Implicit Differentiation

Transcription

1 Implicit Differentiation Thus far, the functions we have been concerne with have been efine explicitly. A function is efine explicitly if the output is given irectly in terms of the input. For instance, in the function f = 4x 2 the value of f is given explicitly or irectly in terms of the input. Just by knowing the input we can immeiately fin the output. A secon type of function that is also useful for us to consier is an implicitly efine function. A function is efine implicitly if the output cannot be foun irectly from the input. For instance, f = 2x is an implicitly efine function, because for each positive x value there is a corresponing f value, but we cannot fin it irectly from the function. We woul nee to square both sies, an then we woul have the explicitly efine function f = 4x 2. It is also possible for us to have implictly efine functions that we cannot rewrite as an explicitly efine function. For instance we might have the function sin(f) + f = x. For a given x value, there may be a corresponing output value f which makes this a true statement. In this way a function f woul be efine for all such x where there is a solution. For instance, we have f(0) = 0, because setting x = f = 0 in the above equation is a true statement. Right now we on t have the proper tools to solve such an equation, but the important concept here is that we can have a function efine in such a way. When we speak of functions, we mean that we have a rule which provies us with at most one output for a given input (there is no output for inputs at which the function is not efine). In a more general sense we might want to look at rules that provie us with multiple outputs for a given input. Such an example woul be the equation x 2 + y 2 = 1, which is the equation of the unit circle. It turns out that this object consists of two functions, namely y = ± 1 x 2. The equation of this circle oes not efine a function (because the output is multi-value), but it oes efine some type of curve in the x-y plane. In general, we shoul be able to escribe an arbitrary curve as a combination of some number of functions. It is sensible (an useful) to consier the slope of certain points on the curve, which woul simply correspon to the slope of the specific function that efines that part of the curve. To fin the slope, we simply ifferentitate the appropriate component function of the curve, that efines the particular portion of the curve that is of interest. In terms of the above circle, we can write explicit equations for the two functions that make up the circle. However, in more complicate examples we may not be able to fin explicit equations for the functions that make up the curve. Nevertheless, because a more complicate curve still consists of functions, we shoul be able to fin its erivative at a particular point. In light of consiering the process of ifferentiation as an operator, we can view ifferentiation as an operator that we apply to both sies of the equation. Just like when multiply by a constant, a a constant, etc, ifferentiation shoul not change the truth of the given equation. In this way, we can look at the function f = x 2 an think about fining the erivative of f with respect to x by ifferentiating both sies of the equation. When we o so, we fin x f = x x2 or f x = 2x

2 which is the familiar result. We will use this iea in orer ifferentiate implicitly efine functions. The process works as follows: we ifferentiate both sies of the equation with respect to x (or the inepenent variable), an then we solve for the erivative of the epenent variable. Anywhere we fin the function f (or the epenent variable), we will use the chain rule to fin the erivative. Let us begin with a simple example. Example 1 Fin y/x if y 2 = x. Solution One metho of solving this problem woul be rewriting the equation so that we an explicititly efine functions y 1 = x an y 2 = x, an ifferentiating. We woul fin y 1 2 x an y 2 x = 1 2 x This works sufficiently well in this situation, but what about a function or curve we cannot rewrite explicitly? We woul nee implicit ifferentiation. Let s apply implicit ifferentiation to this situation, as a means of exercise. y 2 = x x y2 = x x 2y y y 2y Were we unable to rewrite y explicitly in terms of x, this is as far as we coul go, but we know that y = ± x, an plugging in this result we fin that y 1 2 x an y 2 x = 1 2 x once again. In this way, we were able to calculuate both erivatives at once. It is interesting to note that when we ifferentiate the above implicitly efine functions, the erivative epens on the value of the epenent variable y. This is an interesting result, but in truth, is the way things must be. Suppose that the erivative i not epen on y. That woul mean for any given x value, the erivatives of both functions ( x an x) woul have to be the same, because simply having information about x is not enough to istinguish between the two functions. Thus, we nee the erivative to epen on the epenent variable y, because we nee information about y to istinguish on which portion of the curve we are on, as ifferent portions of the curve with the same x value have ifferent erivatives. Example 2 Fin f/x for sin(f) + f = x.

3 Solution Here we have no choice but to apply implicit ifferentiation. sin(f) + f = x x (sin(f) + f) = x x cos(f) f x + f f (cos(f) + 1) x = 1 f cos(f) + 1 Before we procee, we like to introuce the notion of a line that is normal to a curve. The wors normal, orthogonal, an perpenicular all have the same meaning, but are generally use in ifferent fiel. A line normal to a curve is a line that crosses the line tangent to the curve at a 90 egree angle. The slope of this line is the negative reciprocal of the slope of the tangent line. Thus, if the slope of the tangent line is m, the slope of the normal will be 1/m. In some fiels, such as geometric optics, the line normal to a curve (or lens) is a particular interest. Example 3 Fin the equation for the lines tangent an normal at the point (2, 4) to the folium of Descartes, given by x 3 + y 3 9xy = x 3 + y 3-9xy = x

4 Solution The first thing we will nee to o is fin the slope of the curve at the point (2,4). x 3 + y 3 9xy = 0 3x 2 + 3y 2 y y 9y 9x x x = 0 (3y 2 9y) y x = 9y 3x2 y 9y 3x2 3y x2 = x 3y 2 = 9y y 2 3y Evaluating the erivative at the point (2,4) we fin y 3y x2 = x (2,4) y 2 = y (2,4) 16 6 = 4 5 Now that we have the slope of the tangent line at the point of interest, we use the point-slope form to fin y t = 4 5 (x 2) + 4 = 4 5 x Now as far as the line normal to the curve at this point is concerne, we nee to fin the line perpenicular to the tangent line. This line will cross through the same point, but the slope will be the negative reciprocal of the slope of the tangent line. It follows that y n = 5 4 (x 2) + 4 = 5 4 x Example 4 Fin 2 y/x 2 for 2x 3 3y 2 = 8. Solution We begin by fining the first erivative Now the secon erivative x (2x3 3y 2 ) = x 8 6x 2 6yy = 0 x 2 yy = 0 y = x2 y y = x 2 x y = 2xy x2 y y 2 = 2x y x2 y y 2 = 2x y x4 y 3 Finally, we can use implicit ifferentiation to fin the erivative of inverse functions. Example 5 Fin the erivative of y = ln(x) using implicit ifferentiation. Solution Presuming that we on t know the erivative of ln(x), we woul rewrite this equation as e y = x using the inverse function. Now we can use implicit ifferentiation (because we know how to ifferentiate both sies of the equation) to fin so e y y y e y = 1 e ln(x) = 1 x

5 which is the familiar result. Example 6 Fin the erivative of y = sin 1 (x) using implicit ifferentiation. Solution Since we on t know how to ifferentiate sin 1 (x), we nee to rewrite the equation to a form in which we can ifferentiate both sies. Writing that sin(y) = x an ifferentiating we fin so that cos(y) y y cos(y) = 1 cos(sin 1 (x)) to simplify we recall the useful ientity sin 2 (x)+cos 2 (x) = 1 which leas to cos(x) = 1 (sin(x)) 2. When we substitute into the above equation, we will apply sine to its inverse (yieling x), an fin x sin 1 (x) = 1 cos(sin 1 (x)) = 1 1 (sin(sin 1 x)) 2 = 1 1 x 2