Calculus problem, explaining the step

[karnten07]

Homework Statement

dy/dx(x+dx) - dy/dx(x-dx)
=> d^2y/dx^2 when dx->0

can someone explain this step to me please?

Homework Equations

The dx in the brackets should be sigma x

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? I am stumped, i don't see how this works at all.

 

[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.

 

[karnten07]

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.

Yes, my notes show that i divide the left hand side by h, but i still don't understand how this makes the left hand side into a 2nd order derivative?

 

[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'.

 

[karnten07]

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'.

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''?

 

[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.

 

[karnten07]

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.

I see, I am sorry if I am missing an obvious point here, but i still can't 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 realize that, but i see how that works now, i think. Thankyou Dick, i got there in the end i guess.

 

[Dick]

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