# What to use and how to best do it?

1. Aug 31, 2014

### SteliosVas

Given: cos(x)$\frac{1}{x}$

Find $\frac{dy}{dx}$:

Now I know... $\frac{dy}{dx}$ of cos(x) = -sin(x)

and $\frac{dy}{dx}$ of $\frac{1}{x}$= $\frac{-1}{x^2}$

But I am unsure what formula to apply, and where to go from here?

2. Aug 31, 2014

### PeroK

Try using a logarithm.

3. Aug 31, 2014

### SteliosVas

I get confused when I convert it to logarithmic.
E.g I get the base as cos(x) and inside the log as y =1/x
So...

4. Aug 31, 2014

### PeroK

$$y = cos(x)^{\frac{1}{x}}$$
$$ln(y) = ln(cos(x)^{\frac{1}{x}})= \frac{1}{x}ln(cos(x))$$
Now differentiate both sides using the chain rule.

5. Aug 31, 2014

### SteliosVas

Thanks very much for your help

Much appreciated !

6. Aug 31, 2014

### Staff: Mentor

Since y doesn't appear in your first line above, it doesn't make much sense to talk about dy/dx. If we know that y = (cos(x))1/x, then dy/dx is meaningful.
I understand what you're trying to say, but you're not saying it correctly. You don't take "dy/dx" of something. dy/dx is already the derivative of y (with respect to x). To indicate that you want to take the derivative with respect to x of cos(x), write d/dx(cos(x)).
You mean d/dx(1/x).

7. Aug 31, 2014

### Shinaolord

&& the quotient rule. Correct?

8. Aug 31, 2014

### PeroK

I never use the quotient rule.

9. Aug 31, 2014

### Shinaolord

Than what of $x^{-1} * ln(cos(x))$??
That's a product. Shouldn't either product or quotient rule be applied?

10. Aug 31, 2014

### PeroK

I always use the product rule and chain rule. The quotient rule is too much to remember. And, unnecessary.

11. Aug 31, 2014

### Shinaolord

I do as well. I just stated it as the quotient rule because of the way you wrote it.