B2.7 – Implicit B2.7 – Implicit DifferentiationDifferentiationIB Math HL/SL & MCB4U - SantowskiIB Math HL/SL & MCB4U - Santowski

(A) Review(A) Review

Up to this point in the course, we have always defined Up to this point in the course, we have always defined functions by expressing one variable functions by expressing one variable explicitlyexplicitly in terms in terms of another i.e. y = f(x) = xof another i.e. y = f(x) = x22 - 1/x + sin(x) - 1/x + sin(x)

In other courses, like the 3U math course, we have In other courses, like the 3U math course, we have also seen functions expressed also seen functions expressed implicitlyimplicitly i.e. in terms of i.e. in terms of both variables i.e. xboth variables i.e. x22 + y + y22 = 25 (a circle). = 25 (a circle).

In simple implicit functions, we can always isolate the In simple implicit functions, we can always isolate the yy term to rewrite the equation in term to rewrite the equation in explicit explicit terms terms i.e. y = i.e. y = ++ (25 – x(25 – x22) )

In other cases, rewriting an implicit relation is not so In other cases, rewriting an implicit relation is not so easy i.e. 2xeasy i.e. 2x55 + x + x44y + yy + y55 = 36 = 36

(B) Implicit (B) Implicit DifferentiationDifferentiation

There are two strategies that must be used There are two strategies that must be used First, the basic rule of equations is that we can do anything to an 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, provided that we do the same thing to both sides of the equation. equation.

So it will relate to taking derivatives So it will relate to taking derivatives we will start by taking the we will start by taking the derivative of both sides. (our eqn is 2xderivative of both sides. (our eqn is 2x55 + x + x44y + yy + y55 = 36) = 36)

How do we take the derivative of yHow do we take the derivative of y55? ? More importantly, how do we take the derivative of yMore importantly, how do we take the derivative of y55 wrt the wrt the

variable variable xx when the expression clearly does not have an when the expression clearly does not have an xx in it? in it? To date, we have always taken the derivative wrt To date, we have always taken the derivative wrt xx because because xx was  was

the only variable in the equation. the only variable in the equation. Now we have 2 variables.Now we have 2 variables.

(B) Implicit (B) Implicit DifferentiationDifferentiation

d/dx (xd/dx (x55) means finding the rate of change of ) means finding the rate of change of xx55 as we change as we change xx (recall our limit and first (recall our limit and first principles work)principles work)

So what does d/dy (ySo what does d/dy (y55) mean? the rate of ) mean? the rate of change of change of yy55 as we change as we change yy

Then what would d/dx (yThen what would d/dx (y55) mean? the rate of ) mean? the rate of change of change of yy55 as we change as we change xx. But one problem . But one problem arises in that arises in that yy55 doesn't have an doesn't have an xx in it. in it.

(B) Implicit (B) Implicit DifferentiationDifferentiation

We apply the chain rule in that we can We apply the chain rule in that we can recognize recognize yy55 as a composed function with as a composed function with the "inner" function being the "inner" function being xx55 and the and the "outer" function being "outer" function being yy

So then according to the chain rule, So then according to the chain rule, derivative of the inner times derivative of derivative of the inner times derivative of the outer = dythe outer = dy55/dy times dy/dx/dy times dy/dx

So then d/dx (ySo then d/dx (y55) = 5y) = 5y44 times dy/dx times dy/dx

(C) Example(C) Example

dx

dy

yx

yxx

dx

dyyxyxx

dx

dyy

dx

dyxyxx

dx

dyy

dx

dyxyxx

dx

dy

dx

dyx

dx

dx

dx

d

yyxx

44

34

4434

4434

4434

545

545

5

410

)5(410

05410

05410

36)(2

362

(D) In Class Examples(D) In Class Examples

ex 2. Find dy/dx if x + ex 2. Find dy/dx if x + y = xy = x22yy33 + 5 + 5 ex 3. Find the slope of the tangent line ex 3. Find the slope of the tangent line

drawn to xdrawn to x22 + 2xy + 3y + 2xy + 3y22 = 27 at x = 0. = 27 at x = 0. ex 4. Determine the equation of the ex 4. Determine the equation of the

tangent line to the ellipse 4xtangent line to the ellipse 4x22 + y + y22 - 8x + - 8x + 6y = 12 at x = 3.6y = 12 at x = 3.

(E) Graphs for Examples(E) Graphs for Examples

(F) Internet Links(F) Internet Links

Visual Calculus - Implicit Differentiation from UVisual Calculus - Implicit Differentiation from UTKTK

Calculus I (Math 2413) - Derivatives - Implicit DCalculus I (Math 2413) - Derivatives - Implicit Differentiation from Paul Dawkinsifferentiation from Paul Dawkins

Examples and Explanations of Implicit DifferentExamples and Explanations of Implicit Differentiation from UC Davisiation from UC Davis

Section 3.5 - Implicit Differentiation from OUSection 3.5 - Implicit Differentiation from OU

(G) Homework(G) Homework

IB Math HL/SL, Stewart, 1989, Chap 2.7, IB Math HL/SL, Stewart, 1989, Chap 2.7, p107, Q1-3eol, 4,5,6,7p107, Q1-3eol, 4,5,6,7

MCB4U, Nelson text, p483, MCB4U, Nelson text, p483, Q5eol,6eol,7,9Q5eol,6eol,7,9