- #1
ultima9999
- 43
- 0
What is the derivative of [tex]cos\frac{1}{x}[/tex]?
Also, would [tex]cos\frac{1}{x}[/tex] equal 0 or infinity?
Also, would [tex]cos\frac{1}{x}[/tex] equal 0 or infinity?
Last edited:
Please post your attempt(s) in finding it.ultima9999 said:What is the derivative of [tex]\textrm(cos)\frac{1}{x}[/tex]?
The cosine function can have values only within the interval [-1,1], and has the value 0 for certain values of x. (an example in this case: [tex] x = \frac{2}{\pi} [/tex])Also, would [tex]\textrm(cos)\frac{1}{x} = 0 \textrm(or) \infty[/tex]?
[tex]\frac{1}{x^2}sin(\frac{1}{x}) [/tex] is correct. The two minus's cancel.ultima9999 said:I attempted to use the chain rule and I got [tex]-\frac{sin}{x^2}\right(\frac{1}{x}\left)[/tex]
EDIT: Stupid me...getting this Latex all messed up! Sorry. Rewrite that asI'm trying to solve a limit where [tex]\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right][/tex].
So I'm using Indeterminate forms and L'Hopital's rule to try and solve it.
I'm really sorry. As I said I'm getting both the logic and the LATEX wrong. I'll let someone better tackle this while I go hide my face from the public.ultima9999 said:Alright, thanks for the first part.
So [tex]\frac{1}{x}, x = 0[/tex], would be infinity and so, [tex]sin(\infty)[/tex] and [tex]cos(\infty)[/tex] would be indeterminate?
Err, sorry to interrupt you guys, but why can't it (i.e the limit) exist?loop quantum gravity said:my mistake it should be:
(1/x^2)sin(1/x)
edit:
about the limit, i don't think it exists, cause as i said it's cos y wher y approaches infinity, which is not defined.
The derivative of cos(1/x) is undefined at x=0, but for all other values of x, it is equal to 0.
The derivative is undefined at x=0 because the function is not continuous at that point. As x approaches 0, the function oscillates infinitely between -1 and 1, making it impossible to determine a single value for the derivative.
The derivative of cos(1/x) is equal to 0 for all values of x except for x=0, where it is undefined. Therefore, it is neither equal to 0 nor infinity for the entire domain of the function.
The derivative of cos(1/x) can be found using the chain rule, which states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In this case, the outer function is cos(x) and the inner function is 1/x.
No, the derivative of cos(1/x) cannot be approximated at x=0 because the function is not continuous at that point, and therefore does not have a well-defined tangent line. Any attempt to approximate the derivative at x=0 would result in an infinite or undefined value.