Implicit differentiation

  • #1
Hey, I found a thread about part of what I'm trying to ask long ago: https://www.physicsforums.com/threads/implicit-differentiation.178328/

Basically, I noticed that if you multiply by x or by y in an equation before implicitly deriving, you get two different answers. Unfortunately their whole discussion about level sets, I'd have to sit down for hours to try and understand since my multivariable calc is not the best, neither is my normal calc but when they started talking about level sets I realized I'd have to sit there really long to understand that thoroughly. Maybe I will if I can't get the response I'm looking for here, but I'm thinking maybe someone can enlighten me quicker.

So I was wondering if anyone can help me with this, why do the derivatives produce different results, specifically, for example, for that problem in the thread? Even further, what about the popular logarithmic differentiation? If you had for example x^3y+y^2=8y, and you do logarithmic differentiation vs. just differentiating it, there is also a different answer for y', and plugging in x,y coordinates give different results for the different y'.

Thanks for any replies!
 

Answers and Replies

  • #2
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If you had for example x^3y+y^2=8y, and you do logarithmic differentiation vs. just differentiating it, there is also a different answer for y', and plugging in x,y coordinates give different results for the different y'.
If you differentiate implicitly, you'll get y' in terms of both x and y. If you divide both sides of the equation in your example by y, you get y = 8 - x3, so y' = -3x2. The point (1, 7) is on the graph of the original equation. At this point I get y' = -3 using either version of the derivative that I calculated.

Can you show what you did in logarithmic differentiation?
 
  • #3
If you differentiate implicitly, you'll get y' in terms of both x and y. If you divide both sides of the equation in your example by y, you get y = 8 - x3, so y' = -3x2. The point (1, 7) is on the graph of the original equation. At this point I get y' = -3 using either version of the derivative that I calculated.

Can you show what you did in logarithmic differentiation?

I have to think about this

OK, your answer does not depend on y at all. Deriving the original equation x^3+y^2=8y, you get y'=(3x^2)/(8-2y-x^3). Try (1,2). On yours it is still y'=-3. On this one it will be y' =2.

Logarithmic differentiation, too, provides y'=-3x^2, which is like your dividing by y. But this is a coincidence, logarithmic differentiation provides a different solution than y'=(3x^2)/(8-2y-x^3)

So my question remains, I understand if there is only a way to explain this by multivariable calculus, and I will try to look into it and understand it. But I want some sort of a lead other than that thread I can go off if you or anyone can help me with that
 
Last edited:
  • #4
35,135
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I have to think about this

OK, your answer does not depend on y at all. Deriving the original equation x^3+y^2=8y, you get y'=(3x^2)/(8-2y-x^3).
The answer I showed doesn't depend on y, but when I differentiated implicitly, y' was in terms of both x and y.

Your answer of ##y' = \frac{3x^2}{8 - 2y - x^3}## is incorrect -- you are missing a factor of y in the numerator.
ilikesoldat said:
Try (1,2). On yours it is still y'=-3. On this one it will be y' =2.
(1, 2) is not a point on the graph of the original relation, ##x^3y + y^2 = 8y##.
(1, 7) is a point on this graph, however. Both forms I calculated for y' gave the same value at this point.
ilikesoldat said:
Logarithmic differentiation, too, provides y'=-3x^2, which is like your dividing by y.
Please show me how you did this by log differentiation. I don't think this problem is amenable to log differentiation because of the terms being added.
ilikesoldat said:
But this is a coincidence, logarithmic differentiation provides a different solution than y'=(3x^2)/(8-2y-x^3)

So my question remains, I understand if there is only a way to explain this by multivariable calculus, and I will try to look into it and understand it. But I want some sort of a lead other than that thread I can go off if you or anyone can help me with that
 
  • #5
Stephen Tashi
Science Advisor
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So my question remains, I understand if there is only a way to explain this by multivariable calculus, and I will try to look into it and understand it. But I want some sort of a lead other than that thread I can go off if you or anyone can help me with that

Discussing the question in generality will get confused with the question of whether you have given a specific example where manipulation of an equation gives two different answers for the derivative of ##y##.

The important generality is that "doing the same thing to both sides of an equation" can change the solutions to an equation - i.e. it can produce a new equation who solutions differ from the original equation. So "doing the same thing to both sides" of an equation that defines a function ##y(x)## implicitly can change the function the equation defines. Hence it can change both ##y(x)## and ##y'(x)## since these functions are defined by the ordered pairs (x,y) that satisfy the equation.

Elementary algebra drills us in the technique of "do the same thing to both sides of the equation". This lulls students into thinking that it is a completely safe and reliable procedure. There are many examples where the technique doesn't change the solutions of equations, but there are exceptions where it does.


For example, adding ##\frac{1}{y-2x}## to both sides of the equation ##y - \frac{1}{y-2x} = 2x - \frac{1}{y-2x} ## produces an equation whose solutions are the ##(x,y)## with ##y = 2x##, but these ordered pairs are not solutions to the original equation.

The relevance of the terminology "level sets" to this phenomena is that it assumes we will write our equations so the right hand size of the equation is zero. For example, ##3xy + x = y \sin(x)## in standard form would be written as ##3xy + x - y \sin(x) = 0##. We regard the left hand side of that equation as a function ##F(x,y)## of two variables.

The values ##(x,y)## that make ##F(x,y)## equal to some constant value are called a "level set" of ##F(x,y)##. The particular values ##(x,y)## that make ##F(x,y) = 0## are thus called the "level set" of ##F(x,y)## that is associated with the value ##0##. (The values ##(x,y)## that make ##F(x,y) = 17 ## are another "level set", but they aren't relevant to how ##F(x,y) = 0## defines ##y## implicitly as a function of ##x##.)

To rephrase the question "Does multiplying both sides of the equation by ##y## change the function ##y##? " in the language of level sets, we say "Does ##F(x,y)## have the same level set for the value 0 as ##yF(x,y)## does?".
 

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