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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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What is $y_{dy/dx}$? –  Pedro Tamaroff Jul 6 '12 at 5:44
    
@PeterTamaroff: Quite possibly an error on my part. I arrived at $y_\frac{dy}{dx}$ by applying the chain rule to y. If I understand correctly, y is considered to be a function of x, so we apply the chain rule to y (when using implicit differentiation). –  Kurt Jul 6 '12 at 5:51
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Explain to me what you mean by "$y_{dx/dy}$". In general, what do you mean by $f_g$, when $f$ and $g$ are functions? –  Pedro Tamaroff Jul 6 '12 at 5:57
    
I meant to express that y is a function of x. My aim was to differentiate the top of bottom of the rational expression. Since the expression is a relation versus a function, I thought I had to differentiate y using the chain rule. –  Kurt Jul 6 '12 at 6:18
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Careful. The top is $y(x)$. The chain rule is used for compositions, and I see none (Do you?). What you ought to be doing is using the quotient rule and treating $y$ as $y(x)$ implicitly. I still can't understand why would you say a function is a functions of its derivative (when the converse might make more sense) or how you arrived to the expression $y'(y y^{-1})$. If you write out your reasoning it might help. –  Pedro Tamaroff Jul 6 '12 at 6:24

3 Answers 3

up vote 2 down vote accepted

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

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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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