What to use and how to best do it?

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SteliosVas
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Given: cos(x)[itex]\frac{1}{x}[/itex]

Find [itex]\frac{dy}{dx}[/itex]:

Now I know... [itex]\frac{dy}{dx}[/itex] of cos(x) = -sin(x)

and [itex]\frac{dy}{dx}[/itex] of [itex]\frac{1}{x}[/itex]= [itex]\frac{-1}{x^2}[/itex]

But I am unsure what formula to apply, and where to go from here?
 
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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...
 
Thanks very much for your help

Much appreciated !
 
SteliosVas said:
Given: cos(x)[itex]\frac{1}{x}[/itex]

Find [itex]\frac{dy}{dx}[/itex]:
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.
SteliosVas said:
Now I know... [itex]\frac{dy}{dx}[/itex] of cos(x) = -sin(x)
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)).
SteliosVas said:
and [itex]\frac{dy}{dx}[/itex] of [itex]\frac{1}{x}[/itex]= [itex]\frac{-1}{x^2}[/itex]
You mean d/dx(1/x).
SteliosVas said:
But I am unsure what formula to apply, and where to go from here?
 
PeroK said:
[tex]y = cos(x)^{\frac{1}{x}}[/tex]
[tex]ln(y) = ln(cos(x)^{\frac{1}{x}})= \frac{1}{x}ln(cos(x))[/tex]
Now differentiate both sides using the chain rule.
&& the quotient rule. Correct?
 
Than what of ##x^{-1} * ln(cos(x))##??
That's a product. Shouldn't either product or quotient rule be applied?
 
Shinaolord said:
Than what of ##x^{-1} * ln(cos(x))##??
That's a product. Shouldn't either product or quotient rule be applied?

I always use the product rule and chain rule. The quotient rule is too much to remember. And, unnecessary.
 
I do as well. I just stated it as the quotient rule because of the way you wrote it.