Implicit Differentiation. Lecture 16. We are use to working only with functions that are efine explicitly. That is, ones like f(x) = 5x 3 + 7x x 2 + 1 or s(t) = e t5 3, in which the function is escribe explicitly by means of a formula for the output. The formula may be pretty messy, but it oes give the value explicitly. In an earlier example we began with an explicitly efine function y = x x, but converte to a function efine implicitly: ln y = x ln x. To etermine the value of the function for some input value x we woul have to solve the resulting equation. More extreme examples are given in equations such as x 2 + xy 3 xy = 7 or y x x 2 + y = 6. Even if we coul solve these for y, they might not actually escribe y as a function of x, For example, we can easily solve for y, but the solution x 2 + y 2 = 25 y = ± 25 x 2 oes not escribe a function; for example, when x = 0, this equation gives two values for y, namely y = ±5. Still if we restrict our attention to just the positive values or just the negative ones, we o have a function. Functions of this sort, efine by equations, are sai to efine functions implicitly. An we can often analyze them successfully just as ones efine explicitly. Here we shall consier their erivatives.
Example 1. Let s begin by assuming that y is some ifferentiable function of x. Thus, y has a erivative, y; so far we on t know x what this erivative is or how to compute it just that it exists. But suppose that we have the aitional information that x an y satisfy some equation, say y = xy 2 + x 3. That equation puts another constraint on y. So ifferentiating both sies of that equation we get x y = x (xy2 + x 3 ) = x (xy2 ) + ( x x3 = x x y2 + y 2 ) x x + x x3 (Prouct Rule) = x x y2 + y 2 + x x3 But y is a function of x, so using the chain rule we can ifferentiate y an hence, y 2. That is, for the two missing pieces x x3 = 3x 2 an So putting it all together, we have y x x y2 = y (y2 ) y x = 2xy y x + y2 + 3x 2. = 2y y x. Still to get y solely in terms of x an y, we must solve this latter x equation. But that clearly gives us (1 2xy) y x = y2 + 3x 2, so y x = y2 + 3x 2 1 2xy. 2
Example 2. Next consier the equation x 2 xy + y 3 = 7. With this we have efine y implicitly as a function of x an, of course, we have also efine x implicitly as a function of y. So let us view y as a function of x, y = y(x). Then just as in the previous example we may view the equation as equating two functions of x, an so we can ifferentiate them an the resulting erivatives will be equal. That is, x 2 xy + y 3 = 7 x (x2 xy + y 3 ) = x (7) x x2 (y x x + x x y) + x y3 = 0 2x (y + x y x ) + ( y y3 2x y x y x + 3y2y x = 0 Solving this last equation for y we get x y x = 2x y x 3y 2. ) y x = 0 For example at the point (1, 2) on the graph of the efining equation we have y = 2 2 x (1,2) 1 12 = 0. 3
Example 3. Let s o this one together. The graph of the equation x 2 + y 2 = 25. is Untitle-1 the circle with center at the origin an raius 5. Let s 1fin the equation of the line tangent to this circle at the point (3, 4). (3, 4) Example 4. Suppose that a point is moving aroun the circle x 2 + y 2 = 25 an we know that when t = 2, the point is at (3, 4) an t x = 2. When t = 2, what is y an what is the rate of change of the point t with respect to t? 4
This implicit ifferentiation provies us with a short-cut that can be quite time saving on occasion an it can bail us out if we forget the Prouct Rule or the Quotient Rule. It s calle Logarithmic Differentiation. First, a couple of very simple examples: Example 1. Suppose that u an v are ifferentiable functions of x, an that y = uv. Then the erivative y follows from a irect x application of the Prouct Rule. But here s an alternative: Take the log of both sies: ln y = ln(uv) = ln u + ln v. Now ifferentiate both sies of this with respect to x to get (ln y) = x x 1 y y x = 1 u u an multiplying by y: (ln u + ln v) = x x + 1 v v x (ln u) + (ln v) x y x = y u u x + y v v x = vu x + uv x, which, of course, is what we woul have gotten using the Prouct Rule! 5
Example 2. This time suppose that y = u. As an alternative v to the Quotient Rule for calculating its erivative, again take the log of both sies: ( u ) ln y = ln = ln u ln v. v Now ifferentiate both sies of this with respect to x to get (ln y) = x x 1 y y x = 1 u u an multiplying by y: y ( y ) u ( y ) v x = u x v x = (ln u ln v) = x x 1 v v x (ln u) (ln v) x ( ) 1 u ( u ) v v x v 2 x = uv vu. v 2 Example 3. If y = u v, then to fin y x of both sies an ifferentiate implicitly: once again take the logs ln(y) = ln(u v ) = v ln u, so x ln y = (v ln u) = v x x 1 y ( v ) u y x = + (ln u)v u x x y ( v ) x = u vu u x (ln u) + (ln u) x v v + (ln u)uv x = vuv 1u x + (ln u)uv v x. 6
Example 4. Let s o a couple together. Say, ifferentiate y = x 4 (x 1) 3 (x 2 + 1) 5. Example 5. Now let s fin y x when y = (x + 5)2 e x (x + 2) 3. 7
Fining Extreme Values. An important practical problem for which ifferentiation can often provie quick an easy answers is that of fining the extreme values, that is maximum an minimum values of a function. To set the stage consier the following graph of a function y = Untitle-1 f(x) efine on the interval a x b y = f(x) F B D E A a C The points A, B, C, D, E, an F are the extrema (singular extreme points) of the function. In particular, f has a relative minimum at A, C an E (i.e., f at that point is less than or equal to all values of f(x) in some interval about that point); f has a relative maximum at B, D, an F (i.e., f at that point is greater than or equal to all values of f(x) in some interval about that point); f has an absolute minimum at C (i.e., at that point f is less than or equal to all values of f on the interval); f has an absolute maximum at F (i.e., at that point f is greater than or equal to all values of f on the interval). b 8
The extrema in this example typify virtually all of the extrema that we shall encounter in this course. For continuous functions extrema occur at only a limite class of points an the five extrema above illustrate each class. For a function y = f(x) a point in its graph is A critical point if either It is a stationary point, that is, its erivative f (x) is zero there; It is a singular point, that is, its erivative oes not exist there; It is an en point, that is, some interval on one sie of the point is not in the omain of f. For example, for the above function, the points B, C an E are stationary an D is singular, so these are the critical points of the function. The points A an F are the en-points. An here is the key fact about extreme points: The extreme points of a continuous function occur only at critical points an en-points. This pretty clearly makes the task of fining all extreme points a much easier task. 9
Example 1. Let s fin all extreme points of f(x) = 12x x 3 on the interval 3 x 5. We begin our search by fining all critical points an that begins with the erivative: f (x) = x (12x x3 ) = 12 3x 2. Since this is efine for all values of x on the interval, there are no singular points. But f (x) = 12 3x 2 = 0 x 2 = 4 x = ±2. So there are only two critical points: ( 2, 16) an (2, 16). Next, there are only two en-points at x = 3 an x = 5. That is the en-points are ( 3, 9) an (5, 65). Since these four points are the only possible extreme points, we nee only compare them to see that so f( 3) f( 2), f( 2) f(2), an f(2) f(5), ( 2, 16) an (5, 65) are relative minima; ( 3, 9) an (2, 16) are relative maxima. Finally just comparing the relative extrema, we see that (5, 65) is the absolute minimum an (2, 16) is the absolute maximum. 10
Example 2. Here is a curious one. Let s fin the extrema of f(x) = x 4 4x. This has no en-points so we nee worry only about critical points. But the erivative is f (x) = 4x 3 4 = 4(x 3 1). So there are no singular points an the only stationary point occurs when f (x) = 0 x 3 1 = 0 x 3 = 1 x = 1. But notice that f(x) < f(1) for x < 1 an f(x) > 1 for x > 1. So (1, 3) is neither a relative maximum nor a relative minimum!! The lesson to be learne here is that Critical points nee not be relative minima or relative maxima. Example 3. Let s try this one together: Fin the extreme values of the function f(x) = x 2 e x. 11