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February 10th, 2015, 05:17 PM   #1
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Exclamation Derivatives

find the second derivative at the indicated value of x and y given
y^2=2xy+3; at x=1 y=3
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February 10th, 2015, 06:04 PM   #2
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Can you post your attempt?
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February 10th, 2015, 06:18 PM   #3
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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.
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February 10th, 2015, 08:34 PM   #4
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One might as well add "$\displaystyle = \frac{3}{(y - x)^3}$" before substituting the given values.
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February 11th, 2015, 10:11 AM   #5
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Indeed we might. Although, if it makes a difference, something is very wrong!
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