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

Question

Differentiate:

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

Preferring to avoid the quotient rule, I take away the fraction:

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

Then:

$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$

If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if I continue the following, I don't. Likely some place I erred, but I cannot figure out where:

Expansion:

$$2x=(x-y)^{-1}+y'(x-y)^{-1}-(x+y)(x-y)^{-2}+y'(x+y)(x-y)^{-2}$$

Preparing to isolate for $y'$:

$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'(x-y)^{-1}+y'(x+y)(x-y)^{-2}$$

$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'[(x-y)^{-1}+(x+y)(x-y)^{-2}]$$

Isolating $y'$:

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

Multiple top and bottom by $(x-y)$:

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

Then, inserting $x^2$ into $(x+y)(x-y)^{-1}$, I get:

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

While the answer states:

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

Which I do get if I multiplied the entire equation by $(x-y)^2$ before. It does not seem to be another form of the answer, as putting $x=2$, the denominator cannot match each other. Where have I gone wrong?

Answer 1

It's worth noting that you haven't actually avoided the quotient rule, at all. Rather, you've simply written out the quotient rule result in a different form. However, we can avoid the quotient rule as follows.

First, we clear the denominator to give us $$x^2(x-y)=x+y,$$ or equivalently, $$x^3-x^2y=x+y.$$ Gathering the $y$ terms on one side gives us $$x^3-x=x^2y+y,$$ or equivalently, $$x^3-x=(x^2+1)y.\tag{$\heartsuit$}$$ Noting that $x^2+1$ cannot be $0$ (assuming that $x$ is supposed to be real), we have $$\frac{x^3-x}{x^2+1}=y.\tag{$\star$}$$ Now, differentiating $(\heartsuit)$ with respect to $x$ gives us $$3x^2-1=2xy+(x^2+1)y',$$ or equivalently $$3x^2-1-2xy=(x^2+1)y'.$$ Using $(\star)$ then gives us $$3x^2-1-2x\cdot\frac{x^3-x}{x^2+1}=(x^2+1)y',$$ which we can readily solve for $y'.$

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As for what you did wrong, the answer is: concluding that different denominators meant different values!

Indeed, if $x=2,$ then solving $x^2=\frac{x+y}{x-y}$ for $y$ means that $y=\frac65.$

Substituting $x=2$ and $y=\frac65$ into $y'=\frac{2x(x-y)-1+x^2}{1+x^2}$ yields $$y'=\cfrac{\frac{31}5}5=\frac{31}{25},$$ while substituting $x=2$ and $y=\frac65$ into $y'=\frac{x(x-y)^2+y}{x}$ yields $$y'=\cfrac{\frac{62}{25}}2=\frac{31}{25}.$$ Hence, the answer is the same in the $x=2$ case! Now, more generally, using $(\star)$ in the equation $y'=\frac{2x(x-y)-1+x^2}{1+x^2}$ yields $$y'=\frac{x^4+4x^2-1}{(x^2+1)^2}.$$ The same result is achieved by using $(\star)$ in the equation $y'=\frac{x(x-y)^2+y}{x}.$ Hence, your answer is the same in both cases, though it doesn't look like it!

Answer 2

You are forgetting that there is a relation between $x$ and $y$. Your answer and the official "correct" answer are the same. One way to see this: make the substitution $x^2=\frac{x+y}{x-y}$ into your answer and:

$$\begin{align} \frac{2x(x-y)-1+x^2}{1+x^2}&=\frac{2x(x-y)-1+\frac{x+y}{x-y}}{1+\frac{x+y}{x-y}}\\ &=\frac{2x(x-y)^2-(x-y)+(x+y)}{(x-y)+(x+y)}\\ &=\frac{2x(x-y)^2+2y}{2x}\\ &=\frac{x(x-y)^2+y}{x}\\ \end{align}$$

Answer 3

You should first find $y$ in terms of $x$ as follows $$x^2=\frac{x+y}{x-y}$$ $$x^3-x^2y=x+y$$$$ (1+x^2)y=x^3-x$$ $$y=\frac{x^3-x}{1+x^2}$$ now, differentiate w.r.t. $x$, $$\frac{dy}{dx}=\frac{(1+x^2)(3x^2-1)-(x^3-x)(2x)}{(1+x^2)^2}$$ $$=\frac{x^4+4x^2-1}{(1+x^2)^2}$$