Dy/dx: must dx be the independent variable?

[Eus]

Hi Ho!

If y=\sin x, \frac{dy}{dx}=\frac{d(\sin x)}{dx}=\cos x.

If x=\sin y, y=\arcsin x, and therefore \frac{dy}{dx}=\frac{d(\arcsin x)}{dx}=\frac{1}{\sqrt{1-x^2}}.

But, if x=\sin y, can \frac{dy}{dx} be done as \frac{dy}{dx}=\frac{dy}{d(\sin y)}=\frac{1}{\frac{d(\sin y)}{dy}}}=\frac{1}{\cos y}=\sec y?

If it cannot, is it because the definition of \frac{dy}{dx} that says that \frac{dy}{dx}=\lim_{\Delta x\rightarrow 0}{\frac{f(x+\Delta x)-f(x)}{\Delta x}} requires that dx must be the independent variable and dy must be the dependent variable? Or, is there any other reason?

Thank you very much.


Eus

 

[HallsofIvy]

Actually, in most mathematics there is little or no distinction between "independent" and "dependent" variables. All we really need is that there be a "relation" between x and y. If y= sin(x), as long as we stay on an interval in which sine is "one-to-one", we can define the inverse function x= arcsin(y). Another way to find the derivative of x= sin(y), with respect to x, is to use "implicit differentiation": take the derivative of both sides with respect to x. 1= cos(y)dy/dx (by the chain rule) so dy/dx= 1/cos(y)= sec(y). If it bothers you that the right side is a function of y instead of x, you can change that by using trig identities, or even simpler thinking of y as an angle in a right triangle. If x= sin(y), then we can take x as the "opposite side" of a right triangle with hypotenuse 1. y is the angle opposite side x of course, so sin(y)= opposite side/near side= x/1= x. By the Pythagorean Theorem, then, the length of the "near side" is \sqrt{1- x^2}. Since secant is defined as "hypotenuse over near side", sec(y)= 1/\sqr{1- x^2}. That is, if x= sin(y), dx/dy= 1/\sqrt{1- x^2} which is precisely the derivative formula for "arcsin(x)".