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

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

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(\partial)/(\partial y)(x/(z-2y))

\frac{\partial}{\partial y} (\frac{x}{z - 2 y})

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

Solution steps

\frac{\partial}{\partial y} (\frac{x}{z - 2 y})

Treat

x, z

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (z - 2 y) = - 2

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

Simplify

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

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

\frac{\partial}{\partial x} (e^{8 x} cos (- 9 y))

e^{- 6 x}

\frac{d}{d x} (2 sin (x) cos (x))

\int \frac{1}{2 + 2 x - x ^{2}} d x

\frac{1}{x + 1}

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