## anonymous one year ago How do you approach a problem like this ? GIVEN f[x] = a E^(r x) Calculate in terms of r the instantaneous growth rate of f[x] at any point x. Do they mean work out the derivative, and then re-arrange for r= ?

1. anonymous

or is it like a limit problem, where I drop the x altogether, with the idea that x is a constant if I am trying to take the derivative in terms of r ?

2. anonymous

Im guessing a chain rule is applied here somehow

3. anonymous

yep its asking for the derivative instantaneous growth of (anything) would be like d(anything) / dt

4. anonymous

I can see how I would get f'[x] but I think it's asking for f'[r] how would things change in that case?

5. anonymous

its asking f'(x) in terms of r f'(x) represent the instantaneous rate of change of f(x) at x whereas f'(r) represent the instantaneous rate of change of f(x) at r

6. anonymous

so you get it?

7. anonymous

so f'(x) in terms of r is different to f'(r) ?

8. anonymous

yes..

9. anonymous

well I understand how to calculate f'[x].. I get f'[x] = a r E^(r x) but I'm not sure what they mean by 'in terms of r..

10. anonymous

in terms of 'r' means as suppose an arbit situation f(x) = r^2-r the f(x) = terms in r

11. Astrophysics

You have the right idea, a simple example is saying solve x in terms of y, this essentially means x = f(y), so if you have something such as 5x-y=12, you would solve for x, x = (12+y)/5

12. anonymous

okay.. so something like r = Log[1/a] / x

13. perl

$\Large{ f(x) = ae^{rx} \\~\\ f'(x) = ae^{rx}\cdot r \\~\\ f'(x) = f(x)\cdot r \\~\\ \frac{f'(x)}{f(x)} = r }$

14. Astrophysics

Not sure, what exactly you did, but perl did it right :)

15. anonymous

thank you perl

16. anonymous

Ahhh ! that's interesting how that works out.

17. Astrophysics

It can be tricky at first with all the wording, but notice it's just a game of substitution in a way :P