I am trying to differentiate the function:
$${\rm ln} \left(\frac{3x \ {\rm tan}(x)}{x^2 + 2}\right)$$
I think step one is to use the quotient rule of natural log expanding the expression.
However doing this would still leave $\ln(3x \tan(x)) – \ln(x^2+2) $. Therefore, I'm confused if I should expand this expression again giving me…..
$$ \ln(3x) + \ln(\tan(x)) – \ln(x^2+2)$$
Now, I would assume I should take the derivative of each term and apply the differentiation rule of natural log which is $u'/u$. However, if I do this I end up with a long answer of fractions that I cannot reduce much.
Where am I going wrong here. The fact that there are two terms in the numerator, "$3x$" and "$\tan(x)$", is confusing me on what to do. I'm lost as to what procedures I should take in approaching problems like these. Can someone help me out or at the very least push me in the right direction?
Best Answer
Hints:
Yes, this can get messy.
You have the derivative of $\ln(u)$.
You have a quotient $\displaystyle \frac{u}{v}$
You have a product rule $u v$.
When you take all of those into account, you should arrive at:
$$\frac{d}{dx} ln\left(\frac{3x \tan x}{x^2+2}\right) = \frac{-x^2+(x^2+2) x \csc(x) \sec(x) + 2}{x(x^2+2)}$$
You can simplify the trig terms a bit, but that is not helpful.