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!

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- Dec 29th 2012, 06:27 AMmarijakopljarDerivative 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! - Dec 29th 2012, 06:42 AMskeeterRe: Derivative of ln(1+cosx)
- Dec 29th 2012, 07:38 AMHallsofIvyRe: 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 . What is the derivative of ln(u) with respect to u? What is the derivative of 1+ cos(x) with respect to x? - Dec 29th 2012, 09:12 AMmarijakopljarRe: Derivative of ln(1+cosx)
**Thank you!**

But what is procedure to get it?

Many thanks - Dec 29th 2012, 09:28 AMTutuRe: Derivative of ln(1+cosx)
Remember, always differentiate from the outside first.

When you differentiate lnx, you'll end up with

So, when you differentiate , you will get

Next, differentiate within the brackets, ie.

Differentiate 1, it will be 0. Differentiate you'll end up with

You have and ,

multiplying,

- Dec 29th 2012, 10:00 AMskeeterRe: Derivative of ln(1+cosx)
is a function of ...

- Dec 29th 2012, 10:08 AMHallsofIvyRe: 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. - Dec 29th 2012, 10:15 AMmarijakopljarRe: 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? - Dec 29th 2012, 11:14 AMPlatoRe: Derivative of ln(1+cosx)
- Dec 30th 2012, 06:31 AMskeeterRe: Derivative of ln(1+cosx)