Section 15.5: The Chain Rule. 1 Objectives. 2 Assignments. 3 Maple Commands. 1. Apply all forms of the chain rule. (3,5,9,11,13,19,21)

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1 Section 155: The Chain Rule 1 Objectives 1 Apply all forms of the chain rule (3,5,9,11,13,19,21) 2 Compute derivatives by implicit differentiation (25,27,29) 2 Assignments 1 Read Section Problems: 1,5,9,11,13,15,19,27,31,37,39 3 Challenge: 23,41,45,47 4 Read Section Maple Commands The Maple commands needed to use the chain rule are not different from the commands that we have alrea studied However, definining variables makes application of the chain rule much easier For instance, to use the chain rule to find when t z x + 3xy 4, x sin(2t), and y cos(t), you would execute the following: > z:x^2+3*x*y^4; :diff(z,x); diff(z,y); x:sin(2*t); ycos(t); :diff(x,t); diff(y,t); :*+*; > _eval:eval(,t0); The last command evaluates the function at the point t 0 To use Maple for implicit differentiation, you have 3 options Consider the function F (x, y) x 3 + y 3 6xy In order for Maple to understand that in this case y is a function of x, you must explicitly type y(x) instead of y when defining F > F:x^3+y(x)^3-6*x*y(x); You can differentiate F 0 on both sides with respect to x, and then have Maple solve for by using the following: > eq1:diff(f0,x); > :solve(eq1,diff(y(x),x)); You can use the Maple implicit differentiation command: 1

2 > F:x^3+y^3-6x*y; > :implicitdiff(f0,y,x); where the the first argument is the dependent variable, the second is the independent variable Note that when using the implicit diff command, you do not need to specify that y is a function of x in the definition of F This is assumed to be the case based on the argument list in the implicitdiff command (See what happens if you swap the y and x arguments) Finally, you can use the book s method and find the ratio of partial derivatives of F > F:x^3+y^3-6*x*y; > :simplify(-diff(f,x)/diff(f,y)); Note that again you do not need to define y explicitly as a function of x here 4 Lecture In this section, we define the chain rule and use it to find partial derivatives of functions and to define implicit differentiation 41 The Chain Rule In Calculus I, you learned that if y f(x), and x g(t), then The expansion of this for Calculus III is to let z f(x, y), x g(t), and y h(t) and get that f + f + Note the use of partials for and, but not for and This is because z is a function of 2 variables, but both x and y are functions of one variable However, in the end, z only depends on t, hence Thus z is the dependent variable, x and y are intermediate variables, and t is the true independent variable 2

3 411 Example 1: Problem 1552 Here, z x 2 + y 2, x e 2t, and y e 2t Then Thus x x2 + y 2 y x2 + y 2 2e 2t 2e 2t 2xe2t 2ye 2t x2 + y Example 2: Problem 1554 Here, z x ln(x + 2y), x sin(t), and y cos(t) Then and x + ln(x + 2y) x + 2y 2x x + 2y cos(t) sin(t) ( ) ( ) x 2x + ln(x + 2y) cos(t) + ( sin(t)) x + 2y x + 2y When t 0, ln(2) 413 Example 2: Problem 1554 We are asked to find the change in wheat production with respect to time given changes in temperature and rainfall We know that W f(t, R), W W 2, T R dr 015, and 01 Thus 8, dt dw W dt T + W dr R ( 2)(015) + (8)( 01) 11units/year 3

4 42 z f(x, y), x g(s, t), y h(s, t) Now, if every function is a function of two variables, all derivatives become partial derivatives, but the chain rule goes forward as you would expect Thus 421 Example 3: Problem 1558 f + f + Now, z x/y, x se t, and y 1 + se t Since s e t, we have 1 Find t et + xe t s y y 2 1/y, x/y2, s et, and 43 In general The chain rule can be extended for functions with any number of variables If u f(x 1, x 2,, x n ), where x i g i (t 1, t 2,, t m ), then u u 1 + u u n t i 1 t i 2 t i n t i 431 Example 4: Problem Given z x/y, x re st, and y rse t, we find that 1 r y est + x se t y 2 1 Find s 2 Find t 44 Implicit Differentiation A function F (x, y) 0 defines y implicitly as a function of x This means that we cannot solve for y to get an expression y f(x), but we can still use partial derivatives to find Given F (x, y) 0, we use the chain rule to differentiate both sides with respect to x to get F + F F F (1) + F x F y 4 0 0

5 Also, if z is defined implicitly as a function of x and y using the function F (x, y, z) 0, then F + F + F 0 F x F z We get the last equation because here; it is an independent variable 1 and 0 since y is not a function of x 441 Example 5: Problem If y 5 +3x 2 y 2 +5x 4 12, then F (x, y) y 5 +3x 2 y 2 +5x 4 12 Thus F x 6xy 2 +20x 3 and F y 5y 4 + 6x 2 y, so that 6xy2 +20x 3 5y 4 +6x 2 y 442 Example 6: Problem Given xyz cos(x + y + z), F (x, y, z) cos(x + y + z) xyz Thus, Fx F z sin(x+y+z)+yz sin(x+y+z)+xy 1 Find 5