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 May 13th, 2010, 05:00 PM #1 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 1st and 2nd Derivatives .. HI, I have: e^(x) / x I need to find the 1st and 2nd derivatives. -- My first question is, when finding the first derivative, I converted it from the quotient representation to the product representation for ease. e^(x)/x == e^(x)(x^(-1/2)) Finding the derivative of this (using the product rule), I have: (e^(x)(x^-1)) + (e^(x)-1x^(-1/2)) = (e^(x)(x^-1)) - 1(e^(x)x^(-1/2)) = (e^(x)(x^-1)) - (e^(x)x^(-1/2)) = (e^(x)/(x)) - (e^(x)/x^(1/2)) Now, Wolfram Alpha is reporting: = (e^(x)/(x)) - (e^(x)/x^(2)) ... how does (e^(x)x^(-1/2)) become: (e^(x)/x^(2)) ?? wulfgarpro. May 13th, 2010, 05:40 PM #2 Global Moderator   Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: 1st and 2nd Derivatives .. Something's fishy... (e^x)/x = (x^-1)(e^x) So you can differentiate this using the product rule... (x^-1) *(e^x) + (x^-1)(e^x) (I'm using ` to represent the derivative...) (-1x^-2)(e^x) + (x^-1)(e^x) (-1x^-2)(e^x) + x(x^-2)(e^x) (Multiply and "divide" by x to get a "common denominator", of sorts) factor out... (e^x)(x^-2)[-1 + x] Got it from here? May 13th, 2010, 06:59 PM #3 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: 1st and 2nd Derivatives .. Wow .. I lost the plot. Would be great if someone could confirm my ANS: Find the 2nd derivative of e(^x)/x. -- e(^x)/x is equivalent to: e^(x)x^(-1) Using the product rule: (x^(-1)e^(x)) + (-1x^(-2)e^(x)) = e^(x)/x - e^(x)/x^(2) :: 1ST Derivative -- Now, finding the 2nd derivative: e^(x)/x - e^(x)/x^(2) is equivalent to: (e^(x)x^(-1)) - (e^(x)x^(-2)) Using the product rule: (e^(x)x^(-1)) = (e^(x)x^(-1)) + (e^(x)-1x^(-2)) = (e^(x)/x) - (e^(x)/x^(2)) (e^(x)x^(-2)) = (e^(x)x^(-2)) + (e^(x)-2x^(-3)) = (e^(x)/x^(2)) - (2e^(x)/x^(3)) Adding back these derivatives back into the 1st derivative we have: = (e^(x)/x) - (e^(x)/x^(2)) - (e^(x)/x^(2)) - (2e^(x)/x^(3)) = (e^(x)/x) - (2e^(x)/x^(2)) - (2e^(x)/x^(3)) :: 2ND Derivative -- wulfgarpro. May 13th, 2010, 07:06 PM #4 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: 1st and 2nd Derivatives .. Everything looks good to me except (e^(x)/x) - (2e^(x)/x^(2)) + (2e^(x)/x^(3)). May 13th, 2010, 08:39 PM #5 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: 1st and 2nd Derivatives .. Hm, From: (e^x)(x^-1)+(e^x)(-1x^-2)-(e^x)(-1x^-2)+(e^x)(2x^-3) I know the correct answer is: {e^x} / {x} - {e^x} / {x^2} - {e^x} / {x^2} + {2e^x} / {x^3} But from what I can see, does not -(e^x)(-1x^-2) become +(e^x)(-1x^-2) ? wulfgarpro. May 13th, 2010, 08:44 PM #6 Senior Member   Joined: Nov 2008 Posts: 265 Thanks: 0 Re: 1st and 2nd Derivatives .. Using the quotient rule, I get [e^x(x^2 - 2x + 2)]/x^3 for the second derivative. In situations like this, I've always found it easier to use the quotient rule for some reason. May 13th, 2010, 08:49 PM #7 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: 1st and 2nd Derivatives .. I'm so close though. One sign out. 0.o. wulfgarpro. May 13th, 2010, 09:15 PM #8 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: 1st and 2nd Derivatives .. If it's any help: May 13th, 2010, 09:50 PM #9 Senior Member   Joined: Mar 2010 From: Melbourne Posts: 178 Thanks: 0 Re: 1st and 2nd Derivatives .. Hm, Still cant see how it works. From my answer: (e^x)(x^-1)+(e^x)(-1x^-2)-(e^x)(-1x^-2)+(e^x)(2x^-3) Broken into two: (e^x)(x^-1)+(e^x)(-1x^-2) (i) '-' (e^x)(-1x^-2)+(e^x)(2x^-3) (ii) (i) (e^x)(x^-1) - (e^x)(x^-2) (ii) -(e^x)(x^-2) + (2e^x)(x^-3) -- Rejoining them: (e^x)(x^-1)-(e^x)(x^-2)--(e^x)(x^-2)+(2e^x)(x^-3) (e^x)(x^-1)-(e^x)(x^-2)+(e^x)(x^-2)+(2e^x)(x^-3) -- Which does not work out ? Any idea where I'm going wrong ? wulfgarpro. May 13th, 2010, 10:22 PM #10 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: 1st and 2nd Derivatives .. Start with: (e^x)(x^-1) Differentiate: (e^x)(x^-1)-(e^x)(x^-2) -- differentiate again: (e^x)(x^-1)-(e^x)(x^-2) - ((e^x)(x^-2) - (e^x)(2x^-3)) Watch the signs! Tags 1st, 2nd, derivatives Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johngalt47 Calculus 2 November 23rd, 2013 11:44 AM fantom2012 Calculus 3 May 6th, 2012 07:53 PM Arley Calculus 5 April 13th, 2012 06:56 PM r-soy Calculus 2 November 27th, 2010 08:26 AM Calavan11 Calculus 2 December 11th, 2009 04:42 AM

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