I'm studying computer vision, and one of the problems in my book is to prove that $\partial f/ \partial x = f(x+1) - f(x)$
It's been a while since I've touched Taylor Series, and so I'm not sure of the approach for this general form. I've found lots of reference online to the special properties/method of calculus of the derivatives OF the Taylor Series expansion, but how does one take the Taylor Series expansion of the above in order to satisfy the proof?
Well, as I suppose that the book is not asking you to prove the equality $\partial f/\partial x = f(x+1)-f(x)$ in general, which is obviously incorrect from the mathematical point of view. However, in the context of image processing, $f(x+1)-f(x)$ can be a reasonably good approximation for $\partial f/\partial x$. An image can be viewed as a two-dimensional function defined over a discrete domain, which is the set of pixels of the image. In that context, $f(x,y)$ is the image intensity at the pixel on row $x$ and column $y$, and $f(x,y+1)$ would be the image intensity of the next pixel to the right.
Let us now assume that we have fixed a column, say $y$, and we want to compute the partial derivative along that column. Then, using the definition of derivative
$$ \partial f/\partial x \approx \frac{f(x+1,y)-f(x,y)}{x+1-x} = f(x+1,y)-f(x,y). $$
This is exactly what you have, except that you have omitted the constant variable $y$.
I think the best place to ask this kind of question is the Computer Science Stack Exchange.