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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{1}^{\infty} x e^{- 7 x} d x

y^{^{'}} + 4 y = 6 e^{- 3 t}

y \to - 3 lim (\frac{y ^{2} - 9}{2 y ^{2} + 7 y + 3})

16 y^{^{''}} - y = 0, y (- 7) = 1, y^{^{'}} (- 7) = - 1

\frac{d}{d x} (x^{\frac{5}{3}} - 5 x^{\frac{2}{3}})

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