# Implicit Differentiation

by dekoi
Tags: differentiation, implicit
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 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. $$\sqrt{x} + \sqrt{y} = 1$$ $$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}}(y') = 0$$ $$y' = (\frac{-1}{2\sqrt{x}})(\frac{2\sqrt{y}}{1}) = \frac{-2\sqrt{y}}{2\sqrt{x}} = \frac{-\sqrt{y}}{\sqrt{x}}$$ $$y'' = \frac{(\frac{-y'}{2\sqrt{y}})(\sqrt{x}) + (\sqrt{y})(\frac{1}{2\sqrt{x}})}{(\sqrt{x})^2}$$ $$y'' = \frac{\frac{-y'\sqrt{x}}{2\sqrt{y}} + \frac{\sqrt{y}}{2\sqrt{x}}}{x}$$ $$y'' = \frac{-y'\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$ $$y'' = \frac{-(\frac{-\sqrt{y}}{\sqrt{x}})\sqrt{x}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$ $$y'' = \frac{\sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y}}{2x\sqrt{x}\sqrt{y}}$$ $$y'' = (\frac{\sqrt{x}\sqrt{y}+y}{2x\sqrt{x}\sqrt{y}})(\frac{\sqrt{x}\sqrt{y}} {\sqrt{x}\sqrt{y}})$$ $$y'' = \frac{xy + y\sqrt{x}\sqrt{y}}{2x^2y}$$ $$y'' = \frac{x + \sqrt{x}\sqrt{y}}{2x^2}$$
 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: $$\left( {\sqrt x + \sqrt y } \right)^\prime = \frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}$$ Now, the second derivative is doing just the same, just a bit longer: $$\left( {\frac{1}{{2\sqrt x }} + \frac{{y'}}{{2\sqrt y }}} \right)^\prime = \left( {\frac{1}{{2\sqrt x }}} \right)^\prime + \left( {\frac{{y'}}{{2\sqrt y }}} \right)^\prime = \cdots$$
 P: n/a I don't understand your first function. Is that supposed to be the second derivative?
 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.
 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.
 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?
 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: $$y'' = \frac{{ - y'\sqrt x \sqrt x + \sqrt y \sqrt y }}{{2x\sqrt x \sqrt y }} = \frac{{y - y'x}}{{2x^{3/2} \sqrt y }}$$
 P: n/a How about after line 6? Is the whole thing correct? Right up to the last line?
 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: $$y'' = \frac{{x + \sqrt x \sqrt y }}{{2x^2 }} = \frac{{\sqrt x + \sqrt y }}{{2x^{3/2} }}$$
 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?
 HW Helper P: 1,021 You realise that in line 4, the first derivative y' is still present in the expression for y''?
 P: n/a nevermind..
 P: n/a Yes I know.
 P: n/a I used my values as: x = 4 y = 1 y' = 1/2
 HW Helper P: 1,021 Huh? Your initial equation was $\sqrt x + \sqrt y = 1$ but [itex]\sqrt 4 + \sqrt 1 \ne 1[/tex]...
 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?
 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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