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

The answer to (\partial)/(\partial x)(-1/((x-y)^2)) is 2/((x-y)^3)

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(\partial)/(\partial x)(-1/((x-y)^2))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{2}{( x - y ) ^{3}} \frac{\partial}{\partial x} (x - y)

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

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

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

Simplify

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

f (y) = cos (y)

\int_{0}^{5} (\frac{x}{x + 4}) d x

y^{^{'}} + y + 1 = 0

\frac{d}{d x} ((1 + x) cos (x))

f (t) = cos^{2} (e^{c o s^{2} (t)})

What is (\partial)/(\partial x)(-1/((x-y)^2)) ?
The answer to (\partial)/(\partial x)(-1/((x-y)^2)) is 2/((x-y)^3)