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= ?

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- anonymous

- schrodinger

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- 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 ?

- anonymous

Im guessing a chain rule is applied here somehow

- anonymous

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

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- 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?

- 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

- anonymous

so you get it?

- anonymous

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

- anonymous

yes..

- 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..

- anonymous

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

- 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

- anonymous

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

- perl

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

- Astrophysics

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

- anonymous

thank you perl

- anonymous

Ahhh ! that's interesting how that works out.

- Astrophysics

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

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