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B2.7 – Implicit Differentiation

This review explores the concept of implicit differentiation, expanding on how to handle functions expressed implicitly, such as equations where both variables are involved. It covers the basic rules of derivatives and the chain rule as they apply to implicit functions. The document includes several examples to clarify the process, from finding derivatives to determining equations of tangent lines. It is an essential resource for IB Math HL/SL students needing to grasp implicit relationships and their derivatives effectively.

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B2.7 – Implicit Differentiation

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  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) = x2 - 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. x2 + y2 = 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 – x2) • In other cases, rewriting an implicit relation is not so easy i.e. 2x5 + x4y + y5 = 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 2x5 + x4y + y5 = 36) • How do we take the derivative of y5? • More importantly, how do we take the derivative of y5 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 (x5) means finding the rate of change of x5 as we change x (recall our limit and first principles work) • So what does d/dy (y5) mean? the rate of change of y5 as we change y • Then what would d/dx (y5) mean? the rate of change of y5 as we change x. But one problem arises in that y5 doesn't have an x in it.

  5. (B) Implicit Differentiation • We apply the chain rule in that we can recognize y5 as a composed function with the "inner" function being x5 and the "outer" function being y • So then according to the chain rule, derivative of the inner times derivative of the outer = dy5/dy times dy/dx • So then d/dx (y5) = 5y4 times dy/dx

  6. (C) Example

  7. (D) In Class Examples • ex 2. Find dy/dx if x + y = x2y3 + 5 • ex 3. Find the slope of the tangent line drawn to x2 + 2xy + 3y2 = 27 at x = 0. • ex 4. Determine the equation of the tangent line to the ellipse 4x2 + y2 - 8x + 6y = 12 at x = 3.

  8. (E) Graphs for Examples

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

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