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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!

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