1 / 10

110 likes | 228 Vues

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)

Télécharger la présentation
## B2.7 – Implicit Differentiation

**An Image/Link below is provided (as is) to download presentation**
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.
Content is provided to you AS IS for your information and personal use only.
Download presentation by click this link.
While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

**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) = 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**(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.**(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.**(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**(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.**(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**(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

More Related