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Differantiation question ..by transgalactic
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#55
Feb2309, 02:16 AM

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I think what Dick is getting at is, what is the definition of f'(0)?



#56
Feb2309, 05:10 AM

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i dont know what is the value of f'(0)
i know that f(0)=0 [tex] f'(x)=\lim _{x>0}\frac{f(x)f(0)}{x0}=\lim _{x>0}\frac{f(x)0}{x0} [/tex] this is the definition of the derivative i dont know how to continue you said also "Difference quotient" so i used [tex] f'(x)=\lim _{h>0}\frac{f(x+h)f(x)}{h} [/tex] but i dont have any values for it. 


#57
Feb2309, 05:12 AM

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i dont know. 


#58
Feb2309, 05:37 AM

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[tex]f'(0)=\lim _{x>0}\frac{f(x)f(0)}{x0}=\lim _{x>0}\frac{f(x)0}{x0}=\lim _{x>0}\frac{f(x)}{x}[/tex] ok, so now you have … [tex]\lim _{x>0}\frac{cos(f(x))  1}{x^2}\ =[/tex] = … ? 


#59
Feb2309, 05:51 AM

P: 1,398

ok i am doing that as a shot in the dark
inspite of the fact the f(x>0) differs f(0) and i put the given f(0)=0 [tex] \ \frac{1}{2}\,\left(\lim _{x>0}\frac{f(x) }{x}\right)^2=\ \frac{1}{2}\,\left(\lim _{x>0}\frac{0 }{0}\right)^2= [/tex] so i dont know how to solve it. 


#60
Feb2409, 03:06 AM

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how to get the last part?



#61
Feb2409, 03:42 AM

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Look, transgalactic, this is screamingly obvious …
since f(0) = 0, what is [tex]\lim _{x>0}\frac{f(x) }{x}[/tex] the definition of? 


#62
Feb2409, 04:08 AM

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i dont know
i get 0 n the numerator and 0 in the denominator i can see it in another way [tex] f'(0)=\lim _{x>0}\frac{f(x)f(0) }{x0} [/tex] but i dont get a value ?? 


#63
Feb2409, 04:21 AM

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Why do you have a mental block about these things? As you say, that limit is f'(0) … ok, now go back to posts #4950 … 


#64
Feb2409, 04:39 AM

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[tex]
\lim _{x>0}\frac{1\ \ \frac{(f(x))^2}{2!}  1}{x^2}=\lim _{x>0}\frac{\ \ (f(x))^2 }{x^22!}=\frac{f'(0)^2}{2!} [/tex] but i was asked to calculate and it doesnt give me a result ?? 


#65
Feb2409, 05:02 AM

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[tex]\lim _{x>0}\frac{(cos(f(x))  1)  (cos(g(x))  1)}{x^2}[/tex] which is … ? 


#66
Feb2409, 11:24 AM

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i think
[tex] \frac{f'(0)^2}{2!}+\frac{g'(0)^2}{2!} [/tex] but its not a result ?? 


#67
Feb2409, 11:53 AM

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… but why do you think that's not a result? 


#68
Feb2409, 12:45 PM

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because i was told
"calculate" i here i have only an expression ?? 


#69
Feb2409, 01:09 PM

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no, an expression is ok … i admit "calculate" usually means a number … but it isn't an official word, it's just another way of saying "work out" is that all that was bothering you? 


#70
Feb2409, 01:16 PM

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i havent been given f'(0)
i was told that it was differentiable on point 0. but i dont know what thing are given so i can use them into the solution expression ?? 


#72
Feb2409, 01:31 PM

P: 1,398

thanks:)



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