Derivatives of $\frac{\csc x}{e^{-x}}$ and $\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$ I have tried to mainly ask thoughtful conceptual questions here, but now I am reduced to asking for help on a specific problem that I have been wrestling with for over an hour.    
Disclaimer: I am not a lazy student trying to get free internet homework help. I am an adult who is learning Calculus from a textbook.  I am deeply grateful to the members of this community for their time. 
$$f(x)= \frac{\csc(x)}{e^{-x}}$$
$$f'(x)= ?$$
Answer key says choice (1): $$e^x\csc2x(1-2\cot2x)$$

2) My answer does not match the answer key.   Is there a typo?
Any guidance is enjoyed...it's driving me nuts!
$$\frac{d}{dx}\ln\left(\frac{3x^2}{\sqrt{3+x^2}}\right)$$
Answer key says choice (2): $$\frac{x^2+6}{x^3+3x}$$

 A: For 1, I would recommend re-writing it as:
$f(x)=\frac{e^x}{\sin{x}}$. 
Then using the quotient rule with $u=e^x, v=\sin{x},$
we obtain:
$f'(x)=\frac{e^x\sin{x}-e^x\cos{x}}{\sin^2{x}}=\frac{e^x(\sin{x}-\cos{x})}{\sin^2{x}}=\frac{e^x}{\sin{x}}\frac{\sin{x}-\cos{x}}{\sin{x}}=e^x\csc{x}(1-\cot{x}).$ 
I do not think this is equivalent to what was given in the textbook, although Wolfram Alpha agrees with my answer so I am inclined to think it is the correct one.
As for 2, it makes it considerably simpler to use log laws first, to show that:
$\ln{\frac{3x^2}{\sqrt{3+x^2}}}=\ln{3}+2\ln{x}-\frac{1}{2}\ln{(x^2+3)}=g(x)$, say.
Then using the standard derivative rules for logs we get:
$g'(x)=\frac{2}{x}-\frac{x}{x^2+3}=\frac{x^2+6}{x^3+3x}$, as required.
A: Thank you for the replies.
For #1, it looks like the question meant to say $csc(2x)$
Looks like I solved the original question correctly, 
and I redid the revised problem to match the given solution.  Check!
For #2, yes, the log rules make it a lot easier.  I was able to do it correctly that way.
I also fixed my chain rule error and solved it correctly the "hard way".  Damn, that took me a few tries and quite some time.  But, I am stronger for it.   
Thanks!!
