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Derivative of e^(x-2)

by quasar987
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quasar987
#1
Nov13-04, 04:10 PM
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I am asked to find the derivative function of [itex]f(x)=e^{x-2}[/itex] using the definition. That is to say, I have to evaluate this limit, if it exists:

[tex]\lim_{x\rightarrow x_0}\frac{e^{x-2} - e^{x_0-2}}{x-x_0} = \lim_{x\rightarrow x_0}\frac{e^{x} - e^{x_0}}{e^2 (x-x_0)}[/tex]

How can this undeterminate form be simplified? Thanks.

(The answer is [itex]f'(x_0)=e^{x_0-2}[/itex].)
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mathman
#2
Nov13-04, 04:14 PM
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Factor out exp(x0) from the numerator. Then expand exp(x-x0) in a power series. The rest is obvious.
quasar987
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Nov13-04, 04:15 PM
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Power serie not allowed, sorry.

StatusX
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Nov13-04, 06:09 PM
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Derivative of e^(x-2)

im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:

[tex]\frac{d}{dx}(e^{(x-2)}) = \lim_{h \rightarrow 0} \frac{e^{(x-2)+h} - e^{(x-2)}}{h}[/tex]

[tex]= \lim_{h \rightarrow 0} e^{(x-2)} \frac{e^{h} -1}{h}[/tex]

now noting that:

[tex]e = \lim_{h \rightarrow 0} (1+h)^{1/h}[/tex]

raising both sides to h (this is the only step I'm not sure about, but it gives the right answer):

[tex]\lim_{h \rightarrow 0} e^h = \lim_{h \rightarrow 0} ((1+h)^{1/h})^h = \lim_{h \rightarrow 0} 1+h [/tex]

so:
[tex]\lim_{h \rightarrow 0} \frac{e^{h} -1}{h}= \lim_{h \rightarrow 0} \frac{1 + h -1}{h}= 1[/tex]

which gives the answer.
cepheid
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Nov13-04, 07:12 PM
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Quote Quote by StatusX
im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:
lol, nobody is defining derivatives in different ways...only the notation differs

[tex] h \equiv \Delta x \equiv x - x_0 [/tex]
StatusX
#6
Nov13-04, 07:17 PM
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yea, thats all i meant.
quasar987
#7
Nov13-04, 07:57 PM
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This looks nice Status, but isn't there a [itex]e^{(x-2)}[/itex] remaining?
DeadWolfe
#8
Nov13-04, 08:58 PM
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Yes, which is multiplied by 1, giving the answer, as is easily verified using the chain rule.
cepheid
#9
Nov14-04, 01:56 AM
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Quote Quote by StatusX
yea, thats all i meant.
oh ok, sorry


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