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 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: 12,163 Thanks: 472 Math Focus: Calculus/ODEs Re: another mind twister If $f'(x)=xe^x-e^x$ then: $f''(x)=xe^x+(1)e^x-e^x=xe^x$
 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: 12,163 Thanks: 472 Math Focus: Calculus/ODEs Re: another mind twister What is $\frac{d}{dx}$$k\cdot e^x$$$ where $k\in\mathbb{R}$ ?
 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: 12,163 Thanks: 472 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): $\frac{d}{dx}$$k e^x$$=k\frac{d}{dx}$$e^x$$=ke^x$ 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: 12,163 Thanks: 472 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: 19,047 Thanks: 1618 The symbol ? means "is in" (in the sense of "is an element of") and $^{^{\mathbb{R}}}$ is the set of real numbers, so k ? $^{^{\mathbb{R}}}$ 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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