1. ## Delta?

what does the symbol delta(the triangle symbol)x mean?

2. It is the 4th Greek letter and usually represents 'change in' x

3. so what would I do if the problem is:

(f(2+delta(x))+ f(2)) / delta(x)

the equation:x^2 - 2x + 1

4. The problem is most likely to find the derivative of f in the point x=2.

5. Originally Posted by CalcGeek31
so what would I do if the problem is:

(f(2+delta(x))+ f(2)) / delta(x)

the equation:x^2 - 2x + 1
Delta, as Galactus said, represents change in x. In your case, it represents the change of x as it approaches zero. This is the derivative at the point x = 2. Also, you made a small typo:

$\lim_{\Delta x \to 0} \frac{f(2+\Delta x ) \color{red}- f(2)}{\Delta x}$

6. Im not sure I understand...

my problem is f(x) is equal to the second equation... now evaluate the first equation...

7. Let $h = \Delta x$

$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

Our objective here is to get rid of the h in the denominator so we can find the limit. Now, let's evaluate f(2+h) and f(2)

$f(2+h) = (2+h)^2 - 2(2+h) - 1$

$f(2) = (2)^2 - 2(2) - 1$

Now simply replace and simplify. You'll find that you can cancel the h in the denominator after all.

8. ohhh... now that makes perfect sense... it takes the form of f(x+h) thank you

I also have a similar problem in which there is no change in x. does this mean I work it all the way up until I do the limit?

9. wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h
then how do I cancel the denominator?

10. Originally Posted by CalcGeek31
wait... I get down to lim(h-> 0) (2h + h^2 - 2) / h
then how do I cancel the denominator?
You made a mistake while simplifying:
$\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

$\lim_{h \to 0} \frac{(2+h)^2 - 2(2+h) - 1 - (-1)}{h}$

$\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}$

...

11. ahhh... k so my final answer is 2?

12. Originally Posted by CalcGeek31
ahhh... k so my final answer is 2?
EDIT: Yes, I made trivial mistake. You're correct.

13. $
f(2+h) = (2+h)^2 - 2(2+h) - 1
$

from that... -2(h) = -2 h + 4h

14. $\lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$

$\lim_{h \to 0} \frac{(2+h)^2 - 2(2+h) - 1 - (-1)}{h}$

$\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 -2h - 1 +1}{h}$

$\lim_{h \to 0} \frac{4h + h^2 -2h}{h}$

$\lim_{h \to 0} \frac{h^2 +2h}{h}$

$\lim_{h \to 0} \frac{h(h+2)}{h}$

$\lim_{h \to 0} (h+2) = 2$

15. $
\lim_{h \to 0} \frac{4 + 4h + h^2 - 4 +2h - 1 +1}{h}
$
[
your +2h in here should be -2h...

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