# Implicit Differentiation

1. Oct 25, 2005

### 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.
$$\sqrt{x} + \sqrt{y} = 1$$
$$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}(y') = 0$$
$$y' = (\frac{-1}{2\sqrt{x}})(\frac{2\sqrt{y}}{1}) = \frac{-2\sqrt{y}}{2\sqrt{x}} = \frac{-\sqrt{y}}{\sqrt{x}}$$
$$y'' = \frac{(\frac{-y'}{2\sqrt{y}})(\sqrt{x}) + (\sqrt{y})(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}$$
$$y'' = \frac{\frac{-y'\sqrt{x}}{2\sqrt{y}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x}$$
$$y'' = \frac{-y'\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$
$$y'' = \frac{-(\frac{-\sqrt{y}}{\sqrt{x}})\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$
$$y'' = \frac{\sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$
$$y'' = (\frac{\sqrt{x}\sqrt{y}+y}{2x\sqrt{x}\sqrt{y}})(\frac{\sqrt{x}\sqrt{y}}{\sqrt{x}\sqrt{y}})$$
$$y'' = \frac{xy + y\sqrt{x}\sqrt{y}}{2x^2y}$$
$$y'' = \frac{x + \sqrt{x}\sqrt{y}}{2x^2}$$

Last edited by a moderator: Oct 25, 2005
2. Oct 25, 2005

### 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:
$$\left( {\sqrt x + \sqrt y } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}$$

Now, the second derivative is doing just the same, just a bit longer:
$$\left( {\frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}} \right)^\prime = \left( {\frac{1}{{2\sqrt x }}} \right)^\prime + \left( {\frac{{y'}}{{2\sqrt y }}} \right)^\prime = \cdots$$

3. Oct 25, 2005

### dekoi

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

4. Oct 25, 2005

### 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.

5. Oct 25, 2005

### 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.

6. Oct 25, 2005

### 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?

7. Oct 25, 2005

### TD

Ok, I wasn't aware of the fact it was an equation at first.

Line 6 seems to be correct, then:

$$y'' = \frac{{ - y'\sqrt x \sqrt x + \sqrt y \sqrt y }}{{2x\sqrt x \sqrt y }} = \frac{{y - y'x}}{{2x^{3/2} \sqrt y }}$$

8. Oct 25, 2005

### dekoi

Is the whole thing correct? Right up to the last line?

9. Oct 25, 2005

### 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:

$$y'' = \frac{{x + \sqrt x \sqrt y }}{{2x^2 }} = \frac{{\sqrt x + \sqrt y }}{{2x^{3/2} }}$$

10. Oct 25, 2005

### 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?

11. Oct 25, 2005

### TD

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

12. Oct 25, 2005

### dekoi

nevermind..

13. Oct 25, 2005

### dekoi

Yes I know.

14. Oct 25, 2005

### dekoi

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

15. Oct 25, 2005

### TD

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

16. Oct 25, 2005

### 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?

17. Oct 25, 2005

### 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.....