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>> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j) > Ar
by ColinWright 5y ago
>> you can discover that for f(x)=x^2 you will find that when the jiggling (call it j) is small, around a value x, the amount of change is roughly (2x)(j)
> Are you implying that the change corresponds to the derivative of the function, or is that just a coincidence in this case?
When you have a function f(x), the amount that f(x) changes as you jiggle the value x is roughly given by the amount of jiggle times the derivative f'(x) [also known by many other names]. That's more-or-less what "derivative" means in this context.
So under appropriate and rather generous conditions:
f(x+j) ~ f(x) + j.f'(x)
Does that answer your question? I've answered the question I think you asked, but that's not always the the question you intended.
- coolgeek 5y agoI guess the question I was really asking was is the gradient of a function equivalent to the derivative of the function? But a quick look at the wiki page for gradient suggests that my calc I/II intuition of derivatives is insufficient here
- ColinWright 5y agoFor reasonable functions f:R->R, yes, the gradient of the graph of the function is the value of the derivative at that point. But "Gradient" is a term that's used in wider contexts, so the real answer is "it depends". But I've had a look at this page: https://en.wikipedia.org/wiki/Gradient https://en.wikipedia.org/wiki/Gradient And that really does say that the "Gradient" is basically the derivative, suitably generalised for larger dimensions.