How is implicit differentiation performed in calculus?

[bugatti79]

Folks,

Differentiate implicitly [tex]\phi(x,y)=0[/tex] I get:

wrt to x [tex]\phi_x+\phi_y \frac{dy}{dx}[/tex] and
wrt to y [tex]\phi_y+\phi_x \frac{dx}{dy}[/tex]

however I don't know how this is derived

[tex]\phi_x dx+\phi_y dy=0[/tex]

 

[Svein]

If φ(x, y) = 0 (constant), then both [itex]\frac{\partial\phi}{\partial x}=0 [/itex] and [itex]\frac{\partial\phi}{\partial y}=0 [/itex]. Therefore [itex]\frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial y}dy = 0 [/itex].

 

[bugatti79]

If φ(x, y) = 0 (constant), then both [itex]\frac{\partial\phi}{\partial x}=0 [/itex] and [itex]\frac{\partial\phi}{\partial y}=0 [/itex]. Therefore [itex]\frac{\partial\phi}{\partial x}dx + \frac{\partial\phi}{\partial y}dy = 0 [/itex].

Sorry, I understand the derivative of a constant is 0 but I still don't see how you get the last term where you have dx and dy by themselves...even if i try
[tex]\phi_x dx=-\phi_y dy[/tex]= [tex]\frac{dy}{dx}=-\frac{\phi_x}{\phi_y}[/tex]

 

[Svein]

As I said: Both [itex]\frac{\partial\phi}{\partial x} [/itex] and [itex]\frac{\partial\phi}{\partial y} [/itex] are 0. Therefore any linear combination of them is 0.

 

[bugatti79]

Ah ok,

So I add in an extra step

wrt to x [tex]\phi_x+\phi_y \frac{dy}{dx}=0[/tex] *by dx to give [tex]\phi_x dx+\phi_y dy=0[/tex]or
wrt to y [tex]\phi_y+\phi_x \frac{dx}{dy}=0[/tex] *by dy to give [tex]\phi_y dy+\phi_x dx=0[/tex]

 

[HallsofIvy]

"Multiply by dx" is alright as a "mnemonic" but I hope you understand that "derivatives are not fractions" and there is a lot more involved in converting derivatives to differentials.

 

[bugatti79]

I can say I fully understand but perhaps some one can enlighten me regarding what is involved converting derivatives to differentials...for my own understanding and information

 

[mathwonk]

a differential is a linear operator on tangent vectors. the differential of a function f is such an operator constructed from that function, and called df. since x and y are both functions, they also have differentials, dx and dy. since R^2 is two dimensional, so is its tangent space and also its space of linear operators on that tangent space. in that "dual space" dx and dy form a basis, and in that basis we can express df as a linear combination of dx and dy, with coefficients ∂f/∂x and ∂f/∂y. so df = ∂f/∂x dx + ∂f/∂y dy.

by the way it is not true that both ∂f/∂x and ∂f/∂y are zero where f = 0. that is only true if the curve f=0 is singular, i.e. crosses itself at the given point or has a kink. i.e. at (0,0) the curve y^2 = x^2 + x^3 does have a singular point, but y^2 = x^3 -x does not. the problem is that when we write f=0 we do not mean this is true everywhere in the plane, rather we are using this notation to specify a (usually one dimensional) locus in the plane where it is true.

In general given an f, the three equations f = 0, ∂f/∂x = 0, and ∂f/∂y=0, define three different curves, usually with no point in common. I.e. usually ∂f/∂x and ∂f/∂y are not both zero at any point where f=0. Points where all thre do equal zero are called singular points and imply the curve f = 0 is not smooth there.

furthermore, you can divide two vectors v,w if you are in a one dimensional vector space since it makes sense to ask if there is a number c such that w = cv, if so, then w/v = c. The point is that in a one dimensional space this is always possible when v ≠ 0, since any vector along a line is a constant multiple of any non zero vector. This is just the same principle that says any non zero interval on the line can be chosen as a unit, i.e. that every other interval is a constant multiple of it.

On a smooth curve, the tangent and cotangent spaces are one dimensional, so along the smooth curve f=0, the functions x and y do have differentials considered only along that curve, so dy/dx does make sense as a fraction, on the curve where f=0, provided dx is not zero. This happens if the curve is not vertical at the desired point. In that sense, along the curve f=0, you can multiply the chain rule equation by dx, and get df = ∂f/∂x dx + ∂f/∂y dy, perfectly rigorously.

 

[mathwonk]

A brief course on differentials in the plane. Since they are functions on tangent vectors we first discuss tangent vectors. The tangent space to the plane R^2 can be regarded as R^2 itself, because of the availability of translations to identify the tangent spaces at any two points, i.e. we might as well always take tangent vectors to be based at the origin. In this sense a tangent vector at a point of R^2 is given by a pair of numbers (c,d). If we take advantage of the notion of dot product, a linear function on tangent vectors is also given by two numbers say (a,b), where the linear function (a,b) acts on the vector (c,d) with value ac+bd. To keep them apart we could consider the vectors as column vectors of length 2, and the linear functions as row vectors of length 2.

Then if f is a smooth function near p, the partials of f at p, (∂f/ ∂x(p), ∂f/∂y(p)) give us a linear function df(p) on tangent vectors at p. These vary with p, and such a family of varying dual vectors is called a differential (one form). Thus df is a one form, i.e. df is a function whose value at each p is a linear function on tangent vectors at p.

More geometrically, a tangent vector at p is represented by a smooth curve
c:(-e,e)-->R^2 with c(0) = p. Then the action of df(p) on this tangent vector is the value of the derivative of the composition of f and c at zero, i.e. (foc)’(0). According to the chain rule, this value equals ∂f/∂x(p) dx/dt (0) + ∂f/∂y(p) dy/dt (0), where (x(t), y(t)) = c(t) are the component functions of the curve c. I.e. the curve c represents the tangent vector v = c’(0) = (dx/dt (0), dy/dt (0)), and the number
∂f/∂x(p) dx/dt (0) + ∂f/∂y(p) dy/dt (0) = df(p)(v), is the value of df at v.

Now df(p) measures the infinitesimal change in f at p, i.e. evaluating df(p) on tangent vectors at p tells us how f is changing in the direction of that tangent vector. Recall that if f is a smooth function on R^2 then f partitions the plane into level curves where f is constant, the constant levels of f, so that f is not changing along one of these curves, and changes fastest perpendicular to these curves. Of course there are sometimes constant levels of f that are just points, or that are curves but not smooth curves. Such special points are signaled by the fact that df = 0 at those points, i.e. both ∂f/∂x and ∂f/∂y are zero at such points.

But along a smooth level curve, say where f=0, we expect df to have value zero on tangent vectors which are tangent to that level curve, (since f does not change in that direction), and to have non zero value on tangent vectors transverse to that curve, so traveling transverse to a level curve takes us to a different level curve where f has a different value.

To say that df is zero in directions tangent to a given level curve means that if we consider only the curve f=0 for instance then the equation df = 0 will describe the tangent directions along that curve. Since df = ∂f/∂x dx + ∂f/∂y dy, this means that at a point p of the curve f=0, the tangent line at that point to that curve will dot to zero with the coefficients of df, i.e. a vector tangent to the curve f=0 at p, will be perpendicular to the dual vector (∂f/∂x(p), ∂f/∂y(p)). Thus when restricted to the curve f=0, the differential one form ∂f/∂x dx + ∂f/∂y dy, although not identically zero on all vectors in the plane, will have zero value at every tangent vector along the curve, so as a differential one form ON THE CURVE, i.e. as a linear function acting on the tangent spaces to the curve, we do have ∂f/∂x dx + ∂f/∂y dy = 0.

If the f=0 curve is not vertical at the given point, i.e. when f is changing in the y direction, i.e. when ∂f/∂y ≠ 0, then locally y is a function of x, and x is changing as we move along the curve. Hence dx ≠ 0 near that point, and speaking strictly of the action on vectors tangent to that one dimensional curve, we do have that dy is a constant multiple of dx. Thus we can divide dy by dx, where these one forms are thought of after restricting to the tangent space of the one dimensional curve, instead of as one forms on the plane.

Then we do have ∂f/∂x + ∂f/∂y (dy/dx) dx = 0, where dy/dx is really dy divided by dx. Moreover also dy/dx = y’(x), where y(x) is the function defined by considering the curve f=0 locally as the graph of a function.

Thus as a one form on the curve itself, we have ∂f/∂x dx + ∂f/∂y dy/dx dx = 0,
hence ∂f/∂x + ∂f/∂y dy/dx = 0, and then we also get the odd looking equation
dy/dx = (-∂f/∂x)/(∂f/∂y).Oh well, I apologize if maybe this story is too long.

 

[mathwonk]

can't stop, i love this topic so, and have thought about and taught it so long.

consider a curve in the plane and its tangent line. at a given point, the linear function dx looks at a tangent vector, i.e. a vector along the tangent line, and spits out as value the change in x along that vector, i.e. the x coordinate of that vector. similarly dy spits out the y coordinate. now if we divide the y coordinate by the x coordinate, for any vector in that tangent line, we always get just the slope of the tangent line, i.e. dy/dx, where we consider the curve as the graph of a function y(x). in that sense for any smooth curve which is not vertical, we do have at each point, the equation y'(x) = dy/dx, the actual quotient of two linear functions on the tangent space.

 

[WWGD]

An obvious point that must still be made is that you

Folks,

Differentiate implicitly [tex]\phi(x,y)=0[/tex] I get:

wrt to x [tex]\phi_x+\phi_y \frac{dy}{dx}[/tex] and
wrt to y [tex]\phi_y+\phi_x \frac{dx}{dy}[/tex]

however I don't know how this is derived

[tex]\phi_x dx+\phi_y dy=0[/tex]

Maybe another way of looking at it, along what Mathwonk said, is that , geometrically, ## \phi_x dx + \phi_y dy ## is the tangent plane (local) approximation to the change of the function ## \phi(x,y) ## (meaning you evaluate the values of the function at points in the tangent place near a point, instead of at the ( graph of ) the function) which is identically ##0 ##. The best local linear approximation to the change of the zero function is the zero function.

 

[HallsofIvy]

By the way, this is "Calculus", not "Differential Equations".