Concerning differentials, when is 1/dy/dx not equal to dx/dy?

When can the derivative not be thought of as a ratio?

The only time that is guaranteed to work is for a continuously differentiable, strictly monotonic function.

 

If a function is not injective, then it does not have an inverse function so the derivative cannot be found.

I see. So with ordinary "high school" mathematics, the derivative as a ratio holds just fine. Which also would explain how the founding fathers (before the derivative as a limit) were able to use and deduce stuff.

The only time that is guaranteed to work is for a continuously differentiable, strictly monotonic function.

 

If a function is not injective, then it does not have an inverse function so the derivative cannot be found.

[math]x^2+y^2=a^2[/math] is not a function of y in terms of x or of x in terms of y, and yet at all but four points on the circle one can compute dy/dx and dx/dy -- and dx/dy = 1/(dy/dx).

 

Things become problematic in higher dimensions working with partial derivatives.

[math]x^2+y^2=a^2[/math] is not a function of y in terms of x or of x in terms of y, and yet at all but four points on the circle one can compute dy/dx and dx/dy -- and dx/dy = 1/(dy/dx).

dy/dx is multivalued at nearly all points, and there exists points where dy/dx is 0 so it doesn't even have a reciprocal.

But it can be chopped up so as to be piecewise differentiable and strictly monotonic which is why taking the reciprocal still works for most of the function.

 

Same as if you were looking at sin(x), it works just fine but only over a small range.

No. It works fine. Use implicit differentiation.

I did say that it'd work.

If a function is not injective, then it does not have an inverse function so the derivative cannot be found.

Surely an injective function can still has a derivative? You mean as a ratio?

Surely an injective function can still has a derivative? You mean as a ratio?

Sorry, I meant the derivative of it's inverse - what with its inverse not existing.