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.

## An annoying derivative

**Moderators:** gmalivuk, Moderators General, Prelates

### An annoying derivative

wee free kings

### Re: An annoying derivative

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).

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).

### Re: An annoying derivative

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) = ∞.

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

wee free kings

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