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4.1 implicit diff.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Wed 10/1/08 4.1 - Implicit Differentiation - Exam 2 in two weeks! In case you’ve been on top of the stuff on the website, you might notice that today was scheduled as related rates. We’re just going to skip that. Today we are going to do 4.1: Implicit Differentiation. For some reason, the book decided it would be a good idea to do related rates first, which is just stupid. So, this means today we will be going over 4.1, tomorrow, you’ll be briefly introduced to related rates with your worksheet, and then on Friday and Monday, we shall work on problem after problem of related rates questions. Your homework will, of course, be affected, and so the homework assigned for tonight will be due next tuesday. Okay, up until now, we’ve basically dealt with functions where we have y = f(x). y is on one side of the equation and a whole bunch of x’s are on the other. No mixing and matching involved. Well, this isn’t a perfect world, and sometimes we have both x’s and y’s on the same side. For example, 2 xy − x y + 1 = x Well, sometimes we can actually turn that equation into a function y = f(x). In this case, we have 2 xy − x y = x − 1 so factor out a y on the left and we get 2 y(x − x ) = x − 1 or x − 1 y = x − x2 Okay, that’s pretty sweet. The problem is, we can’t always solve explicitely for y. Or, perhaps we can, but it’s just so difficult or perhaps once we do solve for y it’s so difficult to take the derivative, that it’s just not worth it! Sometimes when we solve for y, we have to actually split it up into two different functions in order to pass the vertical line test. For example, consider the circle with center at the origin and radius 3: x + y = 9 If we were to solve for y, we split it into two functions p y = 9 − x2 p y = − 9 − x2 The first graph is the upper semicircle, while the second is the lower semicircle. Together, they form the entire circle (ie, just graphing both functions on the same axes). So, a function f is defined implicitly by some equation in x and y if the graph y = f(x) coincides with a portion of the graph of the equation. 1 Example: let x = y . Note this is not a function because it fails the vertical line test! But, if we √ √ solve for y, we get y = x and y = − x, and thinking of those two as separate functions, they each pass the vertical line test individually. So, we say x = y implicitly defines the functions √ √ f1(x) = x and f 2x) = − x. So, why are we bothering to think about this stuff if we can just solve for y? Well, sometimes you can’t. For example, it is impossible to solve for y in this equation: x + y = 3xy Go ahead and try it! I doubt you’ll get very far... The graph of this equation is called the Folium of Descartes, by the way, and its graph looks like this: 1 x –2 –1 1 2 3 0 –1 y –2 –3 –4 Okay, so now let’s move on to differentiating these implicit functions. What exactly do we do? Well, you basically just differentiate as you normally would, just keep in mind that y is a function of x! Example: find the derivative of y + xy = 1. Solution: So, there are two different ways we can do this problem. One way is to solve for y and then take the derivative as normal. This is how it works out: y(x + 1) = 1 1 y = x + 1 dy = −1 dx (x + 1)2 By the way, you shouldn’t have to use quotient rule here! You certainly can, but it’s a bit more work than necessary. Just use the chain rule with y = (x + 1) −1. If there’s just a constant in the numerator or in the denominator, you don’t need the quotient rule! It’s just a waste of time because the derivative of that constant will turn out to be zero! Anywho, here’s how to go about solving this implicitly. Just go across and differentiate everything with respect to x. Also keep in mind that y is a function of x (i guess you can think of it as f(x) if 2 you’d like to), so if you’re differentiating xy, you need to use the product rule! y + xy = 1 dy + d [x]y + xdy = d [1] dx dx dx dx dy + y + xdy = 0 dx dx dy And from here, just solve
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