Register to reply

Differantiation question ..

by transgalactic
Tags: differantiation
Share this thread:
tiny-tim
#55
Feb23-09, 02:16 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
I think what Dick is getting at is, what is the definition of f'(0)?
transgalactic
#56
Feb23-09, 05:10 AM
P: 1,398
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)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}

[/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.
transgalactic
#57
Feb23-09, 05:12 AM
P: 1,398
Quote Quote by D H View Post
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.
tiny-tim
#58
Feb23-09, 05:37 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by transgalactic View Post
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)}{x-0}=\lim _{x->0}\frac{f(x)-0}{x-0}[/tex]
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):

[tex]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}[/tex]

ok, so now you have
[tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}\ =[/tex]
Quote Quote by transgalactic View Post
[tex]\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}[/tex]
[tex]=\ \frac{-1}{2}\,\left(\lim _{x->0}\frac{f(x) }{x}\right)^2[/tex]

= ?
transgalactic
#59
Feb23-09, 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.
transgalactic
#60
Feb24-09, 03:06 AM
P: 1,398
how to get the last part?
tiny-tim
#61
Feb24-09, 03:42 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Look, transgalactic, this is screamingly obvious

since f(0) = 0, what is [tex]\lim _{x->0}\frac{f(x) }{x}[/tex] the definition of?
transgalactic
#62
Feb24-09, 04:08 AM
P: 1,398
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) }{x-0}
[/tex]
but i dont get a value
??
tiny-tim
#63
Feb24-09, 04:21 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by transgalactic View Post
i can see it in another way
[tex] f'(0)=\lim _{x->0}\frac{f(x)-f(0) }{x-0}[/tex]
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 Quote by tiny-tim View Post
ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ?
Quote Quote by transgalactic View Post
[tex]
\lim _{x->0}\frac{1\ -\ \frac{(f(x))^2}{2!} - 1}{x^2}=\lim _{x->0}\frac{\ -\ (f(x))^2 }{x^22!}
[/tex]
which = ?
transgalactic
#64
Feb24-09, 04:39 AM
P: 1,398
[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
??
tiny-tim
#65
Feb24-09, 05:02 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by tiny-tim View Post
ok, so what can you say about [tex]\lim _{x->0}\frac{cos(f(x)) - 1}{x^2}[/tex] ?
Quote Quote by transgalactic View Post
[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
Yes
Quote Quote by transgalactic View Post
f(x) and g(x) are differentiable on 0
f(0)=g(0)=0

calculate
[tex]\lim _{x->0}\frac{cosf(x)-cosg(x)}{x^2} [/tex]
which is the same as
[tex]\lim _{x->0}\frac{(cos(f(x)) - 1) - (cos(g(x)) - 1)}{x^2}[/tex]

which is ?
transgalactic
#66
Feb24-09, 11:24 AM
P: 1,398
i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result

??
tiny-tim
#67
Feb24-09, 11:53 AM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by transgalactic View Post
i think
[tex]
\frac{-f'(0)^2}{2!}+\frac{-g'(0)^2}{2!}
[/tex]
but its not a result

??
[tex]\frac{-f'(0)^2}{2!}+\frac{g'(0)^2}{2!}[/tex] actually


but why do you think that's not a result?
transgalactic
#68
Feb24-09, 12:45 PM
P: 1,398
because i was told
"calculate"
i here i have only an expression

??
tiny-tim
#69
Feb24-09, 01:09 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by transgalactic View Post
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?
transgalactic
#70
Feb24-09, 01:16 PM
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
??
tiny-tim
#71
Feb24-09, 01:22 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
Quote Quote by transgalactic View Post
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
??
yes yes yes!
no problemo!
go for it!
transgalactic
#72
Feb24-09, 01:31 PM
P: 1,398
thanks:)


Register to reply

Related Discussions
Differantiation proof question .. Calculus & Beyond Homework 11
Differantiation proof question .. Calculus & Beyond Homework 5
Differantiation proof question exlanation problem.. Calculus & Beyond Homework 2
Differantiation proof question.. Calculus & Beyond Homework 1
Differantiation proof question.. Calculus & Beyond Homework 3