- #1

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## Homework Statement

[itex]\dfrac{x^2}{x+y}=y^2+8[/itex]

## Homework Equations

Quotient Rule: [itex]\dfrac{g(x)\cdot f'(x)-g'(x)\cdot f(x)}{(g(x))^2}[/itex]

Product Rule: [itex]f(x)\cdot g'(x)+g(x)\cdot f'(x)[/itex]

## The Attempt at a Solution

[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

I feel like there are a couple of ways to go about this. Would it be easier to flip the denominator and use the product rule? I just used the quotient from here on out.

What you see my trying to do is get [itex]\dfrac{dy}{dx}[/itex] on the left side and everything else on the right, then factor [itex]\dfrac{dy}{dx}[/itex] out. This is my first encounter with this type of problem so I get confused very fast.

[itex]\dfrac{(x+y\cdot\dfrac{dy}{dx})(2x)-(1\cdot\dfrac{dy}{dx})(x^2)}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

[itex]\dfrac{(2x^2+2xy\cdot\dfrac{dy}{dx})-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot \dfrac{dy}{dx})^2} = 2y\cdot\dfrac{dy}{dx}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy\cdot\dfrac{dy}{dx})}[/itex]

[itex]\dfrac{-1}{x^2}\cdot\dfrac{\dfrac{dy}{dx}-(x^2\cdot\dfrac{dy}{dx})}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{(2x^2+2xy)}\cdot\dfrac{-1}{x^2}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}[/itex]

[itex]\dfrac{1}{\dfrac{dy}{dx}}\cdot\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y\cdot\dfrac{dy}{dx}}{-x^2(2x^2+2xy)}\cdot\dfrac{1}{\dfrac{dy}{dx}}[/itex]

[itex]\dfrac{\dfrac{dy}{dx}-\dfrac{dy}{dx}}{\dfrac{dy}{dx}\cdot(x+y\cdot\dfrac{dy}{dx})^2}=\dfrac{2y}{-x^2(2x^2+2xy)}[/itex]

I don't want to go much farther because I could be doing this wrong. On the left side, I want to factor, but I'm curious if this is right so far or have I made any errors?