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 February 10th, 2015, 05:17 PM #1 Newbie   Joined: Oct 2014 From: Phil Posts: 11 Thanks: 0 Derivatives find the second derivative at the indicated value of x and y given y^2=2xy+3; at x=1 y=3 February 10th, 2015, 06:04 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Can you post your attempt? Thanks from topsquark February 10th, 2015, 06:18 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra We use implicit differentiation for this type of problem \begin{aligned} && y^2 &= 2xy + 3 \\ &\text{differentiating with respect to x} & 2y {\mathrm d y \over \mathrm d x} &= 2y + 2x {\mathrm d y \over \mathrm d x} \\ && (y-x) {\mathrm d y \over \mathrm d x} &= y \\ && {\mathrm d y \over \mathrm d x} &= {y \over y-x} \\[8pt] && {\mathrm d^2 y \over \mathrm d x^2} &= {\mathrm d \over \mathrm d x} \left({\mathrm d y \over \mathrm d x}\right) \\ && &= {\mathrm d \over \mathrm d x} \left({y \over y-x} \right) \\ && &= {(y-x){\mathrm d y \over \mathrm d x} - y({\mathrm d y \over \mathrm d x} - 1) \over (y-x)^2} \\ && &= {y - x{\mathrm d y \over \mathrm d x} \over (y-x)^2} \\ &\text{substituting our formula for {\mathrm d y \over \mathrm d x}} & &= {y(y-x) - xy \over (y-x)^3} \\ && &= {y^2 - 2xy \over (y-x)^3}\end{aligned} All you need to do after that is to plug the given values into the formula. Thanks from topsquark February 10th, 2015, 08:34 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 One might as well add "$\displaystyle = \frac{3}{(y - x)^3}$" before substituting the given values. Thanks from topsquark and v8archie February 11th, 2015, 10:11 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2654 Math Focus: Mainly analysis and algebra Indeed we might. Although, if it makes a difference, something is very wrong! Tags 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 math221 Calculus 7 February 3rd, 2013 07:23 PM bilano99 Calculus 2 December 18th, 2012 07:50 AM Arley Calculus 3 April 9th, 2012 04:18 PM shango Calculus 2 October 21st, 2009 06:38 PM

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