It seems to me that should be at least

It seems you wish to integrate the function,

Let use find whether

The only possible solution is but thus, .

In that case,

If, then,

Thus,

Therefore, there exists a point in the interval

Using the integral subdivision rule,

Thus,

Let,

Therefore, those two functions can be expressed as composition functions,

But by the Fundamental theorem,

And since are both differenciable on we can use the chain rule.

Something interesting to note.

of this expression if not equal to which means the function you mentioned is not everywhere differenciable.

The other part of the problem is to consider the case . Since it is late now I cannot complete the solution but I think it will be the same as for