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Rate of change of x wrt sin(x)

  1. Apr 9, 2014 #1
    Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
  2. jcsd
  3. Apr 9, 2014 #2


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    Have you tried just plotting it in Excel? You will get some 1/0 inflection points that you will need to deal with...
  4. Apr 9, 2014 #3


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    Compare with ##\frac{dsin^{-1}(x)}{dx}##

    Or, turn the graph of sin(x) on its side.
  5. Apr 9, 2014 #4


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    Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
  6. Apr 9, 2014 #5
    Why isn't it relevant?

    If we have dx/d(sinx), why is it not reasonable for me to let sinx = y and therefore have that x = arcsin(y)?

    Then I have d(arcsin(y))/y = 1 / sqrt(1-y²)

    And thus dx/d(sinx) = 1/sqrt(1-sin²(x)) = 1/sqrt(cos²(x)) = 1/|cos(x)|

    Just curious, so where does this break down? This is a very unique question so I'm sure there is an error in my logic. One may note that this comes out to be only slightly different than the reciprocal method (no absolute value). I suspect this is because in assigning these variables in the way I did, I restricted/changed domain?
    Last edited: Apr 9, 2014
  7. Apr 9, 2014 #6
    Due to the inverse function theorem, it is very relevant.
  8. Apr 10, 2014 #7
    After a quick Google, it looks like I followed this theorem with f(a) = sin(x). So why does f'(b) evaluate to 1/|cos(x)| according to my work rather than 1/cos(x) as the inverse function theorem claims it should?

    I mean, hopefully we can agree that the answer to the OP's question is actually 1/cos(x) and not 1/|cos(x)|.
  9. Apr 10, 2014 #8


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    There are two ways to "differentiate x with respect to sin(x)". The first is to use the fact that
    [tex]\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex].
    Here y= sin(x) so dy/dx= cos(x). Then dx/dy= 1/cos(x) is a perfectly good answer.

    The other way is to say that if y= sin(x) then [itex]x= sin^{-1}(y)[/itex] so that the derivative of "x with respect to sin(x)" is [itex]dx/dy= d(sin^{-1}(y))/dy= 1/\sqrt{1- y^2}[/itex].

    It's not at all difficult to prove that those are the same. If y= sin(x) then [itex]1- y^2= 1- sin^2(x)= cos^2(x)[/itex] so [itex]1/\sqrt{1- y^2}= 1/cos(x)[/itex].
  10. Apr 10, 2014 #9
    So, I can differentiate any function f(x) wrt another any function g(x).

    If df = f'(x) dx and dg = g'(x) dx, thus df/dg = f'/g' ...
  11. Apr 10, 2014 #10
    Our work is the same, but I don't understand how you conclude that sqrt(cos²(x)) = cos(x). This is |cos(x)|, don't you agree?
  12. Apr 10, 2014 #11
    Well, the sine function is not invertible, since it's not injective and not surjective. This is of course no problem for the inverse function theorem, since it will take a "local inverse".
    You worked with the arcsine, which is indeed a local inverse. But not all local inverses are like the arcsine. That is, if we restrict the sine to ##(-\pi/2,\pi/2)## then the arcsine is an inverse. But guess what? The cosine function is actually positive on that said, so the absolute values drop.

    If you apply the inverse function theorem on some other domain, then you can't use the arcsine anymore in that form.

    Does that make sense?
  13. Apr 10, 2014 #12
    Yes, thank you.
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