mopit_011 said:
Homework Statement:: N/A
Relevant Equations:: N/A
Doesn’t this proof not apply to any function that isn’t one-to-one them as delta u will be 0 when delta x is not equal to 0 in a one-to-one function?
To address the actual question asked, as shown above:
The theorem can apply to most non-one-to-one functions because, provided the function ##g## is differentiable at x and has nonzero derivative there, we can
always find an open interval around x in which ##g## is one-to-one. Since we are only aiming to find the derivative of the composed function at x, we can apply the theorem within that open interval to get the result.
For a function that is not differentiable at x, such as g(z) = sin 1/(z - x), we may not be able to find such an open interval where it is one-to-one.
So the key property is being differentiable. Being one-to-one doesn't matter. Accordingly, it's also unnecessary to have a nonzero derivative, as the composed function, if differentiable, will have zero derivative at ##x## if ##g## does,
By the way, ##g(z) = z^3 \sin \frac1z## is a function that is not one-to-one on any interval including 0, but is differentiable at 0, with derivative 0.