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

Printable View

- Nov 10th 2011, 02:47 PMBarthaynDerivative of cos(ln(x))
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 - Nov 10th 2011, 02:50 PMskeeterRe: How to take the derivative of ln(lnx)?
again,

**the chain rule**...

$\displaystyle \frac{d}{dx} \cos(u) = -sin(u) \cdot \frac{du}{dx}$ - Nov 10th 2011, 02:55 PMBarthaynRe: How to take the derivative of ln(lnx)?

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 $\displaystyle f'(x) = \frac{-sin(lnx)}{x}$ - Nov 10th 2011, 03:01 PMQuackyRe: How to take the derivative of ln(lnx)?
Your derivative is fine. More likely the error is with what you're entering into the graphing technology.

- Nov 10th 2011, 03:03 PMskeeterRe: How to take the derivative of ln(lnx)?
mode switch in radians?

- Nov 10th 2011, 03:15 PMBarthaynRe: How to take the derivative of ln(lnx)?
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. - Nov 10th 2011, 03:33 PMQuackyRe: How to take the derivative of ln(lnx)?
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. - Nov 10th 2011, 04:26 PMskeeterRe: How to take the derivative of ln(lnx)?
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.