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?

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 $$\begin{aligned} && y^2 &= 2xy + 3 \\ &\text{differentiating with respect to $x$} & 2y {\mathrm d y \over \mathrm d x} &= 2y + 2x {\mathrm d y \over \mathrm d x} \\ && (yx) {\mathrm d y \over \mathrm d x} &= y \\ && {\mathrm d y \over \mathrm d x} &= {y \over yx} \\[8pt] && {\mathrm d^2 y \over \mathrm d x^2} &= {\mathrm d \over \mathrm d x} \left({\mathrm d y \over \mathrm d x}\right) \\ && &= {\mathrm d \over \mathrm d x} \left({y \over yx} \right) \\ && &= {(yx){\mathrm d y \over \mathrm d x}  y({\mathrm d y \over \mathrm d x}  1) \over (yx)^2} \\ && &= {y  x{\mathrm d y \over \mathrm d x} \over (yx)^2} \\ &\text{substituting our formula for ${\mathrm d y \over \mathrm d x}$} & &= {y(yx)  xy \over (yx)^3} \\ && &= {y^2  2xy \over (yx)^3}\end{aligned}$$ All you need to do after that is to plug the given values into the formula. 
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.

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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