- #1
Jhenrique
- 685
- 4
Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Jhenrique said:Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
Jhenrique said:Is possible to compute the rate of change of x wrt sin(x)? ##\frac{dx}{d \sin(x)}##
PeroK said:Compare with ##\frac{dsin^{-1}(x)}{dx}##
Or, turn the graph of sin(x) on its side.
mathman said:Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
mathman said:Compare to reciprocal dsin(x)/dx. Sin-1(x) is not relevant.
micromass said:Due to the inverse function theorem, it is very relevant.
HallsofIvy said: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].
1MileCrash said: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)|.
micromass said: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?
The rate of change of x with respect to sin(x) is the instantaneous change in the value of x with respect to the change in the value of sin(x). In other words, it measures how much x changes for every unit change in sin(x).
The rate of change of x with respect to sin(x) can be calculated using the derivative formula, which is the limit of the change in x over the change in sin(x) as the change in sin(x) approaches 0.
A positive rate of change of x with respect to sin(x) indicates that as sin(x) increases, x also increases. This means that the graph of x with respect to sin(x) has a positive slope. On the other hand, a negative rate of change of x with respect to sin(x) indicates that as sin(x) increases, x decreases, and the graph has a negative slope.
The rate of change of x with respect to sin(x) is related to the slope of the graph of x and sin(x). The slope at any point on the graph is equal to the rate of change of x with respect to sin(x) at that point. This means that the steeper the graph, the higher the rate of change of x with respect to sin(x).
Yes, the rate of change of x with respect to sin(x) can be negative infinity or undefined. This happens when the graph of x and sin(x) has a vertical tangent, which means that the slope is undefined or infinite at that point. This can occur at points where sin(x) is equal to 0 or 1, since the derivative of sin(x) is undefined at these points.