# Trig derivative

1. May 27, 2006

### ultima9999

What is the derivative of $$cos\frac{1}{x}$$?

Also, would $$cos\frac{1}{x}$$ equal 0 or infinity?

Last edited: May 27, 2006
2. May 27, 2006

### neutrino

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: $$x = \frac{2}{\pi}$$)

3. May 27, 2006

### MathematicalPhysicist

if you put: x=1/2pi*n
you get cos' (2pi*n)=(-2pi)sin(1/x)

4. May 27, 2006

### ultima9999

I attempted to use the chain rule and I got $$-\frac{sin}{x^2}\left(\frac{1}{x}\right)$$

I'm trying to solve a limit where $$\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right]$$.

So I'm using Indeterminate forms and L'Hopital's rule to try and solve it.

Last edited: May 27, 2006
5. May 27, 2006

### MathematicalPhysicist

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.

Last edited: May 27, 2006
6. May 27, 2006

### neutrino

$$\frac{1}{x^2}sin(\frac{1}{x})$$ is correct. The two minus's cancel.

EDIT: Stupid me...getting this Latex all messed up! Sorry. Rewrite that as
$$\lim_{x \rightarrow 0} \left[3x^4\frac{cos\left{(\frac{1}{x}\right)\right}{\frac{1}{x}}]$$

Last edited: May 27, 2006
7. May 27, 2006

### ultima9999

Alright, thanks for the first part.

So $$\frac{1}{x}, x = 0$$, would be infinity and so, $$sin(\infty)$$ and $$cos(\infty)$$ would be indeterminate?

8. May 27, 2006

### neutrino

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.

9. May 27, 2006

### VietDao29

Err, sorry to interrupt you guys, but why can't it (i.e the limit) exist?

10. May 27, 2006

### neutrino

I'm back to embarrass myself, again. The limit exists and, if I'm correct, it is zero.