Definitions of Derivative: $h \to 0$ vs $z \to x$

Question

I have come across two definitions of the derivative. The first is $$f'(x)= \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$

The second is $$f'(x) = \lim_{z\to x} \dfrac{f(z)-f(x)}{z-x}$$

I understand the first equation reflects an arbitrary secant getting closer and closer to a specific point of a function (as h approaches 0) to find the "instantaneous rate of change" at that point.

However, I do not understand where the second definition was derived from and what it represents?

Also, I often find that, practically, it's much easier to calculate derivatives from first principles using the second definition, but I don't understand why it works that way, is there some intuition I'm missing about the second definition.

Answer 1

In either case, the base point for the tangent line is at the point $(x,f(x))$.

In the first definition of the derivative, the nearby point is $(x+h,f(x+h))$. This formulation emphasizes the displacement ($h$ from the base point to the nearby point). As $h\to 0$, the nearby point approaches the base point, and so the slopes approach the slope of the tangent line.

In the second definition of the derivative, the nearby point is $(z,f(z))$. As $z\to x$, again the nearby point gets closer to the base point, and so again the slopes approach the slope of the tangent line.

Answer 2

I think you've noticed that we can write $h$ as $h = z - x$, so $f(x+h)=f(x+z-x)$ therefore equals $f(x)$. And, as we know from geometry, a tangent slope $m$ is definded as $m=\frac{y_2-y_1}{x_2-x_1}$, and a derivative of a function is also defined as a tangent slope. Therefore, in the context of derivation, $f'(x)=\frac{y_2-y_1}{x_2-x_1}=\frac{f(z)-f(x)}{z-x}$.