Differentiating Implicitly Defined Functions

0.1 Differentiating Implicitly Defined Functions Thus far, the functions we have been concerned with have been defined explicitly. A function is defined explicitly if the output is given directly in terms of the input. For instance, for the function f = 4x2 the value of f is given explicitly or directly in terms of the input. Just by knowing the input we can immediately find the output. A second type of function that is also useful for us to consider is an implicitly-defined function. A√function is defined implicitly if the output cannot be found directly from the input. For instance, f = 2x is an implicitly-defined function, because for each positive x value there is a corresponding f value, but we cannot find it directly from the function. We first need to square both sides, and then we find the values of f directly from the explicitly-defined function f = 4x2 . In general, it is not possible to rewrite an implicitly-defined function as an explicitly-defined function. Consider the function f given by sin(f ) + f = x. For a given x value, there may be a corresponding output value f which makes this a true statement. In this way a function f would be defined for all such x where there is a solution. For instance, we have f (0) = 0, because setting x = f = 0 makes the above equation a true statement. As we have defined them, exponential functions are an entire class of implicitly-defined functions. For instance, f given by f = 2x is defined as the number that when raised to the 1/x power gives 2, even if we don’t know how to find that number. A function is a relation that provides us with at most output for a single input. More generally, we can consider relations that provide us with multiple outputs for a single input, such as x2 +y 2 = 1, the equation of the unit circle. √ We can think of this more general relation as consisting of multiple functions, in this case y = ± 1 − x2 . In general, we can describe an arbitrary curve (like a circle) as a combination of some number of functions. It follows that we should be able to find lines tangent to arbitrary curves, simply by differentiating the function that describes the appropriate portion of the curve. In some situations, like a circle, we can find explicit expressions for the multiple functions our curve consists of. However, in general we will not be able to do so. Nevertheless, because a more complicated curve still consists of functions, we should be able to find its derivative at a particular point. This is a situation in which the power of the operator perspective of differentiation shines through. Just like when we multiply by a constant, add a constant, etc, differentiation should not change the truth of a given equation (if we are differentiating the same or equal functions, the result must also be equal). In this way, we can look at the function f = x2 and think about finding the derivative of f with respect to x by differentiating both sides of the equation. When we do so, we find d d 2 f= x , dx dx or df = 2x, dx which is a familiar result. When we differentiate implicitly-defined functions it is essential to keep in mind that f and x are not independent. In the above expression, f has an explicit dependence on x, but x also has an implicit dependence on f . If we change either of these two variables, the other must correspondingly change to make the equation remain true. Now let’s suppose we were interested in looking at d d f = x2 , df df 1 which would yield 1= d 2 x . df It is clear that the right-hand side of the equation must be nonzero, because otherwise the equation would not be a true statement. However, it is not clear exactly how x2 changes with respect to f , so we cannot directly evaluate the derivative on the right-hand side. Nevertheless, we can view x as a some unknown function of f , and apply the chain rule. The chain rule tells us to multiply the rate of change of x2 with respect to x by the rate of change of our intermediate function x(f ) with respect to f . In other words, d 2 d dx d 2 x = x · x = 2x · . df dx df df In the above expression we are left with df /dx, which is an unknown quantity. However, if we equate this expression with the left-hand side of our equation we find 1 = 2x df , dx which allows us to solve for df /dx, finding 1 df = . dx 2x In essence, when differentiating implicitly-defined functions we need to take into account the fact that there is dependence between our two variables of interest. The chain rule gives us a means of taking this dependence into account, by allowing us to differentiate through an invisible intermediate function. Once we differentiate in this way we are left with the unknown derivative, which we are then able to solve for. Example 1. Find dy/dx if y 2 = x. Solution In order to proceed we differentiate both sides of the given equation. y2 = x d d 2 y = x dx dx d 2 dy d y · = x dy dx dx dy 2y = 1 dx dy 1 = . dx 2y Note that our derivative depends on y. If we look at the graph of the curve y 2 = x we can see the √ reason for this. This curve consists of the two functions ± x. Now suppose that we are interested in finding the line tangent to this curve at some point x = 0. If the derivative depended solely on √ √ x, we would have no way of distinguishing whether or not we were on the function x or − x, yet clearly the lines tangent to these two functions are different. Thus, we need the derivative to depend on the dependent variable y, because we need information about y to distinguish on which portion of the curve we are on, as different portions of the curve with the same x value have different derivatives. 2 Example 2. Find df /dx for sin(f ) + f = x. Solution We proceed by differentiating both sides of the given equation. sin(f ) + f d (sin(f ) + f ) dx df df cos(f ) + dx dx df (cos(f ) + 1) dx df dx = x dx = dx = 1 = 1 = 1 . cos(f ) + 1 Before we proceed, we’d like to introduce the notion of a line that is normal to a curve. The words normal, orthogonal, and perpendicular all have the same meaning, but are generally used in different fields. A line normal to a curve is a line that crosses the line tangent to the curve at a 90 degree 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 fields, such as geometric optics, the line normal to a curve (or lens) is a particular interest. Example 3. Find the equation for the lines tangent and normal at the point (2, 4) to the folium of Descartes (see figure 1), given by x3 + y 3 − 9xy = 0. x3 + y 3 − 9xy = 0 4 2 0 −2 −4 −4 0 x −2 2 4 Figure 1: The Folium of Descartes. Solution The first thing we will need to do is find the slope of the curve at the point (2,4). x3 + y 3 − 9xy dy dy − 9y − 9x 3x2 + 3y 2 dx dx dy 2 (3y − 9y) dx dy 9y − 3x2 = 2 dx 3y − 9y = 0 = 0 = 9y − 3x2 = 3y − x2 . y 2 − 3y Evaluating the derivative at the point (2,4) we find dy dx (2,4) = 3y − x2 y 2 − 3y (2,4) 3 = 12 − 4 4 = . 16 − 6 5 Now that we have the slope of the tangent line at the point of interest, we use the point-slope form to find 4 4 12 yt = (x − 2) + 4 = x + . 5 5 5 Now as far as the line normal to the curve at this point is concerned, we need to find the line perpendicular 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 5 13 5 yn = − (x − 2) + 4 = − x + . 4 4 2 Example 4. Find d2 y/dx2 for 2x3 − 3y 2 = 8. Solution We begin by finding the first derivative d d (2x3 − 3y 2 ) = 8 dx dx 6x2 − 6yy 0 = 0 x2 − yy 0 = 0 x2 . y0 = y Now the second derivative y 00 = d x2 2xy − x2 y 0 2x x2 y 0 2x x4 = = − = − 3. dx y y2 y y2 y y Finally, we can use implicit differentiation to find the derivative of inverse functions. Example 5. Find the derivative of y = ln(x) using implicit differentiation. Solution Presuming that we don’t know the derivative of ln(x), we would rewrite this equation as ey = x using the inverse function. Now we can use implicit differentiation (because we know how to differentiate both sides of the equation) to find ey so dy = 1, dx dy 1 1 1 = y = ln(x) = , dx e x e which is the familiar result. Example 6. Find the derivative of y = sin−1 (x) using implicit differentiation. Solution Writing sin(y) = x and differentiating we find cos(y) so that dy = 1, dx dy 1 1 = = . dx cos(y) cos(sin−1 (x)) p To simplify, we recall the useful identity sin2 (x)+cos2 (x) = 1 which leads to cos(x) = 1 − (sin(x))2 . When we substitute into the above equation, we will apply sine to its inverse (yielding x), and find 1 d 1 1 =p sin−1 (x) = =√ . −1 2 dx cos(sin−1 (x)) 1 − x2 1 − (sin(sin x)) 4