# Rate of change of x wrt sin(x)

by Jhenrique
Tags: rate, sinx
 P: 686 Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
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P: 40,649
 Quote by Jhenrique Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Have you tried just plotting it in Excel? You will get some 1/0 inflection points that you will need to deal with...
P: 354
 Quote by Jhenrique Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Compare with ##\frac{dsin^{-1}(x)}{dx}##

Or, turn the graph of sin(x) on its side.

P: 6,030
Rate of change of x wrt sin(x)

 Quote by PeroK Compare with ##\frac{dsin^{-1}(x)}{dx}## Or, turn the graph of sin(x) on its side.
Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
P: 1,274
 Quote by mathman Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
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?
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P: 18,019
 Quote by mathman Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
Due to the inverse function theorem, it is very relevant.
P: 1,274
 Quote by micromass Due to the inverse function theorem, it is very relevant.
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)|.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,304 There are two ways to "differentiate x with respect to sin(x)". The first is to use the fact that $$\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}$$. 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 $x= sin^{-1}(y)$ so that the derivative of "x with respect to sin(x)" is $dx/dy= d(sin^{-1}(y))/dy= 1/\sqrt{1- y^2}$. It's not at all difficult to prove that those are the same. If y= sin(x) then $1- y^2= 1- sin^2(x)= cos^2(x)$ so $1/\sqrt{1- y^2}= 1/cos(x)$.
 P: 686 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' ...
P: 1,274
 Quote by HallsofIvy There are two ways to "differentiate x with respect to sin(x)". The first is to use the fact that $$\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}$$. 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 $x= sin^{-1}(y)$ so that the derivative of "x with respect to sin(x)" is $dx/dy= d(sin^{-1}(y))/dy= 1/\sqrt{1- y^2}$. It's not at all difficult to prove that those are the same. If y= sin(x) then $1- y^2= 1- sin^2(x)= cos^2(x)$ so $1/\sqrt{1- y^2}= 1/cos(x)$.
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?
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P: 18,019
 Quote by 1MileCrash 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)|.
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?
P: 1,274
 Quote by micromass 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?
Yes, thank you.

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