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

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.

• Define z:=x+h in the first one. – Tito Eliatron Nov 1 '18 at 18:18

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.
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, $$f'(x)=\frac{y_2-y_1}{x_2-x_1}=\frac{f(z)-f(x)}{z-x}$$.