Derivative help

• Jun 11th 2010, 05:43 AM
Glitch
Derivative help
Question:

f(x) = ln(cosx)

I need to find the derivative.

I'm unsure where to start. I tried the chain rule, and I get stuck, as follows:

$\displaystyle \frac {1}{x} (cosx)^0 (-sinx)$

$\displaystyle \frac {-sinx}{x}$

The answer is supposed to be -tanx
• Jun 11th 2010, 05:48 AM
pickslides
Quote:

Originally Posted by Glitch
f(x) = ln(cosx)

I need to find the derivative.

make $\displaystyle u = \cos x \implies y = \ln u$

$\displaystyle \frac{du}{dx} = -\sin x$ and $\displaystyle \frac{dy}{du}= \frac{1}{u}$

$\displaystyle \frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}= \frac{1}{u}\times -\sin x = \frac{-\sin x}{\cos x} = -\tan x$
• Jun 11th 2010, 05:52 AM
Glitch
Interesting. So that's the chain rule? Your working out isn't very familiar to me.
• Jun 11th 2010, 06:04 AM
Raoh
Quote:

Originally Posted by Glitch
Interesting. So that's the chain rule? Your working out isn't very familiar to me.

hi (Happy)
if u have two functions of $\displaystyle x$,$\displaystyle f$ and $\displaystyle g$,how do you find the derivative of $\displaystyle f(g(x))$ ?
• Jun 11th 2010, 06:11 AM
Glitch
I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.
• Jun 11th 2010, 06:22 AM
Raoh
Quote:

Originally Posted by Glitch
I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

well ..
$\displaystyle \left (\ln f(x) \right )'=\frac{f'(x)}{f(x)}$
$\displaystyle \left (\ln \cos(x) \right )'=\frac{-\sin(x)}{\cos(x)}$
$\displaystyle (\cos(x))'=-\sin(x)$
• Jun 11th 2010, 06:26 AM
Glitch
Yeah, I made an error. Thanks for the help!
• Jun 18th 2010, 06:48 AM
chisigma
A general formula for the 'logarithmic derivative' of a function is...

d/dx ln f(x) = f'(x)/f(x) (1)

... and here is f(x) = cos x , so that...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Jun 18th 2010, 07:58 AM
TheCoffeeMachine
Quote:

Originally Posted by Glitch
I find the derivative of g(x), then the derivative of f(g(x)), then multiply them. I think the bit that confuses me is since the function is the natural logarithm, you don't lower the power like I did in the OP. Or maybe you do, and I'm going nuts.

Why were you lowering the power, though? The derivative of ln{x} is 1/{x} and the derivative of cosx is -sinx. That should be sufficient:

http://latex.codecogs.com/gif.latex?...0=%20-\tan{x}.