Derivative of ln(1+cosx)

• December 29th 2012, 05:27 AM
marijakopljar
Derivative of ln(1+cosx)
Dear All!
Please, if the f'(1+cosx)=1/(1+cosx) or sinx/(1+cosx)?
I would say the first one, but got the second one in a book. Is that a typo?
Many thanks!
• December 29th 2012, 05:42 AM
skeeter
Re: Derivative of ln(1+cosx)
$\frac{d}{dx} \left[\ln(1+\cos{x})\right] = -\frac{\sin{x}}{1+\cos{x}}$
• December 29th 2012, 06:38 AM
HallsofIvy
Re: Derivative of ln(1+cosx)
What kind of courses are you taking or have taken? If you are asked about derivatives, you should be taking Calculus- but then you should surely know that "f'(ln(1+ cos(x))" is just bad notation. You mean "if f(x)= ln(1+ cos(x)), what is f'(x)?" Now, do you know the chain rule? Clearly, you are expected to because it is needed here. You have f(u)= ln(u) with u= 1+ cos(x). Then $\frac{df}{dx}= \frac{df}{du}\frac{du}{dx}$. What is the derivative of ln(u) with respect to u? What is the derivative of 1+ cos(x) with respect to x?
• December 29th 2012, 08:12 AM
marijakopljar
Re: Derivative of ln(1+cosx)
Thank you!
But what is procedure to get it?
Many thanks
• December 29th 2012, 08:28 AM
Tutu
Re: Derivative of ln(1+cosx)
Remember, always differentiate from the outside first.
When you differentiate lnx, you'll end up with $\frac{1}{x}$
So, when you differentiate $ln(1+\cos\theta)$, you will get $\frac{1}{1+\cos\theta}$
Next, differentiate within the brackets, ie. $(1+\cos\theta)$
Differentiate 1, it will be 0. Differentiate $\cos\theta$ you'll end up with $\-sin\theta$
You have $-\sin\theta$ and $\frac{1}{1+\cos\theta}$,
multiplying,
$\frac{-\sin\theta}{1+\cos\theta}$
• December 29th 2012, 09:00 AM
skeeter
Re: Derivative of ln(1+cosx)
$u$ is a function of $x$ ...

$\frac{d}{dx} [\ln(u)] = \frac{u'}{u}$
• December 29th 2012, 09:08 AM
HallsofIvy
Re: Derivative of ln(1+cosx)
Tutu and skeeter are both referring to the chain rule. That is a very important rule for differentiation- look it up in your Calculus text.
• December 29th 2012, 09:15 AM
marijakopljar
Re: Derivative of ln(1+cosx)
Thank you very muc-it become clear in just a few minutes!
Please, what if we have derivative of ln(sinx+cosx)?

I supose, we first have 1/(sinx+cosx)
then we have cosx and (-sinx)
so we multiply 1/(sinx+cosx) with (cosx-sinx) and we get (cosx-sinx)/(sinx+cosx)
Am I right?
• December 29th 2012, 10:14 AM
Plato
Re: Derivative of ln(1+cosx)
Quote:

Originally Posted by marijakopljar

I supose, we first have 1/(sinx+cosx)
then we have cosx and (-sinx)
so we multiply 1/(sinx+cosx) with (cosx-sinx) and we get (cosx-sinx)/(sinx+cosx)
Am I right?

YES!
• December 30th 2012, 05:31 AM
skeeter
Re: Derivative of ln(1+cosx)
Quote:

Originally Posted by marijakopljar
if we have:
lim(x to 1) ((x/(x-1)-1/lnx)
then we get:

lim(x to 1) ((xlnx-x+1)/(xlnx-lnx))
then we take x=1:

lim(x to1)((ln(1)+1-1) / (ln(1)-ln(1))
we get 0/0
and can apply LH:

but NOW:
How do we get NOW lim(x to1)((xlnx)/(xlnx+x-1)???
What is the inversion doing here?
Many thanks!

in future, start a new problem with a new thread ... I have moved this post and re-titled it as Limit Problem