1. B2.7 – Implicit Differentiation IB Math HL/SL & MCB4U - Santowski

2. (A) Review  Up to this point in the course, we have always defined functions by expressing one variable explicitly in terms of another i.e. y = f(x) = x 2 - 1/x + sin(x)  In other courses, like the 3U math course, we have also seen functions expressed implicitly i.e. in terms of both variables i.e. x 2 + y 2 = 25 (a circle).  In simple implicit functions, we can always isolate the y term to rewrite the equation in explicit terms  i.e. y = +  (25 – x 2 )  In other cases, rewriting an implicit relation is not so easy i.e. 2x 5 + x 4 y + y 5 = 36

3. (B) Implicit Differentiation  There are two strategies that must be used  First, the basic rule of equations is that we can do anything to an equation, provided that we do the same thing to both sides of the equation.  So it will relate to taking derivatives  we will start by taking the derivative of both sides. (our eqn is 2x 5 + x 4 y + y 5 = 36)  How do we take the derivative of y 5 ?  More importantly, how do we take the derivative of y 5 wrt the variable x when the expression clearly does not have an x in it?  To date, we have always taken the derivative wrt x because x was the only variable in the equation.  Now we have 2 variables.

4. (B) Implicit Differentiation  d/dx (x 5 ) means finding the rate of change of x 5 as we change x (recall our limit and first principles work)  So what does d/dy (y 5 ) mean? the rate of change of y 5 as we change y  Then what would d/dx (y 5 ) mean? the rate of change of y 5 as we change x. But one problem arises in that y 5 doesn't have an x in it.

5. (B) Implicit Differentiation  We apply the chain rule in that we can recognize y 5 as a composed function with the "inner" function being x 5 and the "outer" function being y  So then according to the chain rule, derivative of the inner times derivative of the outer = dy 5 /dy times dy/dx  So then d/dx (y 5 ) = 5y 4 times dy/dx

6. (C) Example

7. (D) In Class Examples  ex 2. Find dy/dx if x +  y = x 2 y 3 + 5  ex 3. Find the slope of the tangent line drawn to x 2 + 2xy + 3y 2 = 27 at x = 0.  ex 4. Determine the equation of the tangent line to the ellipse 4x 2 + y 2 - 8x + 6y = 12 at x = 3.

8. (E) Graphs for Examples

9. (F) Internet Links  Visual Calculus - Implicit Differentiation from UTK Visual Calculus - Implicit Differentiation from UTK Visual Calculus - Implicit Differentiation from UTK  Calculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkins Calculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkins Calculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkins  Examples and Explanations of Implicit Differentiation from UC Davis Examples and Explanations of Implicit Differentiation from UC Davis Examples and Explanations of Implicit Differentiation from UC Davis  Section 3.5 - Implicit Differentiation from OU Section 3.5 - Implicit Differentiation from OU Section 3.5 - Implicit Differentiation from OU

10. (G) Homework  IB Math HL/SL, Stewart, 1989, Chap 2.7, p107, Q1-3eol, 4,5,6,7  MCB4U, Nelson text, p483, Q5eol,6eol,7,9