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#1
Oct2505, 01:46 PM

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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] 


#2
Oct2505, 01:50 PM

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P: 1,021

Your first derivative is correct, but I don't see what you mean with 0y' then... You also appear to have a lot of equalsigns, it seems a bit weird...
So we have: [tex]\left( {\sqrt x + \sqrt y } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}[/tex] Now, the second derivative is doing just the same, just a bit longer: [tex]\left( {\frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}} \right)^\prime = \left( {\frac{1}{{2\sqrt x }}} \right)^\prime + \left( {\frac{{y'}}{{2\sqrt y }}} \right)^\prime = \cdots [/tex] 


#3
Oct2505, 01:57 PM

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I don't understand your first function. Is that supposed to be the second derivative?



#4
Oct2505, 02:00 PM

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Implicit Differentiation
And I have a lot of equal signs because I'm trying to simplify my second derivative. I think that's where I made the mistake... in simplifying.



#5
Oct2505, 02:01 PM

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P: 1,021

My first line is the first derivative, the second line is the beginning of the second derivative. Since it's linear, I already split it for you. The first term will be easy since it only depends on x, the second will be a bit longer.



#6
Oct2505, 02:04 PM

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Well that's what I did, starting in my fourth line. I used the quotient rule on my simplified form of the first derivative, to find the second derivative. Shouldn't both methods result in the same answer?



#7
Oct2505, 02:17 PM

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P: 1,021

Ok, I wasn't aware of the fact it was an equation at first.
It's much clearer now you've adjusted your initial post. Line 6 seems to be correct, then: [tex]y'' = \frac{{  y'\sqrt x \sqrt x + \sqrt y \sqrt y }}{{2x\sqrt x \sqrt y }} = \frac{{y  y'x}}{{2x^{3/2} \sqrt y }}[/tex] 


#8
Oct2505, 02:22 PM

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How about after line 6?
Is the whole thing correct? Right up to the last line? 


#9
Oct2505, 02:26 PM

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P: 1,021

I took over after line 6 but I didn't realise you substituted y' again.
Anyway, it seems to correct (at least your solution, I didn't carefully check every step). It's possible to 'simplify' it one last bit, although it doesn't really matter a lot anymore: [tex]y'' = \frac{{x + \sqrt x \sqrt y }}{{2x^2 }} = \frac{{\sqrt x + \sqrt y }}{{2x^{3/2} }}[/tex] 


#10
Oct2505, 02:30 PM

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This makes sense, however...
When I substitute a value of x, y, & y' into my equation on line 4 (the first y''), I get a different value for y'' than when I substitute the exact same values into my last equation for y''. Why is that? 


#11
Oct2505, 02:35 PM

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You realise that in line 4, the first derivative y' is still present in the expression for y''?



#12
Oct2505, 02:35 PM

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nevermind..



#13
Oct2505, 02:36 PM

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Yes I know.



#14
Oct2505, 02:38 PM

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I used my values as:
x = 4 y = 1 y' = 1/2 


#15
Oct2505, 02:42 PM

HW Helper
P: 1,021

Huh? Your initial equation was [itex]\sqrt x + \sqrt y = 1[/itex] but [itex]\sqrt 4 + \sqrt 1 \ne 1[/tex]...



#16
Oct2505, 02:51 PM

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You are right.
I know why now... I rearranged that formula to put it in the graphing calculator. But I forgot to consider the fact that squares can be positive or negative. ................ Well at least I think that's my mistake. Do you know what the graph of this function looks like? Is it a circle? Or not? 


#17
Oct2505, 02:56 PM

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Never mind that......
Substitute x = 4 into the original equation and solve for y. You will get the value "1" . That's what I just got. This is odd..... 


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