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!
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 $\displaystyle \frac{df}{dx}= \frac{df}{du}\frac{du}{dx}$. What is the derivative of ln(u) with respect to u? What is the derivative of 1+ cos(x) with respect to x?
Remember, always differentiate from the outside first.
When you differentiate lnx, you'll end up with $\displaystyle \frac{1}{x}$
So, when you differentiate $\displaystyle ln(1+\cos\theta)$, you will get $\displaystyle \frac{1}{1+\cos\theta}$
Next, differentiate within the brackets, ie. $\displaystyle (1+\cos\theta)$
Differentiate 1, it will be 0. Differentiate $\displaystyle \cos\theta$ you'll end up with $\displaystyle \-sin\theta$
You have $\displaystyle -\sin\theta$ and $\displaystyle \frac{1}{1+\cos\theta}$,
multiplying,
$\displaystyle \frac{-\sin\theta}{1+\cos\theta}$
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?