What to use and how to best do it?

[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?

 

[PeroK]

Try using a logarithm.

 

[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...

 

[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.

 

[SteliosVas]

Thanks very much for your help

Much appreciated !

 

[Mark44]

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

Find \frac{dy}{dx}:

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.

Now I know... \frac{dy}{dx} 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)).

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

You mean d/dx(1/x).

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

 

[Shinaolord]

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.

&& the quotient rule. Correct?

 

[PeroK]

&& the quotient rule. Correct?

I never use the quotient rule.

 

[Shinaolord]

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

 

[PeroK]

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.

 

[Shinaolord]

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