November 26th, 2012, 08:20 PM  #1 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  another mind twister
The derV of f(x) = xe^x  e^x f "(x) = [ xe^x + e^x ] e^x  book says we got this via the product & Difference Rules I can't see it; a little help please 
November 26th, 2012, 08:29 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: another mind twister
If then: 
November 26th, 2012, 08:42 PM  #3 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: another mind twister
Mark I am trying to apply the product rule. Would you be so kind to show me the first times the derV of the second + the second times the derV of the first so that I can see what I am not seeing? It is very embarrassing to keep having the trouble in algebra sometimes I think I have a wet brain or something 
November 26th, 2012, 08:48 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: another mind twister
What is where ?

November 26th, 2012, 08:52 PM  #5 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: another mind twister
Not familiar with ER symbols you used on the right, but I do believe that the derV of ke^x is just e^x the k being constant goes away and in finding the derV of exponential fx's we do the chain rule... so derV of e^x is just e^x and the chain rule would be either 1 or zero gosh I'm reaching hard Mark 
November 26th, 2012, 09:06 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: another mind twister
No, the constant doesn't go away, what we have is (and the symbols just mean k is a real constant): So, when you use this rule along with the product rule, what do you get? 
November 26th, 2012, 09:11 PM  #7 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: another mind twister
Hey I don't know if I posted wrong but you have first derV and then 2nd derV; my post is f(x) and then f '(x). 
November 26th, 2012, 09:53 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: another mind twister
Your first post says the derivative is such and such, then the second derivative is...that's what I was going by. But, differentiating a function is the same no matter what we call it.

November 27th, 2012, 01:30 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 17,702 Thanks: 1353 
The symbol ? means "is in" (in the sense of "is an element of") and is the set of real numbers, so k ? effectively means "k is real". However, k isn't needed for this problem. Let u = x and v = e^x, so that uv = xe^x. The derivatives of x and e^x are 1 and e^x respectively, so by the product rule, the derivative of uv = u(dv/dx) + v(du/dx) = x(e^x) + (1)e^x = xe^x + e^x. Using the above, xe^x  e^x can now be differentiated by the difference rule to give (xe^x + e^x)  e^x. 
November 27th, 2012, 07:58 PM  #10 
Senior Member Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0  Re: another mind twister
Mark, this is my mistake. It does say this is derV and then goes on to show the 2nd derV. 

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