# 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): $$\frac{x^2+6}{x^3+3x}$$

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Please, try to make the title of your questions more informative. Note that our maths renderer MathJax also works in titles. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. For more information on choosing a good title, see this post. – Lord_Farin Oct 24 '13 at 17:39
The first is probably a typo, they thought they had $\csc(2x)$ on top. By the way, your work (which is correct for $\frac{\csc x}{e^{-x}}$) would have been somewhat easier if you had made the preliminary simplification to $e^x\csc x$. For the second, I would use "laws of logs" to simplify the expression before differentiating. Lots of work saved. – André Nicolas Oct 24 '13 at 17:44
The textbook answer to the second question is right. You missed a 2 in the first line of your work for the $-u \cdot \frac{dv}{dx} \space$part. It should have been $(-3x^2)\cdot \frac{2x}{2\sqrt{3+x^2}}$ – K. Rmth Oct 24 '13 at 17:51
Andre! Yes, the first question had a typo. Good call, it's very obvious now! – JackOfAll Oct 24 '13 at 18:05
K.Rmth! Yes, I forgot the chain rule at that step. Let me add back the (2x) and see if I can finish it the hard way. (Using Log rules up front sure make it easier) – JackOfAll Oct 24 '13 at 19:06

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.

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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!!

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