# Differantiation question ..

by transgalactic
Tags: differantiation
 Sci Advisor HW Helper Thanks P: 26,160 I think what Dick is getting at is, what is the definition of f'(0)?
 P: 1,398 i dont know what is the value of f'(0) i know that f(0)=0 $$f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}$$ this is the definition of the derivative i dont know how to continue you said also "Difference quotient" so i used $$f'(x)=\lim _{h->0}\frac{f(x+h)-f(x)}{h}$$ but i dont have any values for it.
P: 1,398
 Quote by D H So what is f(x) near zero? Tiny-tim, please do not give this away.
i gave every option i can think of.
i dont know.
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P: 26,160
 Quote by transgalactic i dont know what is the value of f'(0) i know that f(0)=0 $$f'(x)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}$$ this is the definition of the derivative i dont know how to continue
transgalactic, that isn't the definition of f'(x), it's the definition of f'(0) (using x instead of the more usual h, and since f(0) = 0):

$$f'(0)=\lim _{x->0}\frac{f(x)-f(0)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}=\lim _{x->0}\frac{f(x)}{x}$$

ok, so now you have …
$$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}\ =$$
 Quote by transgalactic $$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}$$
$$=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2$$

= … ?
 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 $$\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{0 }{0}\right)^2=$$ so i dont know how to solve it.
 P: 1,398 how to get the last part?
 Sci Advisor HW Helper Thanks P: 26,160 Look, transgalactic, this is screamingly obvious … since f(0) = 0, what is $$\lim _{x->0}\frac{f(x) }{x}$$ the definition of?
 P: 1,398 i dont know i get 0 n the numerator and 0 in the denominator i can see it in another way $$f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}$$ but i dont get a value ??
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P: 26,160
 Quote by transgalactic i can see it in another way $$f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}$$
Yes, that's it!

Why do you have a mental block about these things?

As you say, that limit is f'(0) …

ok, now go back to posts #49-50 …
 Quote by tiny-tim ok, so what can you say about $$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}$$ ?
 Quote by transgalactic $$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}$$
which = … ?
 P: 1,398 $$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}$$ but i was asked to calculate and it doesnt give me a result ??
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P: 26,160
 Quote by tiny-tim ok, so what can you say about $$\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}$$ ?
 Quote by transgalactic $$\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}=\frac{-f'(0)^2}{2!}$$ but i was asked to calculate
Yes …
 Quote by transgalactic f(x) and g(x) are differentiable on 0 f(0)=g(0)=0 calculate $$\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2}$$
which is the same as …
$$\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}$$

which is … ?
 P: 1,398 i think $$\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}$$ but its not a result ??
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P: 26,160
 Quote by transgalactic i think $$\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}$$ but its not a result ??
$$\frac{-f'(0)^2}{2!}+\frac{g'(0)^2}{2!}$$ actually

but why do you think that's not a result?
 P: 1,398 because i was told "calculate" i here i have only an expression ??
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P: 26,160
 Quote by transgalactic because i was told "calculate" i here i have only an expression ??
oh i see!

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?
 P: 1,398 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 ??