# Calculus problem, explaining the step

1. Oct 30, 2007

### karnten07

1. The problem statement, all variables and given/known data
dy/dx(x+dx) - dy/dx(x-dx)
=> d^2y/dx^2 when dx->0

can someone explain this step to me please?

2. Relevant equations

The dx in the brackets should be sigma x

3. The attempt at a solution
I know that dx -> means it tends to zero

i think i might have missed something out that might make a difference to this step;
it may be necessary to divide through the first line by dx (sigma x) to complete the step? Im stumped, i dont see how this works at all.

Last edited: Oct 30, 2007
2. Oct 30, 2007

### Dick

Write h for 'dx' in 'brackets'. Then you are saying lim h->0 y'(x+h)-y'(x-h)=y''(x). Not true. It's ok if you divide the left side by 2*h. It's just a difference quotient representation of a derivative.

3. Oct 31, 2007

### karnten07

Yes, my notes show that i divide the left hand side by h, but i still dont understand how this makes the left hand side into a 2nd order derivative?

4. Oct 31, 2007

### Dick

The derivative of f(x) can be defined by the limit of f(x+h)-f(x-h)/(2h) (yes, 2h). Put f=y and you get a definition of derivative y'. Now put f=y'.

Last edited: Oct 31, 2007
5. Oct 31, 2007

### karnten07

say h = 0.1, and i put f=y', i would get (1.1 - 0.9)/0.2 = 1
how does this mean that it equals f''?

6. Oct 31, 2007

### Dick

You said "I know that dx -> means it tends to zero". You have to take the limit as h->0 to get the true derivative.

7. Oct 31, 2007

### karnten07

I see, im sorry if im missing an obvious point here, but i still cant see how as h tends to zero it makes the formula become a second derivative.

I chose 0.1 because it was a relatively low number to see if i could understand the workings of the equivalence we are studying. So as h gets smaller and smaller i still get the answer of 1, and when h is zero i get:

(f'x - f'x)/0

which equals zero - OOOOOOOHHHHH - and that is what it equals as a second derivative! I didnt realise that, but i see how that works now, i think. Thankyou Dick, i got there in the end i guess.

8. Oct 31, 2007

### Dick

Don't put h=0! The result is undefined! Take the limit as h->0. Look back at the definition of a 'derivative'.