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c3 exponentials.. watch

1. im quite confused about c3 logarithms as i missed all my lessons on it due to a course, and now im having to teach myself.. but the book isnt very explanatory.. im having trouble finding the gradient.. im pretty sure its not just differentiating the expression y = e^x?

Thanks in advance..
2. And by L'Hoptial once the limit will be and thus the derivative of e^x is e^x...

Sorry don't really know what your question was
3. Nope, that was his question

The differential of e^x is e^x. Odd
4. (Original post by nota bene)
And by L'Hoptial once the limit will be and thus the derivative of e^x is e^x...
How can you use l'Hopital's rule when it relies on knowing what the derivative of e^x is in the first place??
5. exactly..

edit: for the post above the one above..

well in the book they just seem to have a gradient already.. they dont explain how they got it..

for example.. the gradient of y = 2^x is 0.7 x 2^x...

why?!?
6. firstly, for A level, you're not going to need to "prove" anything about the derivative of e^x - you just take it as given that its derivative is e^x.

now, if y = a^x, then taking logs gives you ln y = x ln a so you can rewrite the original equation as y = e^(x ln a).

since ln a is just a constant, you can differentiate this to get:

dy/dx = (ln a) e^(x ln a) = (ln a) a^x.

In your case, a= 2 so the constant that appears when you differentiate is just ln 2 which is approx 0.7.
7. (Original post by nota bene)
And by L'Hoptial once the limit will be and thus the derivative of e^x is e^x...

Sorry don't really know what your question was
No. I think the proof of e^x consists of it's maclaurin expansion being differentiated.
8. (Original post by Pencil King)
No. I think the proof of e^x consists of it's maclaurin expansion being differentiated.
But the Maclaurin series still relies on knowing f'(0), f''(0) etc and plugging these into a formula - i.e you have to differentiate your function f!!

A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
9. umm.. thanks for your help.. but im guessing it doesn't yet have anything to do with maclaurin or l'hoptial.. as i have no idea what they are and the book doesnt seem to mention them anyway... ?
10. (Original post by davros)
But the Maclaurin series still relies on knowing f'(0), f''(0) etc and plugging these into a formula - i.e you have to differentiate your function f!!

A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
Agreed.
11. (Original post by n0b0dy)
umm.. thanks for your help.. but im guessing it doesn't yet have anything to do with maclaurin or l'hoptial.. as i have no idea what they are and the book doesnt seem to mention them anyway... ?
That's right - if you look at my post at 11:44, that explains what you need to know / do...

Post back if you're still confused
12. hmm.. uve explained it really well.. but i think i need to look over logarithms again.. summer holidays r not good..

Thank u very much all..
13. (Original post by davros)
A proper proof would start from defining e^x as a power series, proving that it converges, and showing that it can be differentiated / integrated using theorems of analysis. That's why you won't find it in an A level course...
Yeah, I was talking crap up there

and by the ratio test , which converges by comparison with (converges by integral test).

Now, is this along the right lines?
14. (Original post by nota bene)
(converges by integral test).

Now, is this along the right lines?
Er, are you sure that sum converges? I was under the impression that the harmonic series DIVERGED
15. (Original post by davros)
Er, are you sure that sum converges? I was under the impression that the harmonic series DIVERGED
it does......

I'm not having a good day
16. (Original post by nota bene)
it does......

I'm not having a good day
no problem...

the ratio test will give you what you want - you just need to re-read how to use it. Basically, it's the LIMIT of the ratio that tells you whether the original series converges (not the SUM of the ratios)...
17. (Original post by nota bene)
Yeah, I was talking crap up there

and by the ratio test , which converges by comparison with [IMG]http://thestudentroom.co.uk/latexrender/pictures/1651c3444f1b741558fb6****bec9e80 .png[/IMG] (converges by integral test).

Now, is this along the right lines?
the first rule of math club: the harmonic series diverges! ;yes;

i don't think you need the integral test though, because there is a general result about sum (1/n)^p, converging if p is more 1, diverging if p is less than or equal to one, and i don't really remember about zero! (which you can use once proved).
18. (Original post by h.b)
the first rule of math club: the harmonic series diverges! ;yes;
You don't have to rub it in my face!
i don't think you need the integral test though, because there is a general result about sum (1/n)^p, converging if p is more 1, diverging if p is less than or equal to one, and i don't really remember about zero! (which you can use once proved).
You use the integral test to prove for which values of p the 1/n^p series converges, so yes I think I need the integral test unless I can assume the result to be true. About 0, what can we say about n^0 ? (it's 1), hence the convergence/divergence would seemingly only depend on the numerator...?
19. (Original post by nota bene)
You don't have to rub it in my face!

You use the integral test to prove for which values of p the 1/n^p series converges, so yes I think I need the integral test unless I can assume the result to be true. About 0, what can we say about n^0 ? (it's 1), hence the convergence/divergence would seemingly only depend on the numerator...?
the first rule of... ach fine i won't.

(i don't really know much about the integral test so i'll take your word for it. but if we can prove that the harmonic series diverges and that (1/n)^2 converges, then the comparison test can always be used i think).

about the zero thing - i'm not sure. i was told something about this but can't remember.

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