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Implicit Differentiation

  1. Oct 5, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi everyone, name is Ryan. This is my first post here, seems like a very beneficial forum. I look forward to being helped and helping others. Anyway I'm trying to teach myself Implicit Differentiation but there doesn't seem like much useful resources online and I don't quite understand it.

    2. Relevant equations
    I'm trying to produce a degree 3 polynomial for (dy/dx) = 1/(3y^2); y(0)=1 using implicit differentiation.

    3. The attempt at a solution
    Alright I started off by trying to find (d^2)y/d(x^2) and this is how I did it:

    (d^2)y/d(x^2) = -6/[y(0)^3] x [(dy/dx)](0)
    = -6x1x(1/3) = -2

    Now to get (d^3)y/d(x^3) I get a bit confused, as well as just getting y.
    I tried getting (d^3)y/d(x^3) as follows:
    (d^3)y/d(x^3) = 18/[y^4] x [(d^2)y/d(x^2)](0)
    = -18 x1x-2 = 36 (I suspect this is wrong)
    And I have no idea how to get y. If someone has any idea how to di is or just some tips that would be much appreciated. Thanks guys.
  2. jcsd
  3. Oct 5, 2007 #2


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    welcome to world of PF

    firstly, [tex]\frac{d}{dx}\left(\frac{1}{3y^2}\right)=-\frac{2}{3} y^{-3} \frac{dy}{dx}[/tex] and not what you have written down

    other than that I am not entirely sure what is your ultimate goal. Are you trying to solve for y at all?
  4. Oct 5, 2007 #3


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    I don't understand what you are doing. You said "I'm trying to produce a degree 3 polynomial for (dy/dx) = 1/(3y^2); y(0)=1". You do that by integrating not by differentiating further!
  5. Oct 5, 2007 #4
    Hey all, this is also my first post. I'm a student, but I like to solve the problems people post on here. So please correct me if I say anything wrong or do anything embarrassingly stupid, etc.

    Ok first of all if you're solving for y, take the integral, not the derivative. This is an integral problem, not a problem in implicit differentiation.

    dy/dx = 1 / 3y^2


    3y^2 dy = dx

    and take the integral of both sides. I'm going to use the "]" symbol to represent an integral symbol.

    3 ] y^2 dy = ] dx


    3 ( y^3 / 3 ) = x + c where c is a constant

    you can cancel and rearrange from here. Hope this helps!

  6. Oct 5, 2007 #5
    Yeah sorry guys maybe i should state the whole question. Here it is:
    Use implicit differentiation to produce a degree 3 polynomial approximating the solution of the Initial Value Problem (IVP):
    (dy/dx) = 1/(3y^2); y(0)=1.

    I know to get the degree 3 polynomial approx. I will have to use the formula Pn(x) = f(a) + f'(a)(x-a) + [f''(x)/2!](x-a)^2 + [f'''(a)/3!](x-a)^3

    The real issue for me is finding f(a), f''(a) and f'''(a)
  7. Oct 5, 2007 #6
    Oh, ok.

    Well you can find f(a) using what I just gave you, and then use the initial condition given in the problem, namely, that y(0)=1. Plug this in, find the constant, and there is your function. It just so happens that you can get y on one side of the equation, so you don't have to use implicit differentiation.

    y = x^(1/3) + c

    Plug y=1 and x=0 to find c. Differentiate three times, and plug into the third-degree Taylor series expansion you have.
  8. Oct 6, 2007 #7
    Thanks AmaniKaleo but that's the method I will be using to verify my answer. As the question unfortunately states I need to use implicit differentiation and that's where my worries lie.
  9. Oct 6, 2007 #8


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    Okay, I mis-interpreted your first post. Sounds to me like you are constructing a Taylor's polynomial of degree 3 for y. In order to do that, you need to know y(0) (and you are given y(0)= 1), dy/dx at x= 0 (which you can find immediately from the given equation), [itex]d^2y/dx^2[/itex] at x= 0, and [itex]d^3y/dx^3[/itex] at x= 0.

    From [itex]dy/dx= 1/(3y^2)= (1/3)y^{-3}[/itex], you know that [itex]d^2y/dx^2= -1 y^{-4} dy/dx[/itex]. It is the "implicit differentiation" that puts that "dy/dx" in the equation. Of course, you know both y and dy/dx at x= 0 so you can easily find [itex]d^2y/dx^2[/itex] at x= 0. Differentiate [itex]d^2y/dx^2= -1 y^{-4} dy/dx[/itex] to find [itex]d^3y/dx^3[/itex]. It will involve y, dy/dx, and [itex]d^2y/dx^2[/itex], all of which you now know, at x= 0.
  10. Oct 6, 2007 #9
    But as dy/dx=1/(3y^2) doesn't that equal (1/3)y^(-2). Now finding (dy^2)/(dx^2) is confusing. I'm sorry if I seem slow but I just have never done implicit differentiation and I'm about unsure what to do.
  11. Oct 6, 2007 #10
    Ok, here's how to do it. You don't actually need to use implicit differentiation to find [tex]y'[/tex] and [tex]y''[/tex], but the problem wants you to practice so here's what to do.

    When you solve for [tex]y[/tex], you find that


    Now let's differentiate both sides.

    You know that there is a general rule, that the derivative of a function [tex]u[/tex] to the [tex]n[/tex]th power [tex]u^{n}[/tex] with respect to some variable [tex]t[/tex] is

    [tex] du/dt = n u^{n-1} du [/tex]

    Now, using that rule, let's differentiate the function in the problem. Differentiating the right side of the equation, [tex]x[/tex], we get

    [tex]1 x^{0} 1 = 1 [/tex]

    Differentiating the left side, we get

    [tex] 3 y^{2} dy/dx [/tex]

    since (especially if the function was longer and more complex and we REALLY wouldn't be able to solve for [tex]y[/tex]) we don't know what [tex]y[/tex] is and so we don't know what [tex]y'[/tex] is. Now the equation looks like,

    [tex]3 y^{2} dy/dx = 1[/tex]

    Now all you have to do is solve for [tex]dy/dx[/tex]. Rearranging,

    [tex]dy/dx = 1/3y^{2}[/tex]

    but [tex]y = x^{1/3}[/tex], from above. Therefore, plugging in,

    [tex]dy/dx = 1/3 x^{-2/3} [/tex]

    Repeat this entire process, but this time solve for [tex]y''[/tex]. Don't forget to substitute at the end :)
    Last edited: Oct 6, 2007
  12. Oct 6, 2007 #11
    Thanks for that, I understand what you did but a stupid question, is that implicit differentiation? And I'm still unsure about finding y'' and y'''. Thanks again.
  13. Oct 6, 2007 #12
    Yes, that is implicit differentiation. Let's say you had something more complicated, liiiike, idk, how about

    [tex] 3y^{3} + 5y^{2} + y = x^{5} + xy^{3}[/tex].

    If you wanted to solve for [tex]y'[/tex] or further derivatives of [tex]y[/tex], you would HAVE to use implicit differentiation.

    9y^{2}y' + 10yy' + y' = 5x^{4} + y^{3} + 3xy^{2}y'

    And then solve for [tex]y'[/tex]. Here I use [tex]y'[/tex] instead of [tex]dy/dx[/tex] to make the equation more legible.
  14. Oct 6, 2007 #13
    Yes I understand that completely but for that and my equation I wouldn't know how to use implicit differentiation to find y'' and y''. Sorry if I seem stupid.
  15. Oct 6, 2007 #14
    No problem, you're not stupid. We all go through this confusion. :)

    To find [tex]y''[/tex] and [tex]y'''[/tex] just differentiate further. In your problem, we found that

    [tex]dy/dx = (1/3)x^{-2/3}[/tex]

    And so just differentiate again.

    d^{2}y/dx^{2} = (-2/3)(1/3) x^{-5/3}[/tex]

    [tex]d^{2}y/dx^{2} = (-2/9)x^{-5/3}[/tex]

    Differentiating further,

    [tex]d^{3}y/dx^{3} = (10/27)x^{-8/3}[/tex]

    waaa! check to make sure you have the updated edited version of this, I'm always going back and fixing typos
    Last edited: Oct 6, 2007
  16. Oct 6, 2007 #15
    Oh so from there it is just simple differentiation, I finally see. I think you mean (d^2x)/(dy^2) = (-2/9)x^(-5/3). Thank you very much AmaniKaleo and everyone else, I really appreciate the help and patience.
  17. Oct 6, 2007 #16
    oh lol that's right.

    *fixes typo in a hurry*

    I hope that helps, if you have any other questions, just post them. :)
  18. Oct 6, 2007 #17


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    Yes, I made a typo and then cofused myself!

    First, what AmaniKaleo is doing in hjis last couple of posts is NOT "implicit differentiation" because he does not have y "implictely defined" as a function- he has y "explicitely" given: y= x1/3 because he directly solved the differential equation.

    What this problem asks is that you use implicit differentiation in order to find the derivatives of y and use them in a Taylor polynomial.
    The third order Taylor polynomial, around x=0, of a function y, is [itex]y(0)+ y'(0)x+ (1/2)y"(0)x2+ (1/6)y"'(0)x^3[/itex].
    You are told that y(0)= 1 so that's easy.
    You are told that [itex]dy/dx= y'= 1/(3y^2)[/itex] so y'(0)= 1/(3y(0)^2)= 1/3.

    Now, from [itex]y'= (1/3)y^{-2}[/itex] (correcting my error) and differentiating both sides of the equation with respect to x, you have [itex]y"= (-2/3)y^{-3}y'[/itex]. You know that y(0)= 1 and y'(0)= 1/3 so y"(0)= (-2/3)(1)(1/3)= -2/9.

    Differentiate both sides of [itex]y"= (-2/3)y^{-3}y'[/itex] to get [itex]y"'= ((2)y^{-4}y')- (2/3)y^{-3}y"y'[/itex] (Do you see that y' in each term- that's the chain rule in action). Now put in the values you have for y(0), y'(0), and y"(0) to find y"' (0) and use those values in the Taylor's series.

    As you have seen (I pointed it out in my first post, before I understood exactly what this question was), this equation was easily integrable. The reason they want you to do it by differentiating (and then writing a Taylor polynomial) is that it applies to much more complicated equations that CAN'T be integrated. This is really a very powerful method for producing approximate solutions to equations like that.
  19. Oct 6, 2007 #18
    That's right, I posted both explicitly (solving for y) and implicitly (pretending we can't solve for y and differentiating implicitly), but I guess I failed to be clear as to which was implicit and which was explicit.
    Or perhaps that wasn't your point, Ivy? Is there something I'm completely missing?
  20. Oct 6, 2007 #19
    I kind of see where you are getting at HallsofIvy, except for y'''(0) and those Latex graphics are new to me and a bit confusing.
  21. Oct 6, 2007 #20


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    No, that was my whole point. I thought it was a little bit confusing (I confuse easily!) about what was or was not an implicit derivative. The crucial point was that this exercise is expecting the student to use the derivatives to construct a Taylor's polynomial.
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