Question. Suppose that is such that is continuous and is differentiable. Should also be differentiable?
Proof. Let . Since satisfies the differentiable equation
is differentiable at all points with by the implicit function theorem and
Where , we note that has a minimum and . This gives
and is differentiable by the definition.
Comment. The only points of interests are the zeros of since has the same sign in some neighborhood of points which are not roots. So there goes half the work. Geometrically it doesn’t make sense for to have anything but derivative 0 at roots of and this can be verified by taking left hand and right hand limits of the quotient.