An annoying derivative

[Qaanol]

I’m trying to take the derivative of a messy function at x=1. I’m pretty sure that y′(1) = 1, but I’m having a world of difficulty proving it. Here’s the setup:

t(x) = (2/π) · arcsin( (2/π) · ( arcsin(x) + x · √(1 - x²) ) )
y(x) = t · √(1 - t²) + x · (1 - √(1 - t²))

Now, I’m reasonably confident that the following are correct:

dt/dx = (8 · √(1 - x²)) / (π² · cos((π/2) · t))
dy/dx = 1 - √(1 - t²) + (dt/dx) · (1 + q·t - 2t²) / √(1 - t²)

It’s the next bit, trying to evaluate dy/dx at x=1, where I keep getting tangled up. When I applied L’Hôpital’s rule it just seemed to open a rabbit hole.

[Nitrodon]

The proof of the product rule can also be used to show that if y(x) = u(x) · v(x) where u(c) = 0, u is differentiable at c, and v is continuous (not necessarily differentiable) at c, then y is differentiable at c, and y'(c) = u'(c) · v(c). This can be used to get rid of the annoying 0/0 terms obtained from the standard product rule.

By adding and subtracting some terms, we can write y(x) = (t - 1) · √(1 - t²) + (x - 1) · (1 - √(1 - t²)) + 1. After proving that t is (left) continuous and (left) differentiable at x=1, we can use the above to evaluate y'(1).

[Qaanol]

Thanks, I ended up using L’Hôpital’s rule with respect to t, and the fact that the limit of a product is the product of the limits.

I’m not quite sure your approach works, since t′(1) = ∞.

[Nitrodon]

Oops. I saw that the inner function (2/π) · ( arcsin(x) + x · √(1 - x²) ) was differentiable at 1, then got stupid and assumed that meant t was also differentiable.

I have still proven that y'(1) = 1 is equivalent to "the derivative of (1-t)^3/2 at x=1 is zero". It is possible to prove the latter statement:

1. By either Taylor series (with the half angle formula) or L'Hôpital, we can prove that (2/π) · ( arcsin(x) + x · √(1 - x²) ) = 1 - O((1-x)^3/2).
2. By the same method, we see that t(x) = 1 - O((1-x)^3/4).
3. This shows that (1-t)^3/2 = O((1-x)^9/8) = o(1-x), so the statement is proven.

[Spacemoss]

Not sure if you were still interested in pursuing this derivative, however, unless I'm interpreting your input and work so far incorrectly, then you are making an error somewhere along the way. Using Alpha we have:

dt/dx =

y =

and dy/dt =

Now you can check dy/dt at x=1. The first 3 terms are 0/0 form so you will have to try l'hospital's rule, but the last term clearly evaluates to 1. I'm assuming once you check the first 3 terms that they will all limit to 0.