February 11th, 2016, 11:57 AM  #1 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1  Euler again!
Ok here's sort of a problem, actually I am asking this: I have this function: e^(1lnx)*lnx And have to graph it, graphic calculator not allowed! So first I multiply this, use logarithm properties and I get this: x * e^(lnx)^2 Now I know the procedure, with extrema and so on, but this function looks really hard to do so like 35 minutes later I finally derive it, get e, 0 and 1 as critical points, leave out 1 as ln is not defined there anyway. Then I test the critical point, something should be at the point (1,1). I take the second derivative again 10 minutes later this time and I get 0. So if it's less than zero it's maxima if it's higher it's minima, and now this looks like I have to take the third derivative, my question is where is this going? Am I making this harder than it should be, or the problem really wants me to derive the function for the third time? What if I get another zero? Last edited by skipjack; February 11th, 2016 at 07:54 PM. 
February 11th, 2016, 12:53 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra 
$$\mathrm e ^{(1\ln x)\ln x}=\left(\mathrm e ^{\ln x}\right)^{(1\ln x)}=x^{(1\ln x)}$$

February 11th, 2016, 01:10 PM  #3 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Now tell me how am I supposed to see that! And it's so simple, but I always somehow miss these things... 
February 11th, 2016, 01:17 PM  #4  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
e^(1  lnx)*lnx = $\displaystyle \ e^{1  \ln x}*\ln x \ \ \ or \ \ \ \ln x*e^{1  \ln x} \ \ $ by the Order of Operations. If it were e^[(1  lnx)*lnx], then that would equal $$ \ \mathrm e ^{(1\ln x)\ln x}=\left(\mathrm e ^{\ln x}\right)^{(1\ln x)}=x^{(1\ln x)}$$ . Last edited by skipjack; February 11th, 2016 at 07:58 PM.  
February 11th, 2016, 01:27 PM  #5 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Sure, I know that! Just wrote it wrong, but all this in the exponent is the right form so : e^((1lnx)*lnx) 
February 11th, 2016, 01:42 PM  #6 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 
Let y = x^(1  lnx) lny = ln[x^(1  lnx)] lny = (1  lnx)*lnx Take derivatives: (1/y)(y') = (1/x)(lnx) + (1/x)(1  lnx) (1/y)(y') = (1/x)(lnx + 1  lnx) (1/y)(y') = (1/x)(1  2lnx) Multiply each side by what y equals: y' = [x^(1  lnx)]*[(1/x)(1  2lnx)] Can you continue? 
February 12th, 2016, 08:54 AM  #7 
Member Joined: Jan 2016 From: / Posts: 43 Thanks: 1 
Hopefully! thank you! 

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