial)/(\partial y)(ln(x-y)) is -1/(x-y)

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

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

- \frac{1}{x - y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= - \frac{1}{x - y}

\frac{d}{d x} (x (e^{x} - 1))

\frac{d}{d x} (- ln (x^{4}))

\int 1 + cos (π θ) d θ

\frac{x}{2}

\frac{2 s - 3}{2 s ^{2} - 8 s + 16}

What is (\partial)/(\partial y)(ln(x-y)) ?
The answer to (\partial)/(\partial y)(ln(x-y)) is -1/(x-y)