In More Derivative Example video lecture, at 4:44, this statement { you find that the amount that f(a) goes out is exactly equal to the derivative times the amount that you nudge a to the right } , this statement we cannot understand, can u give example how if we nudge very much small value to the right then how it becomes f(a) exactly equal to 2a ? so how f(a) = 2a for very much small value nudge to the right ?

If we nudge very much small instead of 0.001 how that little term get ignored ? can u give example in terms of math please ?

At 5.11 , this statement f(a) went up isn’t exactly given by the formula but it’s only a kind of approximately given by the derivative, here what does it mean kind of approximately given by the derivative ? can u please clarify ?

Hi @Anbu the derivative represent the istantaneus rate of change of a function. From a mathematical stand point is based on the concept of limit:

The function needs to be differentiable in a in order to calculate the derivative.
If the the derivate needs to be calculated by a computer you need a numerical approximation: taking a small increment and calculate the ratio is one possible method and is sometimes called finite difference method. But of course numerical differentiation is much more than that same for numerical derivative approximation. Maybe a good start could be from wiki
Hope this clarify your points

@crisrise sorry wiki hard to understaable , can u please help to answer to those three points please ? We would like to know how the derivative value time amount we nudge a to the right become exactly equal to the f(a) if we do nudge very small value a to the right ?

Hi @Anbu I try to explain it differently: the 0.001 in the video from Andrew is the h in the limit formula above. Assume you are at a = 2, the derivative from the calculus formula will tell that the derivative is 2 * a = 2 * 2 = 4.
Then if we try to calculate derivatives numerically (like in the video):

if the delta is 0.001 the approximated value of the derivative will be 4.001

if the delta is 0.000001 the approximated value of the derivative will be 4.000001
The message is: the smaller is the delta the lower the derivative approximation error will be.
Hope this can help you out.

Respected sir, I think there is mistake while calculating second order derivatives. f(a) = 4 for a=2 and f(a) = 4.002 for a=2.001 but it is mentioned that f(a)=4.004. And my major concern is that the derivatives of the second order equation is first order equation. And the first order equation is supposed to give same slope for every steps. So, isn’t it conflicting ? Thank you. Regards.