Expert: Scotto - 11/20/2009

Question
How do I determine the fourth derivative of (x^2+1)^1/2. I can get (x^2+1)for the first derivative, but I am having trouble from there.  I really appreciate any help I can get with this.

Thank you

Answer
After looking through it, I found an error.  After checking it again, it looked fine.
Here is the corrected copy.

It can be said that f²(x) = x^2 + 1, so take the derivative.  This gives 2f(x)f'(x) = 2x.
Solve for f'(x) and get f'(x) = x/f(x).

f"(x) = (f(x) - xf'(x))/f^2(x).
At this point I'll drop the x's and say f, f', etc and not f(x), f'(x), etc.

I still know that f and its derivatives are functions of x
in that f' = f/x, so I'll put that in and get f" = (f - (x^2)/f)/f^2.
Multiply the whole fraction by f/f, getting f" = (f^2 - x^2)/f^3.

Now to compute f"'(x).  This is again a chain rule, (lo d hi - hi d lo)/lo^2.
f'" = ((f^3)(2ff'-2x)- (f^2 - x^2)3(f^2)f')/f^6.
First, lets cancel f^2 from the two terms in the numerator and also in the denominator.

This gives f'"(x) = (f(2ff'-2x)- (f^2 - x^2)3f')/f^4.
Multiplying out the top gives f'" = (2f²f' - 2xf - 3(f^2)f' + 3(x^2)f')/f^4.
Note that f'f = x, so we have f'" = (2xf - 2xf - 3xf + 3x²f')/f^4.
This reduces to f'" = (-3xf + 3x²f')/f^3.
Again, noting f' = x/f, this is f'" = -3(x/f^3) + 3(x^3/f^3).

Using this, we can then find f""(x).
That is f"" = -3(f^3 - x^3f^2f')/f^6 + 3((f^53x^2 - x^35f^4f')/f^10).
Splitting this into separate terms and cancelling gives
f"" = -3/f^3 + 9xf'/f^4 + 9x^2/f^5 - 15x^3f'/f^6.

Again, since f' = x/f, we get f"" = -3/f^3 + 9x^2/f^5 + 9x^2/f^5 - 15x^4/f^7.
Combining the two center terms gives
f"" = -3/f^3 + 18x^2/f^5 - 15x^4/f^7.
Rewrite this as f"" = -3(1/f^2 - x^2/f^4 + 5x^4/f^6)/f.
Note that f^2 = (x^2 + 1), so f^4 = (x^2 + 1)^2.
The denominator is f, which is (x^2 + 1)^0.5.