Given a function <em>g(x)</em>, its derivative, if it exists, is equal to the limit
![g'(x) = \displaystyle\lim_{h\to0}\frac{g(x+h)-g(x)}h](https://tex.z-dn.net/?f=g%27%28x%29%20%3D%20%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7Dh)
The limit is some expression that is itself a function of <em>x</em>. Then the derivative of <em>g(x)</em> at <em>x</em> = 1 is obtained by just plugging <em>x</em> = 1. In other words, find <em>g'(x)</em> - and this can be done with or without taking a limit - then evaluate <em>g'</em> (1).
Alternatively, you can directly find the derivative at a point by computing the limit
![g'(1) = \displaystyle\lim_{h\to0}\frac{g(1+h)-g(1)}h](https://tex.z-dn.net/?f=g%27%281%29%20%3D%20%5Cdisplaystyle%5Clim_%7Bh%5Cto0%7D%5Cfrac%7Bg%281%2Bh%29-g%281%29%7Dh)
But this is essentially the same as the first method, we're just replacing <em>x</em> with 1.
Yet another way is to compute the limit
![g'(1) = \displaystyle\lim_{x\to1}\frac{g(x)-g(1)}{x-1}](https://tex.z-dn.net/?f=g%27%281%29%20%3D%20%5Cdisplaystyle%5Clim_%7Bx%5Cto1%7D%5Cfrac%7Bg%28x%29-g%281%29%7D%7Bx-1%7D)
but this is really the same limit with <em>h</em> = <em>x</em> - 1.
You do not compute <em>g</em> (1) first, because as you say, that's just a constant, so its derivative is zero. But you're not concerned with the derivative of some <em>number</em>, you care about the derivative of a function that depends on a <em>variable.</em>