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Chain rule

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    Snap2.jpg
    Isn't the derivative of y with respect to x Defined as ##~\frac{dy}{dx}##?
    What and how do i have to prove?

    2. Relevant equations
    The chain rule:
    $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

    3. The attempt at a solution
    $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$
     
  2. jcsd
  3. Aug 8, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    What a horrible question. dy and dx are not numbers, and dx/dy is not a ratio.

    What you can probably do is studying d y(x(t))/dt and then abusing the notation sufficiently until you get dy/dx, showing that t does not matter.
     
  4. Aug 8, 2017 #3
    What is y[x(t)]? isn't it just y(t)?
    For example, the example given in the book:
    $$x=t+\frac{1}{t},~~y=t-\frac{1}{t}~\rightarrow~x^2-y^2=4$$
    isn't ##~y[x(t)]=t-\frac{1}{t}~##?
     
  5. Aug 8, 2017 #4

    fresh_42

    Staff: Mentor

    No. What is ##y(x)## if ##x## was just a real number?
     
  6. Aug 8, 2017 #5

    Ray Vickson

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    In Nonstandard Analysis, ##dy## and ##dx## are actual infinitesimals and ##dy/dx## really is a ratio; the "standard part" of the ratio equals the ordinary derivative.

    However, I have serious doubts that the OP's course is using full-fledged non-standard analysis, but is instead likely using a form of heuristic argument that has been around since the days of Leibnitz.
     
  7. Aug 9, 2017 #6
    $$y=\sqrt{(x-2)(x+2)}$$
     
  8. Aug 9, 2017 #7

    fresh_42

    Staff: Mentor

    This is not how ##y## is defined.
     
  9. Aug 9, 2017 #8
    If x were a real number then ##~y=x-\frac{1}{x}##
    But what is y[x(t)]?
    $$y[x(t)]=y\left[ t+\frac{1}{t} \right]=\left(t+\frac{1}{t} \right)-\frac{1}{\left(t+\frac{1}{t} \right)}\neq t-\frac{1}{t}$$
    I don't understand
     
    Last edited: Aug 9, 2017
  10. Aug 9, 2017 #9

    fresh_42

    Staff: Mentor

    You have ##y(sth.)= sth. \,- \frac{1}{sth.}##. Next this something is ##x(t) = sth. = t + \frac{1}{t}##. What do we get by substitution of that something?
     
  11. Aug 9, 2017 #10
    We get the value of y, but y isn't a function of x, each (x and y) are functions of t. we just join the points (x,y) produced by t.
     
  12. Aug 9, 2017 #11

    fresh_42

    Staff: Mentor

    You have ##y(t)=t-\frac{1}{t}##. This means, if ##t## is replaced by ##x##, then we get ##y(x)=x-\frac{1}{x}##. And if then the function ##x=x(t)= t + \frac{1}{t}## is parameterized by ##t##, we get
    $$
    (y \circ x)(t)=y(x(t))=x(t) -\frac{1}{x(t)} = (t+\frac{1}{t})- \frac{1}{t+\frac{1}{t}}
    $$
    Now you can calculate either ##\frac{d}{dt}(y\circ x)(t)## as function of ##t## or use the chain rule and calculate
    $$
    \frac{d}{dt}(y\circ x)(t)= \frac{d}{dx}y(x) \cdot \frac{d}{dt}x(t) = \frac{d}{dx} (x-\frac{1}{x}) \cdot \frac{d}{dt} (t+\frac{1}{t})
    $$
    and then substitute ##x=t+\frac{1}{t}## and hope both quotients are equal.
     
  13. Aug 10, 2017 #12
    It is very tiring to prove that both these equations are equal.
    And even so, what did i prove? that ##~\frac{d}{dx}y(x) \cdot \frac{d}{dt}x(t) = \frac{d}{dt}y[x(t)]##
    Is it what the question asked? in this example:
    $$y=\sqrt{(x-2)(x+2)}~\rightarrow~y'(x)=...$$
    And i have to prove it equals dy/dx
    I leave this question since i don't want to tire you and me, i think i won't get further.
    Thank you fresh_42
     
  14. Aug 10, 2017 #13

    Ray Vickson

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    No: there are two solutions
    $$y(x) = \pm \sqrt{(x-2)(x+2)}$$
    This really is true: if you plot the points ##(x(t),y(t))## for ##t> 0## you will find they lie on the curve ##y = -\sqrt{(x-2)(x+2)}## for ##0 < t \leq 1## and on the curve ##y = + \sqrt{(x-2)(x+2)}## for ##1 \leq t < \infty##. This is one branch of the hyperbola ##x^2 - y^2 = 4##. For ##t < 0## you get the opposite branch of the hyperbola, opening out to the left of ##x = -2##.
     
  15. Aug 10, 2017 #14
    Thank you Ray, you all are specialists, they drew the plot in the book and it's as you've said
     
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