Help answering this problem!
- Thread starter Ghost3k
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Unknown008
Junior Member
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- Sep 6, 2011
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Can anyone help me answer this question. My friend needs help with it but it seems I cant get the right answer whenever I put it in on the online homework. He really needs it by tomorrow...
I was getting something like this
Did you use the product rule?
\(\displaystyle y = uv\)
\(\displaystyle y' = uv' + vu'\)
And in this case, u = e^-14x, v = ln(5x)
Which means:
\(\displaystyle f'(x) = \left(e^{-14x}\right)\left(\dfrac{5}{5x}\right) + \left(\ln(5x)\right)\left(-14e^{-14x}\right)\)
Can you complete it?
no, I already got that far. After that I found the slope by plugging in x= 3 and it turned out to be some really big number. Anyways after that I found the why by plugging x= 3 to the original function and I got ln(15)*e^42. Then I set up the equation using y-y1= m(x-x1). This gave me the answer I posted above however, it seems it is not right.
Unknown008
Junior Member
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- Sep 6, 2011
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Really? When I plug x = 3, this is what I get:
\(\displaystyle f'(x) = \left(e^{-14(3)}\right)\left(\dfrac{5}{5(3)}\right) + \left(\ln(5(3))\right)\left(-14e^{-14(3)}\right)\)
\(\displaystyle f'(x) = \dfrac{e^{-42}}{3} - 14e^{-42}\ln(15)\)
\(\displaystyle f'(x) = e^{-42}\left(\dfrac{1}{3} - 14\ln(15)\right)\)
This comes to something like \(\displaystyle -2.16\times 10^{-17}\)