Chapter 8 Multivariable Calculus

1. Chapter 8Multivariable Calculus Section 2 Partial Derivatives

2. Learning Objectives for Section 8.2 Partial Derivatives The student will be able to evaluate partial derivatives and second-order partial derivatives.

3. Introduction to Partial Derivatives We have studied extensively differentiation of functions of one variable. For instance, if C(x) = 100 + 500x, then C ´(x) = 500. If we have more than one independent variable, we can still differentiate the function if we consider one of the variables independent and the others fixed. This is called a partial derivative.

4. Example For a company producing only one type of surfboard, the cost function is C(x) = 500 + 70x, where x is the number of boards produced. Differentiating with respect to x, we obtain the marginal cost function C ´(x) = 70. Since the marginal cost is constant, $70 is the change in cost for a one-unit increase in production at any output level.

5. Example(continued) For a company producing two types of boards, a standard model and a competition model, the cost function is C(x, y) = 700 + 70x + 100y, where x is the number of standard boards, and y is the number of competition boards produced. Now suppose that we differentiate with respect to x, holding y fixed, and denote this by Cx(x, y); or suppose we differentiate with respect to y, holding x fixed, and denote this by Cy(x, y). Differentiating in this way, we obtain Cx(x, y) = 70, and Cy(x, y)= 100.

6. Example(continued) Each of these is called a partial derivative, and in this example, each represents marginal cost. The first is the change in cost due to a one-unit increase in production of the standard board with the production of the competition board held fixed. The second is the change in cost due to a one-unit increase in production of the competition board with the production of the standard board held fixed.

7. Partial Derivatives If z = f (x, y), then the partial derivative of f with respect to x is defined by and is denoted by The partial derivative of f with respect to y is defined by and is denoted by

8. Example • Let f (x, y) = 3x2 + 2xy – y 3 . • Find fx(x, y). [This is the derivative with respect to x, so consider y as a constant.]

9. Example • Let f (x, y) = 3x2 + 2xy – y 3 . • Find fx(x, y). [This is the derivative with respect to x, so consider y as a constant.] • fx(x, y)= 6x + 2y • b. Find fy(x, y).[This is the derivative with respect to y, so consider x as a constant.]

10. Example • Let f (x, y) = 3x2 + 2xy – y 3 . • Find fx(x, y). [This is the derivative with respect to x, so consider y as a constant.] • fx(x, y)= 6x + 2y • b. Find fy(x, y).[This is the derivative with respect to y, so consider x as a constant.] • fy(x, y) = 2x – 3y2 • c. Find fx(2,5).

11. Example • Let f (x, y) = 3x2 + 2xy – y 3 . • Find fx(x, y). [This is the derivative with respect to x, so consider y as a constant.] • fx(x, y)= 6x + 2y • b. Find fy(x, y).[This is the derivative with respect to y, so consider x as a constant.] • fy(x, y) = 2x - 3y2 • c. Find fx(2,5). • fx(2,5) = 6 · 2 + 2 · 5 = 22.

12. Example Using the Chain Rule Let f (x, y) = (5 + 2xy 2) 3. Hint: Think of the problem as z = u3and u = 5 + 2xy2. a. Find fx(x, y)

13. Example Using the Chain Rule Let f (x, y) = (5 + 2xy 2) 3. Hint: Think of the problem as z = u3and u = 5 + 2xy2. a. Find fx(x, y) fx(x, y)= 3 (5 + 2xy2)2 · 2y2 b. Find fy(x, y) The chain.

14. Example Using the Chain Rule Let f (x, y) = (5 + 2xy 2) 3. Hint: Think of the problem as z = u3and u = 5 + 2xy2. a. Find fx(x, y) fx(x, y)= 3 (5 + 2xy2)2 · 2y2 b. Find fy(x, y) fy(x, y)) = 3 (5 + 2xy2)2 · 4xy The chain. The chain.

15. Second-Order Partial Derivatives Taking a second-order partial derivative means taking a partial derivative of the first partial derivative. If z = f (x, y), then

16. Example Let f (x, y) = x3y3 + x + y2. a. Find fx(x, y).

17. Example • Let f (x, y) = x3y3 + x + y2. • a. Find fxx(x, y). • fx(x, y)= 3x2y3 + 1 • fxx(x, y)= 6xy3 • Find fxy(x, y).

18. Example • Let f (x, y) = x3y3 + x + y2. • a. Find fx(x, y). • fx(x, y)= 3x2y3 + 1 • fxx(x, y)= 6xy3 • Find fxy(x, y). • fx(x, y) = 3x2y3 + 1 • fxy(x, y)= 9x2y2

19. Summary • If z = f (x, y), then the partial derivative of f with respect to x is defined by • If z = f (x, y), then the partial derivative of f with respect to y is defined by • We learned how to take second partial derivatives.