Derivative question

1. Jul 30, 2007

daniel_i_l

1. The problem statement, all variables and given/known data
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.

2. Jul 30, 2007

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.

3. Jul 30, 2007

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?

4. Aug 3, 2007

Anyone?
Thanks.

5. Aug 3, 2007

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.

6. Aug 3, 2007

George Jones

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

7. Aug 3, 2007

Dick

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

8. Aug 3, 2007

George Jones

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

9. Aug 3, 2007

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.

Last edited: Aug 3, 2007
10. Aug 4, 2007

George Jones

Staff Emeritus
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

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

Last edited: Aug 4, 2007
11. Aug 4, 2007

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 $x\pm 1$.

12. Aug 4, 2007

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

13. Aug 4, 2007

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.