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Implicit differentiation (beginner)

  1. Oct 5, 2016 #1
    The problem statement, all variables and given/known data

    Find y' ....
    X^2+y^2=25


    I understand (I think) implicit differentiation, but there is one issue which hangs me up. I've done this before and this is just a refresher as my last calculus course was four years ago.
    From what I understand,

    2x+2y(y')=0

    But why isn't it 2x+2y(y^2)=0. ?
    Sorry for not using the notation format but I'm in a bit of a rush.
     
    Last edited: Oct 5, 2016
  2. jcsd
  3. Oct 5, 2016 #2

    Ssnow

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    Hi, because the derivative of ##x^2## is ##2x##. By the derivative of composition ##(f(g(x)))'=f'(g(x))g'(x)## you obtain the first result (##2x+2y(y')=0##) where the second addendum is a composition with ##f(x)=x^{2}## and ##g(x)=y(x)## and you must use the rule before ...

    Ssnow
     
  4. Oct 5, 2016 #3
    I'm sorry, but that didn't clarify anything for me :/
     
  5. Oct 5, 2016 #4

    ehild

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    y is a function of x, so y2 is a composite function, like sin(x2), for example. How would you differentiate it?
     
  6. Oct 5, 2016 #5

    Ssnow

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    In your equation after the first addendum you must derive ## y^2## that is ##(y(x))^2## so a composition of functions ... Using the rule before ##(f(g(x)))'=((y(x))^{2})'=f'(g(x))g'(x)=2(y(x))\cdot (y'(x))=2yy'## ...
     
  7. Oct 5, 2016 #6
    Erm... well here's my thought process:


    Here's the original equation that I want to apply differentiation to:
    x2+y2=25

    I differentiate x2:

    2x+y2=25

    I differentiate 25.

    2x+y2=0

    I now want to differentiate y2
    I suppose this is where I run into trouble as in most previous cases, I would isolate "y" and differentiate from that point. However, implicit differentiation seems to be "x" in terms of "y," and as a helper mentioned above, this is a composition of functions. This is where I lose the thread.
    I want to say:

    2x+2y(y2)=0

    As this is the chain rule as I understand it. I applied the derivative of the term y2 and the result was 2y. I multiply this by the original expression (y^2). The product of 2y and y^2 would be the derivative of y^2 in this case. I'm applying the logic of a composition function in its barest sense here and it may not jive with implicit differentiation.
     
  8. Oct 5, 2016 #7

    Ray Vickson

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    Whoever told you that you differentiate fine ##dy^2/dx## by multiplying ##2y## and ##y^2## was lying to you; certainly, nobody on this Forum ever told you such a thing.

    To get it right, it may be helpful to you to apply the somewhat imperfect (but still useful and revealing) form of the chain rule:
    $$\frac{d y^2}{dx} = \frac{d y^2}{dy} \cdot \frac{dy}{dx}$$.
    More generally, for ##F(x) = f(g(x))##:
    $$\frac{dF}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}$$
    You can think of this as "cancelling the ##dg##s:
    $$ \frac{d f(g)}{dx} = \frac{df}{d\not g} \cdot \frac{d\not g}{dx} $$.

    This is actually shorthand for
    $$ \frac{d}{dx} f(g(x)) = f'(g(x)) g'(x) \equiv \left. \frac{df(u)}{du}\right|_{u = g(x)} \cdot \frac{dg(x)}{dx} $$
     
  9. Oct 5, 2016 #8
    ????
    I thought the chain rule meant


    f(x)=(2x)^1/2
    f'(x)=(1/2(2x')(2x))^-(1/2)
    f'(x)=(1/2(2)(2x))^-(1/2)
    f'(x)=2x^(-1/2)
     
  10. Oct 5, 2016 #9

    Ray Vickson

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    $$\frac{d (2x)^{1/2}}{dx} = \frac{(2x)^{1/2}}{d (2x)} \frac{d (2x)}{dx}$$
    Since ##d u^{1/2}/du = (1/2) u^{-1/2}## we have ##d (2x)^{1/2} /d (2x) = (1/2)(2x)^{-1/2}##. Altogether, we have
    $$d(2x)^{1/2}/dx = (1/2)(2x)^{-1/2} \cdot 2 = (2x)^{-1/2} = (1/\sqrt{2}) x^{-1/2}$$.
     
    Last edited: Oct 6, 2016
  11. Oct 5, 2016 #10

    Mark44

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    Since there are two variables involved in the equation ##x^2 + y^2 = 25## it's important to be clear with respect to which variable you're differentiating.
    The derivative, with respect to x...

    ... with respect to x...

    ... with respect to x.
    If you differentiate y2, with respect to y, you get 2y. However, if you differentiate y2, with respect to x, you have to use the chain rule, as Ray Vickson showed near the beginning of his comments in post #7.
     
  12. Oct 6, 2016 #11

    Erm...

    Chain Rule..

    f(g(x))=(f' * (g(x)) * f'(x)

    where f' is the derivative of the outer function
    (x) is the evaluated inner function
    and g'x is the derivative of the inner function
     
  13. Oct 6, 2016 #12

    Ray Vickson

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    What you wrote is wrong; you are saying that ##f(g(x)) = (f' \times (g(x)) \times f'(x)##, and that cannot possibly be true in general. What I wrote above (as have several others) is the 100% absolutely correct form of the chain rule!
     
  14. Oct 6, 2016 #13

    Mark44

    Staff: Mentor

    This is not quite right, with most of the errors due to the lack of parentheses. I have added line numbers so that I can refer to specific lines in your work here.

    Line 1 -- (2x)^1/2 would usually be interpreted as ##\frac{(2x)^1}{2}##. To fix this, write it as (2x)^(1/2) or as (2x)1/2 using the X2 button in the menu strip, or as ##(2x)^{1/2}## using LaTeX (this is ##(2x)^{1/2}##).
    Line 2 -- On the right side, you have ...(2x')... which is sloppy if you mean (2x)' .
    Line 4 -- You have 2x^(-1/2), but you should have (2x)^(-1/2).
     
  15. Oct 7, 2016 #14

    vela

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    I hope you don't write stuff like this on your exams. You're claiming
    \begin{align*}
    x^2+y^2 &= 25 \\
    2x+y^2 &= 25 \\
    2x+y^2 &= 0
    \end{align*}
    From the last two equations, I'd conclude 0 = 25! You can't differentiate one term at a time and keep the equal signs there. Instead, you should write something like
    \begin{align*}
    x^2+y^2 &= 25 \\
    (x^2+y^2)' &= (25)' \\
    (x^2)'+(y^2)' &= (25)' \\
    2x+(y^2)' &= (25)' \\
    2x+(y^2)' &= 0
    \end{align*}
    Do you see the difference? I bring this up because you're being REALLY sloppy with notation in this thread, and I think that might be contributing to part of your confusion.

    I'm not sure what you mean by "a composition function in its barest sense here."

    Do you understand the following calculation?
    \begin{align*}
    \frac{d}{dx} (\sin x)^2 &= 2(\sin x)^1 \times (\sin x)' \\
    &= 2(\sin x)^1 \cos x \\
    &= 2 \sin x \cos x
    \end{align*}
    You differentiate the "outside" function first, leaving the inside intact, and then you multiply the derivative of the inside. In your problem, the inside function is ##y##.
    \begin{align*}
    \frac{d}{dx} (y)^2 &= 2(y)^1 \times (y)' \\
    &= 2y\,y'
    \end{align*}
     
  16. Oct 12, 2016 #15

    Thank you. I've done a bunch of practice questions on the chain rule and have moved on to implicit differentiation and this post helped the most.

    I apologize for my sloppy notation and confusing statements above, I had skipped ahead to implicit differentiation for some reason while I was doing trig diff. I'll do the rest of the work in order.
    If I have any more questions regarding this topic I'll post in this thread.
     
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