# F(x+dx)=F(x)+F'(x)dx eqn an identity

1. Jun 8, 2005

### isabella

F(x+dx)=F(x)+F'(x)dx

is the above eqn an identity or something?can someone explain to me what is happening in this eqn?

Last edited: Jun 8, 2005
2. Jun 8, 2005

### dextercioby

It's the approximation which follows from the definition of the derivative

$$F'(x)=:\lim_{\Delta x\rightarrow 0} \frac{F(x+\Delta x)-F(x)}{\Delta x}$$

by not considering the limit,but the increment in "x",viz.$\Delta x$,is very,very small (infinitesimal,if you prefer).

Daniel.

3. Jun 9, 2005

### SteveRives

Locate a discussion of Taylor series and the Newton-Rapson algorithm. You'll find a good bit on both in a Numerical Analysis textbook. Now, if you are reading a NA text, and that makes you ask your question, then I am sorry for redirecting you to where you already are. In that case, just notice the obvious.

If dx = 0, then of course f(x+dx) = f(x) + f'(x)dx.

What happens if we move dx slightly away from zero? ...well, see dextercioby's last post.

Last edited: Jun 9, 2005