Now either y' - 1 = 0 or y - x = 0.

I can see the algebra that leads to this, it is not intuitive. Is it conventional to differentiate and the set equal to zero?

But this is how you solve equations. It's a good practice to try the factoring. Also, what is not intuitive in passing from ab = b to b(a - 1) = 0? Why is it more intuitive to divide by b? In algebra one should be weary of division because the divider may accidentally be 0.

So, if you are bent on dividing by (y - x) you should say, "Assume y - x ≠ 0, then ... The case of when y - x = 0 will be considered separately."

Then, again, there is no problem. If y - x = 0 you get a line. If y - x ≠ 0, you get y' = 1. It is here that you may start wondering.

>Is it conventional to differentiate and the set
>equal to zero?

Who did?

What if a student first rewrote your equation as (y - x)² = 0?

Differention then gives 2(y - x)(y' - 1) = 0, with exactly same result.

thanks...

I think there remains a valid point after the following discussion:

>>I can see the algebra that leads to this, it is not
>>intuitive.
>
>But this is how you solve equations. It's a good practice to
>try the factoring. Also, what is not intuitive in passing
>from ab = b to b(a - 1) = 0? Why is it more intuitive to
>divide by b? In algebra one should be weary of division
>because the divider may accidentally be 0.
>
>So, if you are bent on dividing by (y - x) you should say,
>"Assume y - x ≠ 0, then ... The case of when y - x = 0
>will be considered separately."
>
>Then, again, there is no problem. If y - x = 0 you get a
>line. If y - x ≠ 0, you get y' = 1. It is here that you
>may start wondering.
>
>>Is it conventional to differentiate and the set
>>equal to zero?
>
>Who did?
>
>What if a student first rewrote your equation as (y -
>x)² = 0?
>
>Differention then gives 2(y - x)(y' - 1) = 0, with exactly
>same result.

Although Alex's procedure is probably better, in the sense of making good use of the usual tools for either direct or implicit differentiation, and for the ensuing algebraic solution of the equation, both Alex's and John's methods leave the central difficulty unresolved: The structure of the solution produces a formula for the derivative only in the case where y <> x, but the curve which is the solution to the original equation is precisely y=x. Hence in John's formula, y' = (y-x)/(y-x), we have a formula for the derivative which is valid everywhere except on the actual graph; in Alex'x solution, we can only conclude that y' - 1 = 0 if we assume that y - x ≠ 0. This is a slightly more subtle version of the same difficulty: the conlusion that y' = 1 follows except in the case where we are considering a point on the actual solution curve of the original equation.

Obviously there are many ways around this. For one thing, one could insert an additive constant on the RHS of the original equation, take the derivative as usual, and then use L'Hopital's rule to evaluate the derivative in the limit as the constant vanishes. Alternatively, and much better, one could revisit the form in which the original question was posed, and recognize that y^2 + x^2 = 2xy reduces to (y-x)^2 = 0, which is a redundant description of its graph of solutions. Then clearly the question should be reduced to y-x=0 before differentiation is attempted -- after which everything is transparently easy.

So I think there is a useful lession here, which is perhaps the point of this problem being assigned in an online course: the initial formulation is sometimes a crucial thing to investigate, in order to avoid puzzling behavior such as both John's and Alex's solutions exhibit. I think this is really a very good teaching problem, provided some discussion such as the above is provided to underline what is to be learned from it (all too often, such discussion is missing from mathematics courses, even when good problems are assigned).

--Stuart Anderson

1. When you split ab = 0 into a = 0 or b = 0, it's not to the exclusion of one from another. The two may coexist and, as in our case, lead to the same solution.

2. Just to exonerate myself, at some point I drop the sentence "this is where you may start wondering." Just did not pursue it further. The intention is of course to the fact that assuming y &ne ;x one gets to consider y' = 1.

Other than that I am in absolute agreement with you. Quite often the rule "calculate then simplify" should be replaced by "simplify then calculate".

Thank you gentleman for your commentary and clarification on the issue. I posted a similar question on the AP Calculus EDG if you are interested in reading on the forum there.

Regards,