First principles differentiation question

  • Thread starter tunabeast
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  • #1
tunabeast
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Homework Statement


Obtain the derivative of [tex]e^{2x}[/tex] from first principles.


Homework Equations





The Attempt at a Solution


I am upto the point where i have [tex]e^{2x}(\frac{e^{2h}-1}{h})[/tex] Where LIM h->0.

I just don't see how 2 is obtained from what's in the ().
 

Answers and Replies

  • #2
HallsofIvy
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[tex]\frac{e^{2h}-1}{h}= 2\frac{e^{2h}-1}{2h}= 2\frac{e^u -1}{u}[/tex]
if you let u= 2x.

Now, do you know what the limit, as u goes to 0 of
[tex]\frac{e^u- 1}{u}[/tex]
is?
 
  • #3
tunabeast
27
0
It tends to 1? I think i understand this now, thanks very much for your help.
 

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