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Derivative of e^(x2)by quasar987
Tags: derivative 
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#1
Nov1304, 04:10 PM

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I am asked to find the derivative function of [itex]f(x)=e^{x2}[/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^{x2}  e^{x_02}}{xx_0} = \lim_{x\rightarrow x_0}\frac{e^{x}  e^{x_0}}{e^2 (xx_0)}[/tex] How can this undeterminate form be simplified? Thanks. (The answer is [itex]f'(x_0)=e^{x_02}[/itex].) 


#2
Nov1304, 04:14 PM

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Factor out exp(x_{0}) from the numerator. Then expand exp(xx_{0}) in a power series. The rest is obvious.



#4
Nov1304, 06:09 PM

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Derivative of e^(x2)
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^{(x2)}) = \lim_{h \rightarrow 0} \frac{e^{(x2)+h}  e^{(x2)}}{h}[/tex] [tex]= \lim_{h \rightarrow 0} e^{(x2)} \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. 


#5
Nov1304, 07:12 PM

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[tex] h \equiv \Delta x \equiv x  x_0 [/tex] 


#6
Nov1304, 07:17 PM

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yea, thats all i meant.



#7
Nov1304, 07:57 PM

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This looks nice Status, but isn't there a [itex]e^{(x2)}[/itex] remaining?



#8
Nov1304, 08:58 PM

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Yes, which is multiplied by 1, giving the answer, as is easily verified using the chain rule.



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