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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x}{3 + x ^{2}} d x

x y^{^{'}} - 2 y = x^{3}

y = x - 1, y = 0, x = 9

h \to 1 lim (\frac{e ^{- h} - 1}{h})

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