# Implicit vs. partial

• Jun 28th 2008, 10:22 PM
Mathstud28
Implicit vs. partial
Ok, I know the answer is probably no, and I feel stupid asking it, but say we have an equation (I can't say function for reasons that will become obvious), lets say

$f(x,y)=g(x,y)$

Now lets say I differentiate it implicitly and solve for y'. that is the derivative at a point.

Now bear with me if this is really bad but could we do this

Let $z=f(x,y)-g(x,y)$

Then will differentiating implicitly give the same answer as...

$\frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}$?

Maybe I am just really lucky, but after pondering this and thinking about it analytically I tried a couple of problems and it seems to pan out, but I do not think it is true, otherwise my book would have it or I would have heard of it.

Mathstud.

EDIT:

The more I think about it, the more I don't think so.

The reason is the difference in the chain rule

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$

But I am pretty sure

$\frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}\ne\frac{\partial{y}}{\partia l{x}}$

Is that right?

EDIT EDIT:

I have done some more thinking and a little experimentation and I think that I was right

but I think that it is actually

$y'=\frac{-\frac{\partial{z}}{\partial{x}}}{\frac{\partial{z} }{\partial{y}}}$
• Jun 29th 2008, 12:13 AM
Aryth
According to the Implicit Function Theorem:

$\frac{dy}{dx} = -\frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = -\frac{F_x}{F_y}$

So, technically:

$-\frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = -\frac{F_x}{F_y}$

And therefore:

$\frac{\frac{\partial{F}}{\partial{x}}}{\frac{\part ial{F}}{\partial{y}}} = \frac{F_x}{F_y}$

Finally:

$\frac{\frac{\partial{F}}{\partial{y}}}{\frac{\part ial{F}}{\partial{x}}} = \frac{F_y}{F_x}$

So you were right in saying that:

$\frac{\frac{\partial{z}}{\partial{x}}}{\frac{\part ial{z}}{\partial{y}}}\ne\frac{\partial{y}}{\partia l{x}}$

As a matter of fact... The best way to do this would have to be to use the Chain Rule.
• Jun 29th 2008, 07:38 AM
ThePerfectHacker
I am not exactly sure what you are asking but as Aryth said it seems to use the implicit function theorem. In its most basic form it says that if $F:\mathbb{R}^2\mapsto \mathbb{R}$ (this notation it means that $F$ is a function of two variables) is a $\mathcal{C}^1$ function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that $F(a,b) = 0$. The implicit function theorem says that if $\partial_y F(a,b) \not = 0$ then the equation $F(x,y) = 0$ can be solved uniquely (for $y$) in the "neighborhood" of $(a,b)$. The term "in the neighborhood" means in some small disk around $(a,b)$ we can solve this equation uniquely for $y$ (in terms of $x$).

For example, consider the circle $x^2+y^2 - 1= 0$. Where can we solve for $y$ uniquely (in a neighborhood)? Define the function $F(x,y) = x^2+y^2 - 1$. The point $(1,0)$ lies on this circle because $F(1,0) = 1^2 + 0^2 - 1 = 0$. If we try to solve the equation within the small disk $\sqrt{(x-1)^2+y^2} < \epsilon$ we see that it has two solutions, one in the upper half and one in the lower half (see red circle) - in fact those solutions are $y = \pm \sqrt{ 1-x^2}$. And this happens no matter how small we make the circle. Therefore, the equation $F(x,y) = 0$ cannot be solved uniquely. Let us see what happens when we use the implicit function theorem. The theorem says that if $\partial_y F(1,0) \not = 0$ then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that $\partial_y F(1,0) = 0$. Check this: $\partial_y F(x,y) = 2y$ and so $\partial_y F(1,0) = 0$. However, if $F(a,b) = 0$ and $b\not = 0$ i.e. we stay away from the break in the graph where it lies above and below the x-axis then $\partial_y F(a,b) = 2b \not = 0$ and therefore we can solve this equation uniquely. As in the case of the blue circle, the solution is $y=\sqrt{1-x^2}$.

Let us say that $F(x,y)$ is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of $(a,b)$ (where $(a,b)$ is a solution to $F(x,y)=0$) we can solve for $y$ uniquely in terms of $x$. This means we can define a new function $y = g(x)$ in this neighborhood. Remember a function is a set of pairs so that for each first coordinate there is a unique matching coordinate. Therefore, if we let $x$ be the first coordinate then by the theorem there is a unique matching coordinate so that $F(x,y) = 0$. And this would define a function $y=g(x)$.

But the implicit function theorem does not stop here, it says more it says that $g(x)$ is itself $\mathcal{C}^1$ (which is differenciable with some more properties). Now since $F(x,y) = 0 \implies F(x,g(x)) = 0$. The functions $F$ and $g$ are differenciable and so by the chain rule for multivariable functions we get that (differenciating both sides of $F(x,g(x)) = 0$) $\partial_x F + \partial_y F \cdot g'(x) = 0$ and this means $g'(x) = - \frac{\partial_x F}{\partial_y F}$ (the partials are evaluated at $(x,y)$). And that gives you a formula.

If you got that then great. There is just one more point to be addressed. If it confuses you just ignore that. The questionable step is dividing by $\partial_y F$. How do we know it is non-zero? This is where we use two facts. The first one is that $\partial_y F (a,b) \not = 0$. The second one is that $F(x,y)$ is $\mathcal{C}^1$. The first fact is not good enough to say that $\partial_y F\not =0$ for all points around $(a,b)$ because it might be non-zero at $(a,b)$ and yet be zero at some points around $(a,b)$. This is where we need the second fact. The meaning of $\mathcal{C}^1$ is that the function is differenciable and the derivative is continous. Therefore $\partial_y F$ is a continous function. Since $\partial_y (a,b)\not = 0$ it means there is a small enough neighborhood so that $\partial_y (x,y)\not = 0$ for points close to $(x,y)$ by the definition of continuity.
• Jun 29th 2008, 07:56 AM
galactus
I think I know what you're saying, and yes, you can differentiate implicitly that way. I have used it before. As a matter of fact, I wrote a program for implicit diff using that method.

An easy-schmeasy example to be sure.

Say you want to implicitly differentiate $x^{3}+y^{3}=1$

The derivative wrt x is $\frac{\partial}{\partial{x}}=3x^{2}$

wrt y it is $-\frac{\partial}{\partial{x}} = -3y^{2}$

So, by dividing, we get $\frac{-x^{2}}{y^{2}}$

Is that what you mean?. I think it is?. It makes sense to do it that way I always thought.
• Jun 29th 2008, 08:03 AM
Mathstud28
Quote:

Originally Posted by ThePerfectHacker
I am not exactly sure what you are asking but as Aryth said it seems to use the implicit function theorem. In its most basic form it says that if $F:\mathbb{R}^2\mapsto \mathbb{R}$ (this notation it means that $F$ is a function of two variables) is a $\mathcal{C}^1$ function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that $F(a,b) = 0$. The implicit function theorem says that if $\partial_y F(a,b) \not = 0$ then the equation $F(x,y) = 0$ can be solved uniquely (for $y$) in the "neighborhood" of $(a,b)$. The term "in the neighborhood" means in some small disk around $(a,b)$ we can solve this equation uniquely for $y$ (in terms of $x$).

For example, consider the circle $x^2+y^2 - 1= 0$. Where can we solve for $y$ uniquely (in a neighborhood)? Define the function $F(x,y) = x^2+y^2 - 1$. The point $(1,0)$ lies on this circle because $F(1,0) = 1^2 + 0^2 - 1 = 0$. If we try to solve the equation within the small disk $\sqrt{(x-1)^2+y^2} < \epsilon$ we see that it has two solutions, one in the upper half and one in the lower half (see red circle) - in fact those solutions are $y = \pm \sqrt{ 1-x^2}$. And this happens no matter how small we make the circle. Therefore, the equation $F(x,y) = 0$ cannot be solved uniquely. Let us see what happens when we use the implicit function theorem. The theorem says that if $\partial_y F(1,0) \not = 0$ then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that $\partial_y F(1,0) = 0$. Check this: $\partial_y F(x,y) = 2y$ and so $\partial_y F(1,0) = 0$. However, if $F(a,b) = 0$ and $b\not = 0$ i.e. we stay away from the break in the graph where it lies above and below the x-axis then $\partial_y F(a,b) = 2b \not = 0$ and therefore we can solve this equation uniquely. As in the case of the blue circle, the solution is $y=\sqrt{1-x^2}$.

Let us say that $F(x,y)$ is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of $(a,b)$ (where $(a,b)$ is a solution to $F(x,y)=0$) we can solve for $y$ uniquely in terms of $x$. This means we can define a new function $y = g(x)$ in this neighborhood. Remember a function is a set of pairs so that for each first coordinate there is a unique matching coordinate. Therefore, if we let $x$ be the first coordinate then by the theorem there is a unique matching coordinate so that $F(x,y) = 0$. And this would define a function $y=g(x)$.

But the implicit function theorem does not stop here, it says more it says that $g(x)$ is itself $\mathcal{C}^1$ (which is differenciable with some more properties). Now since $F(x,y) = 0 \implies F(x,g(x)) = 0$. The functions $F$ and $g$ are differenciable and so by the chain rule for multivariable functions we get that (differenciating both sides of $F(x,g(x)) = 0$) $\partial_x F + \partial_y F \cdot g'(x) = 0$ and this means $g'(x) = - \frac{\partial_x F}{\partial_y F}$ (the partials are evaluated at $(x,y)$). And that gives you a formula.

If you got that then great. There is just one more point to be addressed. If it confuses you just ignore that. The questionable step is dividing by $\partial_y F$. How do we know it is non-zero? This is where we use two facts. The first one is that $\partial_y F (a,b) \not = 0$. The second one is that $F(x,y)$ is $\mathcal{C}^1$. The first fact is not good enough to say that $\partial_y F\not =0$ for all points around $(a,b)$ because it might be non-zero at $(a,b)$ and yet be zero at some points around $(a,b)$. This is where we need the second fact. The meaning of $\mathcal{C}^1$ is that the function is differenciable and the derivative is continous. Therefore $\partial_y F$ is a continous function. Since $\partial_y (a,b)\not = 0$ it means there is a small enough neighborhood so that $\partial_y (x,y)\not = 0$ for points close to $(x,y)$ by the definition of continuity.

That made perfect sense actually! Thank you very much.

So for the last paragraph since

$F(x,y)$ being $C^{1}$ (a smooth function) this implies that $F_y$ is continous which implies that

$\lim_{(x,y)\to(a,b)}F_y=F_y(a,b)$ from all paths, so this implies that there exists a region containing $(a,b)$ such that $F_y\ne{0}$.

We know this because we have that

$\lim_{(x,y)\to(a,b)}F_y(a,b)$ represents the area immediately surrounding the point $)(a,b)$, or in other words the aforementioned region.

And since as was said earlier

$\lim_{(x,y)\to(a,b)}F_y(x,y)=F_y(a,b)$ (due to its continuity)

and

$F(x,y)\ne{0}$

This implies that

$\lim_{(x,y)\to(a,b)}F_y(x,y)\overbrace{\ne}^{\text {must}}0$

Which gives us the guarantee that dividing by $F_y$ will not produce a division by zero.

Is that right?
• Jun 29th 2008, 08:04 AM
Mathstud28
Quote:

Originally Posted by galactus
I think I know what you're saying, and yes, you can differentiate implicitly that way. I have used it before. As a matter of fact, I wrote a program for implicit diff using that method.

An easy-schmeasy example to be sure.

Say you want to implicitly differentiate $x^{3}+y^{3}=1$

The derivative wrt x is $\frac{\partial}{\partial{x}}=3x^{2}$

wrt y it is $-\frac{\partial}{\partial{x}} = -3y^{2}$

So, by dividing, we get $\frac{-x^{2}}{y^{2}}$

Is that what you mean?. I think it is?. It makes sense to do it that way I always thought.

I was asking both actually, the analytic reason behind and the actual application.

Thanks Galactus!
• Jun 29th 2008, 08:06 AM
galactus
I think PH gave you the in depth of why it works.
• Jun 29th 2008, 08:16 AM
ThePerfectHacker
Quote:

Originally Posted by Mathstud28
Is that right?

Here is what I was thinking when I wrote it. Since $\partial_y F$ is continous at $(a,b)$ it means $\lim_{(x,y)\to (a,b)}\partial_y F(x,y) = \partial_y F(a,b)$. But this means for any $\epsilon > 0$ there exists $\delta > 0$ such that if $\sqrt{(x-a)^2+(y-b)^2} <\delta \implies |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$. Rewrite $|\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$ as $- \epsilon < \partial_y F(x,y) - \partial_y F(a,b) < \epsilon$ and so $\partial_y F(a,b) - \epsilon < \partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ for all $\sqrt{(x-a)^2+(y-b)^2} < \delta$. Now $\partial_y F(a,b) \not = 0$. Thus, either $\partial_y F(a,b) > 0$ or $\partial_y F(a,b) < 0$. If $\partial_y F(a,b) > 0$ choose an $\epsilon$ small enough so that $\partial_y F(a,b) - \epsilon > 0$. And then $\partial_y F(x,y) > 0$ (because $\partial_y F(a,b) - \epsilon < \partial_y F(x,y)$ ) for $\sqrt{(x-a)^2+(y-b)^2}< \delta$ (and this is a neighborhood around $(a,b)$) thus $\partial_y F(x,y) \not = 0$. Now if $\partial_y F(a,b) < 0$ choose $\epsilon >0$ small enough so that $\partial_y F(a,b) + \epsilon < 0$. It will follow from $\partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ that $\partial_y F(x,y) < 0$ in this neighborhood and so $\partial_y F(x,y) \not = 0$.
• Jun 29th 2008, 08:28 AM
Mathstud28
Quote:

Originally Posted by ThePerfectHacker
Here is what I was thinking when I wrote it. Since $\partial_y F$ is continous at $(a,b)$ it means $\lim_{(x,y)\to (a,b)}\partial_y F(x,y) = \partial_y F(a,b)$. But this means for any $\epsilon > 0$ there exists $\delta > 0$ such that if $\sqrt{(x-a)^2+(y-b)^2} <\delta \implies |\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$. Rewrite $|\partial_y F(x,y) - \partial_y F(a,b) | < \epsilon$ as $- \epsilon < \partial_y F(x,y) - \partial_y F(a,b) < \epsilon$ and so $\partial_y F(a,b) - \epsilon < \partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ for all $\sqrt{(x-a)^2+(y-b)^2} < \delta$. Now $\partial_y F(a,b) \not = 0$. Thus, either $\partial_y F(a,b) > 0$ or $\partial_y F(a,b) < 0$. If $\partial_y F(a,b) > 0$ choose an $\epsilon$ small enough so that $\partial_y F(a,b) - \epsilon > 0$. And then $\partial_y F(x,y) > 0$ (because $\partial_y F(a,b) - \epsilon < \partial_y F(x,y)$ ) for $\sqrt{(x-a)^2+(y-b)^2}< \delta$ (and this is a neighborhood around $(a,b)$) thus $\partial_y F(x,y) \not = 0$. Now if $\partial_y F(a,b) < 0$ choose $\epsilon >0$ small enough so that $\partial_y F(a,b) + \epsilon < 0$. It will follow from $\partial_y F(x,y) < \partial_y F(a,b) + \epsilon$ that $\partial_y F(x,y) < 0$ in this neighborhood and so $\partial_y F(x,y) \not = 0$.

Ok got it! Thanks.