# Which expression yields the best approximation to df/dx (h 1)?

1. Nov 29, 2007

### neelakash

Some interesting calculus...

Which of the following expressions yields the best approximation to
df/dx (h<<1)?

A. $$\frac{f(x+h)-f(x)}{h}$$

B. $$\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}$$

C. $$\frac{f(x)-f(x-h)}{h}$$

D. $$\frac{f(x+h)-f(x-h)}{h}$$

From school days I have been taught A which is almost the same as C

I would like B to be the correct.What do other people think?

2. Nov 29, 2007

### Integral

Staff Emeritus
You can use simple substitution to show that A, B and C are equivalent.

3. Nov 29, 2007

### neelakash

You are correct:

I will go for A now...

It looks that all A,B and C are equivalent.But definition of y' is to find the increment in y at x=x.Say,y changes to y+dy and x changes to x+dx.

My point is that to find df/dx at x,we must take increment f(x+h) from f(x)

Remember,by definition, df/dx is evaluated at x.

4. Nov 29, 2007

### mathman

When going to the limit A,B,C give the same result (as long as the derivative is continuous at x). However if you want an approximation where h is fixed and small then B is the best approximation.

D is just wrong - you need to divide by 2 (then identical to B).

5. Nov 29, 2007

### Xevarion

Are we supposed to assume that the same h is used in each expression? In that case there would be a unique right answer to the original question, but you would need to know something about f to say which it is...