My next question is how do you take the derivative of cos(lnx). This is what I do:
let u = lnx
f'(x) = u'(-sinu)
f'(x) = -1/x(sin(lnx))
When I check to see if I am write, I get the wrong answer. :S
I am going to assume that $\displaystyle u = \ln{x}$
$\displaystyle \frac{d}{dx} \cos(u) = -sin(u) \cdot \frac{du}{dx}$[/QUOTE]
$\displaystyle \frac{d}{dx} \cos(u) = -sin(lnx) \cdot \frac{1}{x}$
$\displaystyle \frac{d}{dx} \cos(u) = \frac{-sin(lnx)}{x}$
$\displaystyle \frac{d}{dx} \cos(u) = \frac{-sin(lnx)}{x}$ is the answer that I got, however, when I check it using graphing technology the slope of the tangent does not matches up for the x-values of what I use for the $\displaystyle f'(x) = \frac{-sin(lnx)}{x}$
My mode is in degrees. However, I believe that there is no way that three people can all get the same answer but be proven wrong by a graphing calculator, so I must of entered in something wrong.
How do you know how to use the chain rule correctly when it comes to natural logs? It seems that is my biggest issue at the moment with mathematics.
For calculus, you should be using radians. Things like the chain rule are always a bit of guess and check - just keep attempting questions and you'll get there. You will gradually find that the substitutions required become very straightforward - just think of anything that will make the differentiation easier and go from there.
the rule is stated in post #2; here is why it works ...
let $\displaystyle u$ be a function of $\displaystyle x$ and $\displaystyle y = \ln{u}$ (note this also makes $\displaystyle y$ a function of $\displaystyle x$)
change $\displaystyle y = \ln{u}$ to an exponential equation ...
$\displaystyle e^y = u$
$\displaystyle \frac{d}{dx} (e^y = u)$
$\displaystyle e^y \cdot \frac{dy}{dx}$ = $\displaystyle \frac{du}{dx}$
substitute $\displaystyle u$ for $\displaystyle e^y$ ...
$\displaystyle u \cdot \frac{dy}{dx} = \frac{du}{dx}$
isolate $\displaystyle \frac{dy}{dx}$ ...
$\displaystyle \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}$
$\displaystyle \frac{d}{dx}(y) = \frac{1}{u} \cdot \frac{du}{dx}$
$\displaystyle \frac{d}{dx}(\ln{u}) = \frac{1}{u} \cdot \frac{du}{dx}$
... that's all there is to it.