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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

\int 147 cos^{2} (x) d x

\int (2 x - 5)^{10} d x

\int ln (x + 1) d x

\frac{( 12 + 4 x )}{x - 1}

n = 0 \sum \infty n (\frac{2}{7})^{n}

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