Derivative of f(x) = arcsin((2x/(1+x^2)) at x=1

[daniel_i_l]

Homework Statement

This isn't exactly a homework question, it's more of a general one. Let's say for example that I have the function f(x) = arcsin((2x/(1+x^2)). I know that arcsin has no derivative at x=1. Does that mean that f also doesn't or to I have to check explicity with a limit? I think that I have to do a limit, is that right? Could someone elaborate on this?
Thanks.

 

[Dick]

The chain rule says (f(g(x))'=f'(g(x))*g'(x). f' may not have a limit but the product may. In your specific example it does. So yes, you have to check.

 

[daniel_i_l]

Thanks. But you say that in this case f has a derivative at x=1? I did the limit and got infinity. Is that right?

 

[daniel_i_l]

Anyone?
Thanks.

 

[jostpuur]

What is the derivative function you got? I got

$$f'(x) = \frac{2-2x^2}{(1+x^2)\sqrt{1+2x^2-3x^4}}$$

If that is right, then the limit x->1 is non-trivial, but I think I got, using L'Hospital's rule, that it is zero and not infinity.

 

[George Jones]

What is the derivative function you got? I got

$$f'(x) = \frac{2-2x^2}{(1+x^2)\sqrt{1+2x^2-3x^4}}$$

If that is right, then the limit x->1 is non-trivial, but I think I got, using L'Hospital's rule, that it is zero and not infinity.

Factoring the top and bottom of your expression also shows that the "problem" disappears.

 

[Dick]

What is the derivative function you got? I got

$$f'(x) = \frac{2-2x^2}{(1+x^2)\sqrt{1+2x^2-3x^4}}$$

If that is right, then the limit x->1 is non-trivial, but I think I got, using L'Hospital's rule, that it is zero and not infinity.

I don't think that derivative is quite correct. After factoring what I get I'm just left with 2/(1+x^2).

 

[George Jones]

I don't think that derivative is quite correct. After factoring what I get I'm just left with 2/(1+x^2).

Until now, I didn't actually do the derivative. I get (with a minus sign) what you have.

 

[jostpuur]

I found the mistake, I had calculated $1 + 2x^2 + x^4 - 4x^2 = 1 + 2x^2 - 3x^4$

Now I got
$$\frac{2}{1+x^2}$$
too, without minus sign though.

 

[George Jones]

I found the mistake, I had calculated $1 + 2x^2 + x^4 - 4x^2 = 1 + 2x^2 - 3x^4$

Now I got
$$\frac{2}{1+x^2}$$
too, without minus sign though.

Actually, the sign is bit tricky, which I didn't see until I graphed

$$arcsin\left(\frac{2x}{1+x^2}\right).$$

The derivative of this function has a jump discontinuity at $x=1$.

Hint: consider

$$\frac{-4}{\sqrt{\left(-4\right)^2}}.$$

 

[jostpuur]

I didn't check the domain earlier, and though that the function would be defined on some interval around origo, but actually the expression

$$\frac{2x}{1+x^2}$$

doesn't get any values outside $[-1,1]$. So the function becomes defined for all real numbers.

I took a closer look at the derivative. Isn't it this?

$$\frac{2(1-x^2)}{|1-x^2|(1+x^2)}$$

So there's plus sign when $x\in ]-1,1[$, minus sign when $x\in ]-\infty,-1[\;\cup\; ]1,\infty[$, and the derivative does not exist for $x\pm 1$.

 

[daniel_i_l]

I took the limit of $$\frac{arcsin\left(\frac{2(x+1)}{1+(x+1)^2}\right) - \frac{\pi}{2}}{x}.$$ when x goes to 0 and got
$$\frac{-2}{1+(x+1)^2}$$
By using L'hopital. It looks like what George got except that instead of x I have x+1. What did I do wrong?
Thanks

 

[jostpuur]

That limit should not exist. However if take only left handed or right handed limit then you should get either 1 or -1. Do you know which limit you are taking?

If I got this right, then left handed limit should give 1, and right handed limit -1.

I'm not sure if there is a mistake concerning the x+1 term. You must take the limit x->0 in the end, and then you also have x+1 -> 1. You have now inserted an expression x+1 in the place of the old parameter x, so x+1 -> 1 is precisely what you are supposed to have.