Expert: Scotto - 7/15/2009

Question
Hi Scotto,

I'm a guy from India preparing for IIT(Indian Institute of Technology) entrance exams.

I have a basic knowledge of whole of mathematics at freshman level.

I am familiar with basic ideas of calculus and can do freshamn level calculus.

I was recently solving a problem and had a doubt when I looked up the solution to that.


The problem is as follows:

If f(x)= sin^-1 {2x/(1 + x^2)} then find the interval where f(x) is differentiable.

I tried to use the idea:

lim       f(x) - f(a)
(x --> a) ---------------
             x - a

Where "a" can be any no. in the domain. In this case, I tried it with 'a' as 1 and -1 since these are the likely points where it may not be differentiable.

While solving it this way, I sort of got stuck and then looked up the solution.

But in the solution, they haven't followed the approach of limits, as I did above. Instead they have just differentiated the function directly and then showed that f'(x) doesn't exist for x=1,-1

My question is, that if we apply the same approach to another function, say
mod.(x), where "mod." stands for modulus.

This function can be written down as

      |
f(x)=  | x,       x > 0 (or = 0)
      |-x,       x < 0
      |

now if we directly differentiate the above function, like it has been done in the solution in my book for the previous func., we'll get,


      |
f'(x)= | 1,       x > 0 (or = 0)
      |-1,       x < 0
      |

This shows that this function f'(x) is defined everywhere and thus f(x) = mod.(x) is differentiable everywhere.
But we know that this is not true. We know that it isn't differentiable at x=0.
So this method of directly differentiating a function, without following the idea of limits must be wrong.

Please solve this paradox and also tell me why am I not able to get an answer to the func.
f(x)= sin^-1 {2x/(1 + x^2)}
when I use the limit approach.

The approach of directly differentiating functions has been widely followed throughout the book....is it correct...??? doesn't seem so because if it were, it would have also given us a correct answer for the function mod.(x) which we didn't get.

Thanks,
Shikhin

Answer
Let's take a look at g(x) = 2x/(1+x^2).
By the quotient rule, g'(x) = ((1+x^2)2 - 2x*2x)/(1+x^2)^2 =
(2 + 2x^2 - 4x^2)/(1+x^2)^2 = 2(1-x^2)/(1+x^2)^2.

It is known that f(g) = sin^-1(g).
From here, we can say that f(g(x)) = sin^-1(g(x)).
It can then be said that df/dx = (df/dg)(dg/dx).
It is known that df/dg = 1/sqrt(1+g^2).
g'(x) was given in the proceeding paragraph.

From what I can see the function is differentiable everywhere.

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This looks like a totally separate question.

When the new definition of f(x) is given as being x and -x, that looks the same as |x|.  I don't see how the two functions relate.

The derivative of this f(x) is either -1 or 1, and that is correctly stated what was sent.

However, this does not show that f(x) is differentiable everywhere.
It gives no indication of f'(0).  It only says for x>0 and x<0.
At 0, there is no derivative.
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That is my answer for now, but I think I'll probably get a response from you.

Note it is better to put everything on one line or send as an attachment.  When I print out questions, spaces are a lot shorter than the rest of the characters.