What is (\partial)/(\partial y)(1/(x-y)) ?

The answer to (\partial)/(\partial y)(1/(x-y)) is 1/((x-y)^2)

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(\partial)/(\partial y)(1/(x-y))

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

2

f (x) = \frac{9}{x ^{2} + 5}

cos^{2} (x), 0

sin (x) + \frac{4}{9} cot (x)

\int_{1}^{t} \frac{17}{x ^{3}} d x

What is (\partial)/(\partial y)(1/(x-y)) ?
The answer to (\partial)/(\partial y)(1/(x-y)) is 1/((x-y)^2)