1. Copyright © Cengage Learning. All rights reserved. 4 Techniques of Differentiation with Applications

2. Copyright © Cengage Learning. All rights reserved. 4.4 The Chain Rule

3. 3 We can now find the derivatives of expressions involving powers of x combined using addition, subtraction, multiplication, and division, but we still cannot take the derivative of an expression like (3x + 1) 0.5. For this we need one more rule. The function h(x) = (3x + 1) 0.5 is not a sum, difference, product, or quotient.

4. 4 The Chain Rule To find out what it is, we can use the calculation thought experiment and think about the last operation we would perform in calculating h(x). 1. Calculate 3x + 1. 2. Take the 0.5 power (square root) of the answer. The last operation is “take the 0.5 power.” We do not yet have a rule for finding the derivative of the 0.5 power of a quantity other than x.

5. 5 The Chain Rule There is a way to build h(x) = (3x + 1) 0.5 out of two simpler functions: u(x) = 3x + 1 (the function that corresponds to the first step in the calculation above) and f (x) = x 0.5 (the function that corresponds to the second step): h(x) = (3x + 1) 0.5 = [u(x)] 0.5 = f (u(x)) u(x) = 3x + 1 f (x) = x 0.5

6. 6 The Chain Rule We say that h is the composite of f and u. We read f (u(x)) as “f of u of x.” To compute h(1), say, we first compute 3  1 + 1 = 4 and then take the square root of 4, giving h(1) = 2. To compute f (u(1)) we follow exactly the same steps: First compute u(1) = 4 and then f (u(1)) = f (4) = 2. We always compute f (u(x)) numerically from the inside out: Given x, first compute u(x) and then f (u(x)).

7. 7 The Chain Rule Now, f and u are functions whose derivatives we know. The chain rule allows us to use our knowledge of the derivatives of f and u to find the derivative of f (u(x)). For the purposes of stating the rule, let us avoid some of the nested parentheses by abbreviating u(x) as u. Thus, we write f (u) instead of f (u(x)) and remember that u is a function of x.

8. 8 The Chain Rule Chain Rule If f is a differentiable function of u and u is a differentiable function of x, then the composite f (u) is a differentiable function of x, and In words The derivative of f(quantity) is the derivative of f, evaluated at that quantity, times the derivative of the quantity. Chain Rule

9. 9 The Chain Rule Quick Example Take f (u) = u 2. Then The derivative of a quantity squared is two times the quantity, times the derivative of the quantity. Because f '(u) = 2u

10. 10 The Chain Rule The following table gives more examples.

11. 11 The Chain Rule To motivate the chain rule, let us see why it is true in the special case when f (u) = u 3, where the chain rule tells us that But we could have done this using the product rule instead: which gives us the same result. Generalized Power Rule with n = 3

12. 12 The Chain Rule A similar argument works for f (u) = u n where n = 2, 3, 4, … We can then use the quotient rule and the chain rule for positive powers to verify the generalized power rule for negative powers as well.

13. 13 Example 1 – Using the Chain Rule Compute the following derivatives.

14. 14 Example 1 – Solution a. Using the calculation thought experiment, we see that the last operation we would perform in calculating (2x 2 + x) 3 is that of cubing. Thus we think of (2x 2 + x) 3 as a quantity cubed. There are two similar methods we can use to calculate its derivative. Method 1: Using the formula We think of (2x 2 + x) 3 as u 3, where u = 2x 2 + x.

15. 15 Example 1 – Solution By the formula, Now substitute for u: = 3(2x 2 + x) 2 (4x + 1) cont’d Generalized Power Rule

16. 16 Example 1 – Solution Method 2: Using the verbal form If we prefer to use the verbal form, we get: The derivative of (2x 2 + x) cubed is three times (2x 2 + x) squared, times the derivative of (2x 2 + x). In symbols, as we obtained above. cont’d

17. 17 Example 1 – Solution b. First, the calculation thought experiment: If we were computing (x 3 + x) 100, the last operation we would perform is raising a quantity to the power 100. Thus we are dealing with a quantity raised to the power 100, and so we must again use the generalized power rule. According to the verbal form of the generalized power rule, the derivative of a quantity raised to the power 100 is 100 times that quantity to the power 99, times the derivative of that quantity. cont’d

18. 18 Example 1 – Solution In symbols, cont’d

19. 19 Example 1 – Solution c. We first rewrite the expression as and then use the generalized power rule as in parts (a) and (b): The derivative of a quantity raised to the 0.5 is 0.5 times the quantity raised to the –0.5, times the derivative of the quantity. Thus, cont’d

20. 20 Example 1 – Solution d. The calculation thought experiment tells us that |4x 2 − x| is the absolute value of a quantity, so we use the generalized rule for absolute values (above): The derivative of the absolute value of a quantity is the absolute value of the quantity divided by the quantity, times the derivative of the quantity. cont’d

21. 21 Example 1 – Solution Thus, cont’d

22. 22 Applications

23. 23 Example 4 – Marginal Product A consultant determines that Precision Manufacturers’ annual profit (in dollars) is given by P = –200,000 + 4,000q – 0.46q 2 – 0.00001q 3 where q is the number of surgical lasers it sells each year. The consultant also informs Precision that the number of surgical lasers it can manufacture each year depends on the number n of assembly line workers it employs according to the equation q = 100n Use the chain rule to find the marginal product Each worker contributes 100 lasers per year.

24. 24 Example 4 – Solution We could calculate the marginal product by substituting the expression for q in the expression for P to obtain P as a function of n and then finding dP/dn. Alternatively—and this will simplify the calculation—we can use the chain rule. To see how the chain rule applies, notice that P is a function of q, where q in turn is given as a function of n.

25. 25 Example 4 – Solution By the chain rule, Now we compute and cont’d Chain Rule Notice how the “quantities” dq appear to cancel.

26. 26 Example 4 – Solution Substituting into the equation for gives Notice that the answer has q as a variable. We can express dP/dn as a function of n by substituting 100n for q: cont’d

27. 27 Chain Rule in Differential Notation If y is a differentiable function of u, and u is a differentiable function of x, then Notice how the units cancel: Applications The terms du cancel.

28. 28 Applications Quick Example If y = u 3, where u = 4x + 1, then

29. 29 Applications Manipulating Derivatives in Differential Notation 1. Suppose y is a function of x. Then, thinking of x as a function of y (as, for instance, when we can solve for x) we have Quick Example In the demand equation q = –0.2p – 8, we have Therefore,

30. 30 Applications 2. Suppose x and y are functions of t. Then, thinking of y as a function of x (as, for instance, when we can solve for t as a function of x, and hence obtain y as a function of x) we have Quick Example If x = 3 – 0.2t and y = 6 + 6t, then The terms dt appear to cancel.

31. 31 Applications To see why the above formulas work, notice that the second formula, can be written as which is just the differential form of the chain rule.

32. 32 Applications For the first formula, use the second formula with y playing the role of t: