Implicit Differentiation

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]

TD

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

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

dekoi

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

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

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

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

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

TD

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

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

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

dekoi

nevermind..

dekoi

Yes I know.

dekoi

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

TD

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

dekoi

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

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.....