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Implicit Differentiation

by dekoi
Tags: differentiation, implicit
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dekoi
#1
Oct25-05, 01:46 PM
P: n/a
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]
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TD
#2
Oct25-05, 01:50 PM
HW Helper
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 equal-signs, 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]
dekoi
#3
Oct25-05, 01:57 PM
P: n/a
I don't understand your first function. Is that supposed to be the second derivative?

dekoi
#4
Oct25-05, 02:00 PM
P: n/a
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.
TD
#5
Oct25-05, 02:01 PM
HW Helper
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.
dekoi
#6
Oct25-05, 02:04 PM
P: n/a
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?
TD
#7
Oct25-05, 02:17 PM
HW Helper
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]
dekoi
#8
Oct25-05, 02:22 PM
P: n/a
How about after line 6?
Is the whole thing correct? Right up to the last line?
TD
#9
Oct25-05, 02:26 PM
HW Helper
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]
dekoi
#10
Oct25-05, 02:30 PM
P: n/a
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?
TD
#11
Oct25-05, 02:35 PM
HW Helper
P: 1,021
You realise that in line 4, the first derivative y' is still present in the expression for y''?
dekoi
#12
Oct25-05, 02:35 PM
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nevermind..
dekoi
#13
Oct25-05, 02:36 PM
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Yes I know.
dekoi
#14
Oct25-05, 02:38 PM
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I used my values as:
x = 4
y = 1
y' = 1/2
TD
#15
Oct25-05, 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]...
dekoi
#16
Oct25-05, 02:51 PM
P: n/a
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?
dekoi
#17
Oct25-05, 02:56 PM
P: n/a
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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