Implicit Differentiation

Transcription

1 Implicit Differentiation In principle, under ery general circumstances, systems of equations in any number of ariables can be soled for some of the ariables in terms of the others, i.e. as functions of the others but not necessarily elementary functions. 1 These functional relationships among the ariables are said to be implicitly defined by the equations. A precise statement which discusses the existence, uniqueness, and differentiability of such implicitly defined functions is called the Implicit Function Theorem see below. If there are m equations and m + n ariables, a solution will inole m dependent ariables one for each equation and n independent ariables. The dependent ariables are the ones we sole for and they can, in general, be chosen in an arbitrary manner. Suppose the ariables are x 1,..., x n, y 1,..., y m with x = x 1,..., x n independent and y = y 1,..., y m dependent. Then the equations can be written in the form F 1 x, y = 0,..., F m x, y = 0 and a solution will hae the form y 1 = Y 1 x,..., y m = Y m x. Assuming the existence and differentiability of the solutions, here we are concerned with how to find the arious deriaties i without haing to actually sole the equations. The method is referred to as implicit differentiation. First, as far as notation is concerned, it must be stressed that if, say, w and t are two of the ariables appearing in the system of equations, then the symbol w does not make sense until it is stated which ariables in addition to t hae been chosen as the independent ariables t or equialently, which ariables in addition to w hae been chosen as the dependent ariables since we must know which ariables are being held fixed when we perform the differentiation with respect to t. The result is generally different in different contexts. Example 1. x + y + u = 0, 3x + 3y + u + = 0 If x and y are taken to be independent and we sole for u and, we obtain u 2x 2y, x y. In this context then, 2. Howeer, if x and are taken to be independent and we sole for u and y, then u = 2,y x and now we find that = 0. 1 When we say, In principle,..., these systems of equations can be soled, we simply mean that the solutions described exist: computing them is another generally difficult matter. 1

2 2 A commonly used notation to distinguish the arious contexts is to list the other independent ariables as subscripts after the differentiation symbol, e.g. w. So, in Example 1, 2, = 0. t r,s,... y To explain the method of implicit differentiation, let us consider the case of two equations in four ariables: F x, y, u, = 0, Gx, y, u, = 0, where u, are dependent and x, y independent. To find and, we regard the equations as functional identities where u and are certain differentiable functions of x and y. We assume also that F and G are differentiable. Using the Chain Rule, we differentiate with respect to x holding the other independent ariable y fixed and we obtain F x 1 + F y 0 + F u u x + F x = 0 and G x 1 + G y 0 + G u u x + G x = 0. The result is a system of two linear equations in the two unknowns u x and x, namely F u u x + F x F x, G u u x + G x G x. If we sole this system by using Cramer s Method, then u x = D 1 /D and x = D 2 /D where [ ] [ ] [ ] Fu F D = Fx F, D G u G 1 = Fu F, D G x G 2 = x. G u G x The erminant D is called the acobian of F, G with respect to u, and is denoted by or. In this notation then, if F x, y, u, = 0 and Gx, y, u, = 0 define u and as differentiable functions of x and y, then x, and u,x proiding Similarly, 0. y, and u,y Example 2. With reference to Example 1 aboe, [ ] 1 1 x, 3 1 [ ] 4 y

3 x,y u,y confirming what we found aboe. [ 3 3 [ 1 3 ] ] 0 2 = 0 In the general case we considered at the outset, F 1 x, y = 0,..., F m x, y = 0, we define the acobian of F = F 1,..., F m with respect to y = y 1,..., y m to be m i =.. j m m m 1... m m and denote it by F1,..., F m or 1,..., F m or y 1,..., y m 1,..., y m. Then, proceeding as we did aboe, using Cramer s Method, we obtain the following formula for i : Let z be obtained from y by replacing y i by x j, i.e. z k = y k for k i but z i = x j ; then i z 1,...,F i,...,f m 1,...,x j,...,y m 1,...,F i,...,f m 1,...,y i,...,y m Remark. In the case of one equation, F x 1,..., x n, y = 0, the aboe reduces to /. For example, if x2 +y 2 +z 2 = xyz is written in the form F x, y, z = 0 where F x, y, z = x 2 + y 2 + z 2 xyz, then z F x 2x yz F z 2z xy proiding 2z xy 0 and, of course, F x, y, z = 0. Example 3. The cure gien by the equations x 2 + y 2 + z 2 = 4, x + y + z = 1 is a circle since it is the section of a sphere by a plane. On this circle, if we take x as the parameter, then x is the independent ariable and both y and z are functions of x. To find dy dz and, we write the gien dx dx equations in the form F = 0, G = 0 and obtain 2x 2z dy dx x,z x z 2y 2z y z = z x y z y,z. 3

4 4 for y z and similarly, dz dx y,x for y z. y,z 2y 2x 2y 2z y x y z = x y y z Example 4. The following three equations in fie ariables x = u, y = u 2 + 2, z = u 2 2 can be soled for u,, z as functions of x and y. To find z we let F x, y; u,, z = u x, Gx, y; u,, z = u y, Hx, y; u,, z = u 2 2 z. Then z,h,x,h,z for z 0; Similarly z for z 0.,H,y,H,z 2u 2 0 u 1 2u 2 0 2u 2 0 u 0 2u 2 1 2u 2 1 u 0 2u 2 0 2u 2 0 u 0 2u 2 1 = 4u 2 u 2 4x z 2u u 2 = y z Here is the theorem on which the method discussed aboe is based, the Implicit Function Theorem: Theorem. Let x = x 1,..., x n, y = y 1,..., y m and suppose that 1 F 1 x, y = 0,..., F m x, y = 0 is a system of m equations in the m + n ariables x 1,..., x n, y 1,..., y m and a P 0 = a, b is a solution of 1, b all the first partials of F 1,..., F m exist and are continuous in a neighbourhood of P 0, and c 1,...,F m 1,...,y m 0 at P 0.

5 Then, in a neighbourhood of P 0, the system 1 defines y 1,..., y m as functions of x 1,..., x n or, one could say, y as a function of x; more precisely, there exist neighbourhoods U of a and V of b and functions Y 1 x,..., Y m x defined on U such that, for all x in U, x, y is a solution of 1 with y in V if and only if y = Y 1 x,..., Y m x. So, for x in U and y in V, the system 1 is equialent to y 1 = Y 1 x,..., y m = Y m x. Moreoer, each of the partials Y i x exists and is continuous on U and Y i x, Yx x z x, Yx where = 1,..., F m 1,..., y m, Yx = Y 1x,..., Y m x and z is obtained from y by replacing y i by x j. Exercises: 1 Cramer s Rule Verify the following and express the results in terms of erminants. i The system of 3 linear equations a i x + b i y + c i z = d i i = 1, 2, 3 can be written as a single ector equation, namely * xa + yb + zc = d. When a b c 0, there is a unique solution: x = d b c a d c a b d, y =, z = a b c a b c a b c Hint: For x, dot both sides of * with b c. ii For 2 linear equations, a 1 x + b 1 y = d 1, a 2 x + b 2 y = d 2, we can obtain a similar result by using i with a 3 = b 3 = d 3 = 0 and c 1 = c 2 = 0, c 3 = 1. 2 In the following, compute the indicated deriaties. At any point that satisfies the equations, what condition on the ariables guarantees that the implied functional relationship exists in some neighbourhood of the point? 3 a sinxyz + x 2 + y 2 + z 2 = 2; z. b 2x 2 + 3y 2 + 4z 2 = 12, x yz = 2; dy dz and. dx dx c x = u 2 2, y = 2u; u x, u y, x, y, u xx, where in each case, u and are functions of x and y. d xu 2 + = y 3, 2yu x 3 = 4x; y, xy 2 + zu + 2 = 3, x 3 z + 2y u = 2, xu + y xyz = 1. 5

6 6 Show that these equations define x, y, z as functions of u and in a neighbourhood of the point P 0 x, y, z, u, = 1, 1, 1, 1, 1 and find when u, = 1, 1. 4 a Show that in a neighbourhood of P 0 x, y, u, = 1, 1, 2, 1, the equations xu + y u = 1, x + yu xy = 2 y 1,...,y m x 1,...,x n define u and as functions of x, y and u x 1, 1 2. b Show that in a neighbourhood of P 0 the equations in a define y and as functions of x, u and 1, Gien 4 equations in 7 ariables x, y,... how many possible interpretations are there for? 6 Suppose y 1,..., y m are differentiable functions of x 1,..., x n. Define the acobian matrix of y = y 1,..., y m with respect to x = x 1,..., x n to be the m n matrix whose i, jth entry is i. We denote it by either D or D x y. If now x 1,..., x n are differentiable functions of t = t 1,..., t p, show that a D t y = D x yd t x matrix multiplication b If m = n = p, then. t c If m = n = p and t = y and the functions x y and y x are inerses of each other, then D x y and D y x are inerses of each other and and are reciprocals of each other. t =