My Math Forum 1st and 2nd Derivatives ..

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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: $f(x)\,=\,e^xx^{-1}$ $f'(x)\,=\,e^xx^{-1}\,-\,e^xx^{-2}$ $f''(x)\,=\,e^xx^{-1}\,-\,e^xx^{-2}\,-\,\left(e^xx^{-2}\,-\,2e^xx^{-3}\right)\,=\,\frac{e^x}{x}\,-\frac{2e^x}{x^2}\,+\,\frac{2e^x}{x^3}\,=\,\frac{e^ x\left(x^2\,-\,2x\,+\,2\right)}{x^3}$
 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!

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