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
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
Then will differentiating implicitly give the same answer as...
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.
Can someone please clarify?
The more I think about it, the more I don't think so.
The reason is the difference in the chain rule
But I am pretty sure
Is that right?
I have done some more thinking and a little experimentation and I think that I was right
but I think that it is actually
According to the Implicit Function Theorem:
So you were right in saying that:
As a matter of fact... The best way to do this would have to be to use the Chain Rule.
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 (this notation it means that is a function of two variables) is a function (ignore what this mean - it just means the function is behaved well-enough such as being differenciable). Say that . The implicit function theorem says that if then the equation can be solved uniquely (for ) in the "neighborhood" of . The term "in the neighborhood" means in some small disk around we can solve this equation uniquely for (in terms of ).
For example, consider the circle . Where can we solve for uniquely (in a neighborhood)? Define the function . The point lies on this circle because . If we try to solve the equation within the small disk 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 . And this happens no matter how small we make the circle. Therefore, the equation cannot be solved uniquely. Let us see what happens when we use the implicit function theorem. The theorem says that if then we can solve the equation uniquely. Since we cannot solve this equation uniquely it must mean that . Check this: and so . However, if and i.e. we stay away from the break in the graph where it lies above and below the x-axis then and therefore we can solve this equation uniquely. As in the case of the blue circle, the solution is .
Let us say that is function which satisfies the condition of the implicit function theorem as in the first paragraph. Then in the neighborhood of (where is a solution to ) we can solve for uniquely in terms of . This means we can define a new function 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 be the first coordinate then by the theorem there is a unique matching coordinate so that . And this would define a function .
But the implicit function theorem does not stop here, it says more it says that is itself (which is differenciable with some more properties). Now since . The functions and are differenciable and so by the chain rule for multivariable functions we get that (differenciating both sides of ) and this means (the partials are evaluated at ). 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 . How do we know it is non-zero? This is where we use two facts. The first one is that . The second one is that is . The first fact is not good enough to say that for all points around because it might be non-zero at and yet be zero at some points around . This is where we need the second fact. The meaning of is that the function is differenciable and the derivative is continous. Therefore is a continous function. Since it means there is a small enough neighborhood so that for points close to by the definition of continuity.
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
The derivative wrt x is
wrt y it is
So, by dividing, we get
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.
Originally Posted by galactus
I think PH gave you the in depth of why it works.
Ok got it! Thanks.
Originally Posted by ThePerfectHacker