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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

x^{2} + 3 y \cdot \frac{d y}{d x} = 0

x \to - 1 lim (x^{3} - 3 x^{2} - 2 x + 1)

\int_{0}^{2} x + 2 - x^{2} d x

\frac{d}{d t} (\frac{t ^{3}}{4 - 3 t + 5})

x^{2} (1 - 2 x)

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