# How to obtain the tangent slope of consecutive points.

For example, $f(x) = 3x^2$, and I want to obtain $f '(2)$ using only information I have with the point $x = 1$. Is there an equation I could use to do this which would apply for all functions, not just the one I mentioned above?

-
Why would you want to do this? You have $f$. What prevents you from simply taking the derivative? –  EuYu Oct 11 '12 at 19:55
I know I can do this, I was just wondering if there's a relationship between consecutive points involving higher derivatives (2nd derivative and so on). –  Adi Oct 11 '12 at 19:57
What do you mean by "consecutive points"? For the question you asked in the post, the answer is no. You cannot determine $f'(2)$ knowing only $f(1)$. You can't determine $f'(2)$ knowing every derivative of $f$ at 1. You cannot even determine $f'(2)$ knowing the value of $f(x)$ for all $x \in (-\infty, 3/2)$. If you assume $f$ is analytic or polynomial, things become more interesting. –  Hurkyl Oct 13 '12 at 15:31
A very elementary method would be to take the first difference of the equation $$\Delta f(x)=f(x+1) - f(x)$$ which would give you the secant through the two points. We can make a rough approximation that $$\frac{f'(x+1) + f'(x)}{2} \approx \Delta f(x)$$ which in turn gives $$f'(x+1) \approx 2\Delta f(x) - f'(x)$$ For this particular function, this works out quite well. It approximates $f'(2)$ perfectly.
The above method can be improved quite a bit by something called Taylor Polynomials (or more generally Taylor series). Any sufficiently smooth function can be closely approximated by a polynomial. The higher the degree of the polynomial, the smaller the error term. If we take our point of interest, say $a$, then we can write a Taylor polynomial for $f$ centered at $a$. In the case of your example, we can rewrite the function $f(x) = 3x^2$ centered at $x=1$ as $$f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2}(x-1)^2$$ You can differentiate this series to give $$f'(x) = f'(1) + f''(1)(x-1)$$ This is an expression for the derivative using only information from $x=1$. If you actually evaluate the expression, you will get $6x$ which is exactly as you would expect. Polynomials are a bit too simple too emphasize methods like this. In general things will not work out exactly as they have in this example.