Explore a world of knowledge and get your questions answered on IDNLearn.com. Ask anything and receive prompt, well-informed answers from our community of experienced experts.
Sagot :
Given a function g(x), its derivative, if it exists, is equal to the limit
[tex]g'(x) = \displaystyle\lim_{h\to0}\frac{g(x+h)-g(x)}h[/tex]
The limit is some expression that is itself a function of x. Then the derivative of g(x) at x = 1 is obtained by just plugging x = 1. In other words, find g'(x) - and this can be done with or without taking a limit - then evaluate g' (1).
Alternatively, you can directly find the derivative at a point by computing the limit
[tex]g'(1) = \displaystyle\lim_{h\to0}\frac{g(1+h)-g(1)}h[/tex]
But this is essentially the same as the first method, we're just replacing x with 1.
Yet another way is to compute the limit
[tex]g'(1) = \displaystyle\lim_{x\to1}\frac{g(x)-g(1)}{x-1}[/tex]
but this is really the same limit with h = x - 1.
You do not compute g (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 number, you care about the derivative of a function that depends on a variable.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.