Talk:Implicit curve

Different methods?
In this edit, reverted me, saying that it is equivalent to compute the slope of the tangent line by implicit differentiation and by taking the partial derivatives of the defining function F.  In my opinion, this is not true, for two reasons: (1) there is a large group of people who understand the description "use implicit differentiation" but not the words "partial derivative", e.g., every student in a standard single-variable calculus sequence; and (2) the two actually describe different algorithms. For example, consider the implicit equation $$x^2 + xy + y^2 - 1 = 0$$. If I compute the slope by implicit differentiation, I get the following sequence of calculations: $$ 2x + (y + xy') + 2yy' = 0 $$ $$ (x + 2y)y' = -2x - y $$ $$ y' = \frac{- 2x - y}{x + 2y}$$ On the other hand, if I compute it using partial derivatives I compute $$F_x = 2x + y$$ $$F_y = x + 2y$$ $$ y' = -\frac{F_x}{F_y} = -\frac{2x + y}{x + 2y}$$ It is true that the results of these two algorithms are the same (because they are both valid computations of the same quantity), and we can systematically relate the steps of the two algorithms to each other, but they are unquestionably not the same algorithm, and it's a theorem of multivariable calculus (not just a triviality of symbol-pushing) that they give the same answer. (How much do I really care about this? As long as the words implicit differentiation don't disappear, not too much.) --JBL (talk) 22:18, 25 October 2017 (UTC)


 * Hi, Joel. Sorry I didn’t respond here sooner. I don’t keep a watchlist – I just monitor the pages of my last 150 edits on my contributions page (so this article appears there, but not this talk page). You can ping me if you need to get my attention.
 * I’m happy with the way the paragraph is now. Loraof (talk) 14:40, 31 October 2017 (UTC)
 * , thanks for your response and no worries about the delay -- I tried to ping but I think I messed up the template :/, good to know about your habits for the future. --JBL (talk) 16:13, 31 October 2017 (UTC)


 * Here’s why I viewed the two techniques as identical. $$-F_x/F_y$$ comes from the following passage in Implicit function, which is the way I’ve always thought of it:


 * To differentiate an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y and then differentiate. Instead, one can totally differentiate  R(x, y) = 0 with respect to x and y and then solve the resulting linear equation for dy /dx to explicitly get the derivative in terms of x and y.


 * Loraof (talk) 18:49, 31 October 2017 (UTC)


 * I guess my point is that the words "totally differentiate" in that quote are not necessary to understand implicit differentiation: if y is a function of x and R is a function of two variables then one can differentiate R(x, y) using the chain rule for functions of one variable -- it is not necessary to know anything about derivatives of multivariate functions. And if R and y have the property that R(x, y) = 0 for every x, then this will give a procedure to find dy/dx.  And my calculus 1 students can do this, even though they don't understand what partial derivatives are.  In particular, for any particular R they can find dy/dx even though they are unable to write a general formula for the coefficient of dy/dx after they differentiate (because that general formula involves partial derivatives, which they don't know anything about).  --JBL (talk) 19:37, 31 October 2017 (UTC)


 * But when you say


 * For example, consider the implicit equation $$x^2 + xy + y^2 - 1 = 0$$. If I compute the slope by i mplicit differentiation, I get the following sequence of calculations:
 * $$ 2x + (y + xy') + 2yy' = 0 $$
 * aren’t you taking partial derivatives with respect to [of] the middle term? Loraof (talk) 19:54, 31 October 2017 (UTC)


 * I'm not entirely sure what the words "taking partial derivatives with respect to the middle term" mean, but (taking for granted that y is a function of a single variable x) the expression $$x^2 + xy + y^2 - 1$$ is a function of a single variable x and its derivative can be computed by applications of the product rule and (single-variable) chain rule -- this does not require any notion of partial derivative. --JBL (talk) 21:39, 31 October 2017 (UTC)


 * Sorry about the sloppy wording – I didn’t mean w.r.t. in a technical sense, just “in regard to”. I’ve struck it above. Anyway, I still think that using the product rule to write $$(y + xy')$$ is just an application of partial derivative-taking. Your expression $$(y+xy^\prime)$$ is nothing but $$\tfrac{\partial xy}{\partial x}+ \tfrac{\partial xy}{\partial y}\tfrac{dy}{dx}.$$ But we can agree to disagree on that since I’m accepting your wording. Loraof (talk) 21:58, 31 October 2017 (UTC)


 * No apologies necessary, thanks for the clarification. It is certainly true that the derivative of xy(x) can be computed using the tools and language of multivariable calculus, but every calculus 1 student also knows how to compute this derivative despite not knowing the definition of the words "partial derivative." --JBL (talk) 16:26, 4 November 2017 (UTC)