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Differentiate given

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

Initially I wanted to use the quotient rule to solve this, but then I tried differentiating it as it is:

$\frac {y_\frac{dy}{dx}}{1-y_\frac{dy}{dx}}=2x$

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

$\frac{dy}{dx}=\frac{2x}{yy^{-1}}$

$\frac{dy}{dx}=\frac{2xy}{y}$

$\frac{dy}{dx}=2x$

I am wondering how I can check to see if this is a valid answer?

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    My confusion is over the concept of$y$being treated as y(x), implicitly. Clearly, I don't see why y is treated this way. If I was to differentiate the same expression for x, then I would assume that x would be treated as x(y). That being said, I think that what was missing initially is that the relation between the two variables implies that one is a function of the other, depending on which one you are differentiating for. Additionally, my approach attempted to solve the rational side of the expression with out using the quotient rule. Thanks for your solution, it has cleared up my thinking2012-07-06

3 Answers 3

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Your equation is

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

You claim that

$y'=2x$ so that

$y=x^2+C$

This means

$\frac{x^2+C}{x-x^2-C}=x^2+1 $

This is absurd, since the quotient of two second degree polynomials can't be a second degree polynomial. In fact you get two non vanishing terms $x^3$ and $x^4$ which are off.

I don't understand what your procedure is, also. I would proceed as follows:

$\displaylines{ \frac{y}{{x - y}} = {x^2} + 1 \cr \frac{d}{{dx}}\left( {\frac{y}{{x - y}}} \right) = \frac{d}{{dx}}\left( {{x^2} + 1} \right) \cr \frac{{y'\left( {x - y} \right) - \left( {1 - y'} \right)y}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr \frac{{y'x - yy' - y + yy'}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr \frac{{y'x - y}}{{{{\left( {x - y} \right)}^2}}} = 2x \cr y'x = 2x{\left( {x - y} \right)^2} + y \cr y' = 2{\left( {x - y} \right)^2} + \frac{y}{x} \cr} $

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An explicit approach:

Rewrite as $y = (x-y)(x^2+1)$, and factor out $y$ to get $y = \frac{x^3+x}{x^2+2}$. This is straightforward to differentiate, yielding $\frac{d y}{d x} = \frac{x^4+5 x^2+2}{(x^2+2)^2}$.

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You can integrate your final expression to get $y=x^2+c$ and see if this works in the original equation.