1 §3.1 Implicit Differentiation The student will learn about implicit differentiation.

2 Function Review and a New Notation We have defined a function as y = x 2 – 5x. We have also used the notation f (x) = x 2 – 5x. In both situations y was the dependent variable and x was the independent variable. However, a function may have two (or more) independent variable and is sometimes specified as F (x, y) = x xy - 3 y 2 +7.

3 Explicit Differentiation Consider the equation y = x 2 – 5x. Then y ‘ =2x - 5 This is what we have been doing and is called explicit differentiation. If we rewrite the original equation, y = x 2 – 5x, as x 2 – y – 5x = 0 it is the same equation. We can differentiate this equation implicitly. Note: we normally do implicit differentiation when explicit differentiation is difficult.

4 Implicit Differentiation Again consider the equation x 2 – y – 5x = 0 We will now implicitly differentiate both sides of the equation with respect to x The same answer we got by explicit differentiation on the previous slide. And solving for dy/dx Discuss 0

5 Implicit Differentiation Let’s examine a short cut where we ignore the dx/dx. And solving for dy/dx

6 Explicit Differentiation Be careful! Remember that x and y play different roles; x is the independent variable while y is the dependent variable. Therefore we must include a (from the generalized power rule) when we differentiate y n. We don’t need to include a when differentiating x n since = 1.

7 Example 2 Consider x 2 + y 2 + 3x + 4y = 0 and differentiate implicitly. Solve for dy/dx This equation would be difficult to differentiate explicitly. 1

8 Example 2 continued We just differentiated x 2 + y 2 + 3x + 4y = 0 and got: We could have differentiated with respect to y Indeed, we could have differentiated with respect to t. The last two derivatives are presented to help you understand implicit differentiation.

9 Example 3 Consider x 2 – 3 xy + 4y = 0 and differentiate implicitly. Solve for dy/dx Notice we used the product rule for the - 3xy term. 1

10 Example 3 (Again) Consider x 2 – 3 xy + 4y = 0 and differentiate implicitly. Solve for y’ Notice we used the product rule for the - 3xy term.

11 Example 3 Continued We just differentiate implicitly x 2 – 3 xy + 4y = 0 to get We could evaluate this derivative at a point on the original function, say (1, - 1). That means that the slope of the tangent line (or any of the other meanings of the derivative such as marginal profit) at (1, - 1) is – 5.

12 Example differentiate implicitly 3x 2 + y 2 – 7x + x 3 y 5 = 0 Solve for dy/dx Notice we used the product rule

13 Finding by implicit differentiation. Implicit Differentiation Review 1. Differentiate both sides of the equation with respect to x. when differentiating a y, include 2. Collect all terms involving on one side, and all other terms on the other side. 3. Factor out the and solve for it by dividing.

14 Summary. We learned how to implicitly differentiate in order to find derivatives of difficult functions.

15 ASSIGNMENT §3.1 on my website

Test Review 16 § 2.1 Know the basic derivative formula. If f (x) = C then f ’ (x) = 0. If f (x) = x n then f ’ (x) = n x n – 1. If f (x) = k u (x) then f ’ (x) = k u’ (x) = k u’. If f (x) = u (x) ± v (x), then f ’ (x) = u’ (x) ± v’ (x).

Test Review 17 § 2.2 Know the Product Rule. If f (x) and s (x), then f s ' + s f ' Know theQuotient Rule. If t (x) and b (x), then continued

Test Review 18 § 2.2 Know Marginal average cost Marginal average revenue Marginal average profit

Test Review 19 § 2.2 Know how to find second derivative and the applications associated with them.

Test Review 20 § 2.3 Know the chain rule. Know that some functions are not differentiable.

Test Review 21 § 2.4 Know how to optimize a function including tax revenue.

There are a lot of applied problems on the test. It would be worth your time to go over the ones assigned for homework! 22