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

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

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

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

- \frac{x}{x - y}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

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

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

Simplify

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

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

\int_{- 1}^{2} (2 x + 3) d x

x + 1

x \to - 6 - lim (\frac{- 1}{x + 6})

x - 1

\int e^{x^{3}} x^{2} d x

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