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B2.7 – Implicit Differentiation

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

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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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