UNIT 2 LESSON 9 IMPLICIT DIFFERENTIATION 1

2 So far, we have been differentiating expressions of the form y = f(x), where y is written explicitly in terms of x. It is not always convenient or possible to isolate the y, and in these cases, we must differentiate with respect to x without first isolating y. This is called implicit differentiation. y = 2x 2 + 5x – 7 y = (x – 4) 3 (2x + 5) 5 2y = y 2 + 3x 3

3 IMPLICIT DIFFERENTIATION We represent the derivative of y as The derivative of x is equal to 1; therefore, it is not necessary to write it as That is, if y = x then

4 If y is a function of x then its derivative is y 2 is a function of y, which in turn is a function of x. Using the chain rule : Find the following derivative with respect to x IMPLICIT DIFFERENTIATION

5 METHOD I: Rearrange for y EXAMPLE 1: a) Differentiate y 2 = x METHOD II: Implicitly OR y = - x ½ y = x ½

6 EXAMPLE 1: b) Find the slopes of the tangent lines on y 2 = x at (4, 2) and (4, -2) (4, 2) (4, -2) Using the x value of 4 we need the graph to know which slope goes with which point.

7 EXAMPLE 1 b) Find slopes of the tangents on y 2 = x at (4, 2) and (4, -2) (4, 2) (4, -2) Using the y values of -2 and 2 we know which slope goes with which point.

8 EXAMPLE 1 c) Find slope of the tangent on y 2 = x at (0, 0) (0, 0) Using the x value of 0 Using the y value of 0

9 EXAMPLE 2a: x 2 + y 2 = 25 This is not a function, but it would still be nice to be able to find the slope for any tangent line. In order to graph this on our calculators we have to rearrange and isolate y y 2 = 25 – x 2 OR

10 EXAMPLE 2b: x 2 + y 2 = 25 Take derivative of both sides. 25 We could differentiate each of the above explicitly but it would be more difficult than using implicit differentiation. OR

11 EXAMPLE 2 c) Find the Equation of the tangent line at (3, 4) So at (3, 4) x 2 + y 2 = 25 (3, 4) Slope =

12 EXAMPLE 3: Differentiate 2y = y 2 + 3x 2

13 Assignment Questions Do Questions 1-9 on pages 3 & 4