Functional relation and implicit functions

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1. Nov 18, 2014

trap101

This is more a conceptual question. So i am doing some self review of multi variate calculus and i am looking at functinal relations of the form F(x, y, z,...) = 0

In the book they talk about implicit differentiation. Now i fully understand how to do the mechanics of it, but i was trying to understand why we need implicit differentiation? I get the fact that we may not be able to solve for all our functions in an explicit form say y = g(x, z,w,...)

But when they state something along the lines of we can use the chain rule to compute the partials of g in terms of F where

F(x, y, z,...g(x, y, z))

And the partials would .be of the form dg/dxj = -djF/d(n+1)F

What is the objective of this expression?

Convuluted i know, if clarification is needed please ask. Thanks for help

2. Nov 23, 2014

Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Nov 23, 2014

Staff: Mentor

This is very difficult to follow. You have F(x, y, z, ..., g(x, y, z)). That's just an expression. Typically you have level curves where F(x, y, z, ..., g(x, y, z)) = C for some constant C.

In this equation, dg/dxj = -djF/d(n+1)F, you have switched from variables x, y, z, ... to indexed variables x1, x2, and so on.
I don't know what you're trying to convey with this expression -- -djF/d(n+1)F.

Do you have a specific example in mind?

4. Nov 24, 2014

mathwonk

a function is a curve that satisfies the vertical line test, and when you have a formula for such a function in terms of y = f(x) you know how to find the tangent line at any point.

A curve that is given by a formula like F(x,y) = 0, may not satisfy the vertical line test, hence it may be impossible to solve for it as a formula like y = f(x), but you may still want to know the tangent line at some point.

If you can find a point (x,y) on your curve, i.e. one that satisfies F(x,y) = 0, and if it is a point where there is a unique tangent line, then implicit differentiation lets you solve for the slope of the tangent line at that point, without solving your formula for y. This is often useful.

The hard part however is finding the coordinates of a point on a curve given in the implicit form F(x,y)= 0, but if you know what point you want the tangent line at, you are ok.

e.g. the curve X^3 +X^2 Y - Y^3 + 2Y -3 = 0 has as one of its points the point (1,1). at this point the partial derivatives are not both zero, so there is a unique tangent line whose slope can be computed by implicit differentiation. in this example however the derivative wrt y is zero there, so the tangent line at this point is vertical. thus one can compute dx/dy but not dy/dx. thus at this point y is not determined implicitly as a diff'ble function of x, but x is a diff'ble implicit function of y, which is just as good for finding the tangent line. if this confuses you, exchange the letters x and y in the example and get a more traditional one.