Expert: Paul Klarreich - 3/6/2008

Question
After having found f '(x)= 2(1-x^2) / sqrt 2-x^2 ( from the given function: f(x)= x(sqrt 2-x^2)
I must now find f''(x), In which I am having trouble doing! :(... This will thus give me concavity, which I can use to sketch my graph. Thanks!

Answer
Questioner:   Shawna
Category:  Calculus
Private:  No
 
Subject:  second derivatives
Question:  After having found f '(x)= 2(1-x^2) / sqrt 2-x^2 ( from the given function: f(x)= x(sqrt 2-x^2)
I must now find f''(x), In which I am having trouble doing! :(... This will thus give me concavity, which I can use to sketch my graph. Thanks!
 

       
f'(x) = (x)(-2x/2sqrt(...)) + (1)(sqrt(...))

f'(x) = (x^2/sqrt(...)) + (sqrt(...))
        - x^2 + 2 - x^2
f'(x) = -----------------
           sqrt(...)

        - 2x^2 + 2
f'(x) = -----------------
           sqrt(...)

        2(1 - x^2)
f'(x) = ------------
         sqrt(...)

Yes, f'(x) confirmed.  SFSG. [so far so good]

Now there is nothing for it except to swallow an aspirin, brew a good cup of coffee, and use the quotient rule.  The 2 is a constant coefficient, of course, which you take out:

To retain sanity, do the D[ sqrt(2 - x^2) separately:

That's equal to -x/sqrt(...), from the first derivative.

           (sqrt(...))(-2x) - (1 - x^2)(-x/sqrt(...))
f''(x) = 2 --------------------------------------------
                       (...)  << no sqrt now.


          (2 - x^2)(-2x) - (1 - x^2)(-x)
f''(x) = 2 -------------------------------- << clear; all multiplied by sqrt(...)
                       (...)^3/2

           -4x + 2x^3 +  x - x^3
f''(x) = 2 -------------------------
                 (...)^3/2

            x^3 - 3x
f''(x) = 2 --------------
            (...)^3/2

I think that's as far as we can go.  As far as sketching is concerned, I think you want the concavity at your extreme, which were  x = 1 and x = -1:

             1 - 3
f''(1) = 2 -----------, negative.  1 is a maximum
            (1)^3/2


             -1 + 3
f''(-1) = 2 -----------, positive.  1 is a minimum
            (1)^3/2

A little checking note:

Your original:  f(x) = x sqrt(2 - x^2)  is an example of an ODD function.  That means:
f(-x) = - f(-x), AND you will conclude:

1. the graph has origin symmetry.  
2. f' will be an even function.
3. f'' will be an odd function.
Yes, that's the way it works.