- #1
dekoi
Can someone check my answer (I am trying to find the second derivative) for any mistakes?
I have looked it over many times, and I've realized that my second derivative is not correct, but I cannot figure out why. Thank you.
[tex]\sqrt{x} + \sqrt{y} = 1[/tex]
[tex]\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}(y') = 0[/tex]
[tex]y' = (\frac{-1}{2\sqrt{x}})(\frac{2\sqrt{y}}{1}) = \frac{-2\sqrt{y}}{2\sqrt{x}} = \frac{-\sqrt{y}}{\sqrt{x}}[/tex]
[tex]y'' = \frac{(\frac{-y'}{2\sqrt{y}})(\sqrt{x}) + (\sqrt{y})(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}[/tex]
[tex]y'' = \frac{\frac{-y'\sqrt{x}}{2\sqrt{y}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x}[/tex]
[tex]y'' = \frac{-y'\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = \frac{-(\frac{-\sqrt{y}}{\sqrt{x}})\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = \frac{\sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = (\frac{\sqrt{x}\sqrt{y}+y}{2x\sqrt{x}\sqrt{y}})(\frac{\sqrt{x}\sqrt{y}}{\sqrt{x}\sqrt{y}})[/tex]
[tex]y'' = \frac{xy + y\sqrt{x}\sqrt{y}}{2x^2y}[/tex]
[tex]y'' = \frac{x + \sqrt{x}\sqrt{y}}{2x^2}[/tex]
I have looked it over many times, and I've realized that my second derivative is not correct, but I cannot figure out why. Thank you.
[tex]\sqrt{x} + \sqrt{y} = 1[/tex]
[tex]\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}(y') = 0[/tex]
[tex]y' = (\frac{-1}{2\sqrt{x}})(\frac{2\sqrt{y}}{1}) = \frac{-2\sqrt{y}}{2\sqrt{x}} = \frac{-\sqrt{y}}{\sqrt{x}}[/tex]
[tex]y'' = \frac{(\frac{-y'}{2\sqrt{y}})(\sqrt{x}) + (\sqrt{y})(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}[/tex]
[tex]y'' = \frac{\frac{-y'\sqrt{x}}{2\sqrt{y}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x}[/tex]
[tex]y'' = \frac{-y'\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = \frac{-(\frac{-\sqrt{y}}{\sqrt{x}})\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = \frac{\sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}[/tex]
[tex]y'' = (\frac{\sqrt{x}\sqrt{y}+y}{2x\sqrt{x}\sqrt{y}})(\frac{\sqrt{x}\sqrt{y}}{\sqrt{x}\sqrt{y}})[/tex]
[tex]y'' = \frac{xy + y\sqrt{x}\sqrt{y}}{2x^2y}[/tex]
[tex]y'' = \frac{x + \sqrt{x}\sqrt{y}}{2x^2}[/tex]
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