# Derivatives and Increments -help, again

1. Aug 5, 2010

### Asphyxiated

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

Find f'(c) and the error estimate for:

$$f(x)= \sqrt{x^{2}+1}$$

2. Relevant equations

The error is given by:

$$E(\Delta x) = \frac {1}{2}M \Delta x$$

and

$$f''(c) \leq M$$

3. The attempt at a solution

So the first derivative is:

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

and for the second derivative I got:

$$f''(x) = (x^{2}+1)^{-3/2}$$

So for f'(c) I got:

$$f'(c) = \frac {2}{\sqrt{5}}$$

which is correct in the back of the book but for the error i got:

$$f''(c) = 5^{-3/2}$$

so

$$E(\Delta x) = \frac {1}{2} (5)^{-3/2} \Delta x \; \approx \; .04472 \Delta x$$

but according to the book it is:

$$E(\Delta x) = \frac {1}{4\sqrt{2}} \Delta x \; \approx \; .17678 \Delta x$$

I have done the second derivative over now a couple of times and still get the same answer but here is how i got f''(x):

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

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

which then boils down to:

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

Next lets combine the terms on the top to get something like:

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

so x^2 will cancel on the top fraction to get:

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

and that can be rewritten as:

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

which can further be rewritten as:

$$(x^{2}+1)^{-1/2} * (x^{2}+1)^{-1} = (x^{2}+1)^{-3/2}$$

so:

$$f''(x) = (x^{2}+1)^{-3/2}$$

where did I go wrong?

2. Aug 5, 2010

### Staff: Mentor

Where is M? For your error estimate you need a number M such that f''(x) <= M for all x in some interval.

What is c? You show values for f'(c) and f''(c) without ever saying what c is.
Your result for f''(x) look fine.

3. Aug 5, 2010

### Asphyxiated

Oh my bad, c = 2 and since:

$$f''(x) \leq M$$

f''(c) = M so in this problem (as shown):

$$M = (x^{2}+1)^{-3/2} \; and \; with \; c=2, \; M=(5)^{-3/2}$$

4. Aug 6, 2010

### Staff: Mentor

There are two parts to this problem:
1. Find f'(2)
2. Find the error estimate for f(x) = sqrt(x^2 + 1).

The first part you have already done.
For the second part, we can estimate f(x) by f(2) + f'(2)$\Delta x$. The error in this estimate is E($\Delta x$), the absolute value of which is equal to (1/2)f''($\xi$)$\Delta x$, for some number $\xi$ between 2 and x.

Since we don't know $\xi$ (the Greek letter xi), the best we can do is find the largest value of f''($\xi$), which we'll call M.

As it turns out, the global maximum value for f''(x) = 1/(x^2 + 1)^(3/2) is 1, which is attained when x = 0. Since the answer in your book doesn't seem to be using that value, but seems to be using f''(1) instead, I suspect that the problem is assuming that $\Delta x$ is less than 1.

You neglected to include the fact that c was given as 2. Did you forget to mention another piece of given information?