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I have been trying to solve this problem for hours.

$\dfrac{9e^{2x}}{8x+3}$

I know $u'(x)$ will be $18e^{2x}$ and $v'(x)$ will be $8$

Written out, it will be $\dfrac{(8x+3)(18e^{2x})-(9e^{2x})(8)}{(8x+3)^2}$

I get to the part above^^ and I'm not sure what to do. I know it's probably something simple that I'm over or under thinking, but please help!

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  • $\begingroup$ What are U and v (and it seems strange to use a capital for one and lower case for the other). It sounds like you are integrating by parts, but you didn't say that. What is your question? $\endgroup$ – Ross Millikan Nov 1 '14 at 1:59
  • $\begingroup$ Yes this problem requires you to find the derivative. About the quotient rule, I know it more or less, but I'm just not sure how to apply it to this problem. I'm just lost. $\endgroup$ – AlexM77 Nov 1 '14 at 2:04
  • $\begingroup$ Don't mind the uppercase U, and lower v, it has no significance, it was just a typo @RossMillikan $\endgroup$ – AlexM77 Nov 1 '14 at 2:07
  • $\begingroup$ You have applied the quotient rule correctly, so your answer is correct for the derivative. One thought would be that you can stop here, because you have a correct expression for the derivative. Another thought would be that you should simplify it. I commented on that in my answer edit. $\endgroup$ – Ross Millikan Nov 1 '14 at 2:11
  • $\begingroup$ Thanks @RossMillikan you've been a great help! $\endgroup$ – AlexM77 Nov 1 '14 at 2:20
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What you have done is fine. Now you are probably expected to "simplify" it. Simplification is often in the eye of the beholder, but you should distribute the $18e^{2x}$ in the numerator over the two terms it mulitplies, then combine one of them with the $9e^{2x} \cdot 8$. Whether you expand the denominator or not is definitely in the eye of the beholder. I would not unless it helped some further computation in a problem.

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  • $\begingroup$ I feel that it's usually preferable to simplify by factoring, since there are so many times where you're solving for critical numbers, which means you have to factor it anyway. $\endgroup$ – Daniel Goldman Nov 1 '14 at 2:09
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Assuming you actually meant $\frac{(8x+3)(18e^{2x})-(9e^{2x})(8)}{(8x+3)^2}$ =

$\frac{(8x+3)(18e^{2x})-(18e^{2x})(4)}{(8x+3)^2}$

Factor out $18e^{2x}$ to obtain

$\frac{18e^{2x}((8x+3)-4)}{(8x+3)^2} = $

$\frac{18e^{2x}(8x-1)}{(8x+3)^2}$

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  • $\begingroup$ Thanks! I'm still a little confused on how you got to your final answer, but I'll go back and rework it. I probably just did something careless. Thanks again! $\endgroup$ – AlexM77 Nov 1 '14 at 2:16
  • $\begingroup$ I edited the post to add a few more steps. If you feel it useful, don't forget to click accept. $\endgroup$ – Daniel Goldman Nov 1 '14 at 2:23

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