1. Feb 29, 2012

### mnb96

Hello,
considering the definition of derivative, what would the following quantity be equal to?
$$\lim_{\delta \to 0} \frac{f(x+g(\delta))-f(x)}{g(\delta)}$$

In this case $g(\delta)$ is a monotonic increasing function such that g(0)=0.
For example we might have $g(\delta)=\delta^3$

2. Feb 29, 2012

### A. Bahat

It depends on the function g. For example, if g(δ)→0 as δ→0, then this limit is just f'(x). If, say, g(δ)=1 for all δ, the limit gives the result of a "difference operator": f(x+1)-f(x).

3. Feb 29, 2012

### Strafespar

If you are confused about derivatives in general..All that equation does is tricks the system into finding the tangent of the "original" graph at every single point (f(x) value). If you think about it to find the tangent of a point on an equation, you are basically finding an infinitesimally small secant line between two infinitesimally small changes in the position of the graph. The tangent line at some x value would then be evaluated by substituting that value in the x value of the equation you wrote. But if you keep the x a variable your result is an equation not a single value. The equation you get is actually just the slope of the tangent line at every point of the original function. For example the derivative of the function x^2 is 2x. meaning that in x^2 at x= 3 the slope of the tangent line is going to be some #. The way we can get this number is by substituting 3 into the derivative equation 2x. I would definitely recommend watching the khan academy calculus videos on youtube if you run into any future problems.

4. Mar 1, 2012

### mnb96

Thanks.

I understand the geometrical interpretation on the first derivative. I am still a bit suspicious because I realized I am not able to formally prove that if, for example, $g(\delta)=\delta^3$, the quantity I wrote in the first post is exactly equal to the definition of derivative.

Can anyone help me prove this?

5. Mar 1, 2012

### Char. Limit

If you take the Taylor series of f(x+d^3) around d=0, you get a formula that looks something like this:

$$\sum_{n=0}^\infty \frac{d^{3n}}{n!} \frac{d^n f(x)}{d x^n}$$

Subtract f(x) from this and divide by d^3 to get the following:

$$\sum_{n=1}^\infty \frac{d^{3n-3}}{n!} \frac{d^n f(x)}{d x^n}$$

Note that n=1 is our starting point here, not n=0. From there, if we expand it out for a bit, we get something like this:

$$f'(x) + d^3 f''(x) + d^6 f'''(x) + \dots$$

Taking the limit of this as d goes to zero gives us f(x), thus proving that the equation listed in the OP is the derivative, at least for g(d)=d^3.

6. Mar 1, 2012

### Strafespar

Oh so thats what he/she was asking lol