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Rate of change of x wrt sin(x) 
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#1
Apr914, 12:59 PM

P: 686

Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##



#2
Apr914, 01:18 PM

Mentor
P: 41,122




#3
Apr914, 02:16 PM

P: 413

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


#4
Apr914, 04:04 PM

Sci Advisor
P: 6,080

Rate of change of x wrt sin(x)



#5
Apr914, 04:21 PM

P: 1,302

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(1y²) And thus dx/d(sinx) = 1/sqrt(1sin²(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? 


#7
Apr1014, 12:05 AM

P: 1,302

I mean, hopefully we can agree that the answer to the OP's question is actually 1/cos(x) and not 1/cos(x). 


#8
Apr1014, 08:51 AM

Math
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Thanks
PF Gold
P: 39,568

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]. 


#9
Apr1014, 09:17 AM

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' ... 


#10
Apr1014, 10:48 AM

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#11
Apr1014, 10:54 AM

Mentor
P: 18,334

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? 


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