Lecture 17: Implicit differentiation


 Myles Watts
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1 Lecture 7: Implicit ifferentiation Nathan Pflueger 8 October 203 Introuction Toay we iscuss a technique calle implicit ifferentiation, which provies a quicker an easier way to compute many erivatives we coul alreay o, an also can be use to evaluate some new erivatives. The main example we will see of new erivatives are the erivatives of the inverse trigonometric functions. Implicit ifferentiation is not a new ifferentiation rule; instea, it is a technique that can be applie with the rules we ve alreay learne. The iea is this: instea of trying to tackle the esire function explicitly, instea just fin a simple equation that the function satisfies (calle an implicit equation. If you ifferentiate both sies of this equation, then you can usually recover the erivative of the function you actually care about with just a little algebra. The reference for toay is Stewart section 3.5 (for implicit ifferentiation generally) an 3.6 (for inverse trig functions specifically). 2 Review: Inverse functions Functions like x, ln x, tan x (which is also written arctan(x)) an so forth are examples of what are calle inverse functions. The iea is that they invert to the effect for some other function. In these cases: ( x) 2 = x, so x is the inverse of x 2. e ln x = x, so ln x is the inverse of e x. tan(tan x) = x, so tan x is the inverse of tan x. The general efinition is: g(x) is an inverse function of f(x) if f(g(x)) = x for all x in the omain of g. Note. Many functions have more than one inverse function. For example, x is also an inverse function of x 2. Graphically, you obtain inverse functions by reflecting the graph of the original function across the line y = x; that is, switch the roles of x an y, as in the following pictures.
2 y = e x y = x 2 y = ln x y = x y = x Note in the secon picture that the two possible inverse functions together form the reflection of the original graph, but neither oes iniviually. The main inverse functions we are intereste in are the inverse trigonometric functions. These are labele on calculators by sin, cos, tan, an they are often calle in other places by the names arcsin, arccos, arctan (there are also, of course, inverse functions of sec, csc, an cot, but we won t iscuss these as much). In these cases, flipping the graph of the original functions give plots that have many y values of each x value, so there are many possible inverse functions. By convention, we simply choose one arc of each of these graphs to get the functions that we call sin, cos, tan. The following pictures show the arc chosen in soli re, an the rest of the flippe graph in ashe re. Note in particular that the conventional inverse trigonometric functions have the following ranges. It is confusing that the inverse cosine has a ifferent range from the other two, but you can always visualize these pictures to remember which range shoul be which. Function Range sin (x) [ π/2, π/2] cos (x) [0, π] tan (x) ( π/2, π/2) 2
3 y = sin(x) y = sin (x) y = cos (x) y = cos(x) y = tan x y = tan (x) 3
4 3 Derivatives of inverse functions If you can ifferentiate a function, you can always write an expression for the erivative of its inverse. The technique is the first example of what we ll later call implicit ifferentiation. I ll illustrate the iea first by fining the erivative of the natural log function. First, write the following equation. e ln x = x This equation can be regraing as basically just the efinition of the natural logarithm. Except it oesn t efine it explicitly; it efine it implicitly, but telling an equation it must satisfy. Now the technique is: ifferentiate both sies of this equation. Since the two sies are equal, their erivatives must be equal also. eln x = x Now consier the two sies of this equation separately. Left sie: eln x. This is a composition of two functions: ln x an e x. So you can ifferentiate it with the chain rule, using the fact that ex = x. In Leibniz notation, it looks like this. eln x = (ln x) eln x ln x (chain rule) = e ln x ln x (erivative of ex ) = x ln x (since eln x = x) Right sie: x. This is easy: the erivative of x is just. x = Putting these together: We can now just set these two expressions equal to each other, an then solve the equation. eln x = x (starting point) x ln x = (analysis of the two sies, above) ln x = (ivie both sies by x) x The result of this work is that, as if by magic, we were able to solve for the erivative of ln x. Asie. Here s a quick an easy way to remember the erivative of natural log, if you have a goo visual imagination. If this oesn t make any sense to you, on t worry about; it isn t essential. The erivative of e x is just e x. That means that if (a, b) is any point on the graph y = e x, then the slope of the tangent line is b. Now, flip this picture be switching x an y. That means that (b, a) is a point on the graph of natural log. The slope of the tangent line flips becomes b (because rise an run have trae place). So: the slope of the tangent line to the graph y = ln x is just the reciprocal of the xcoorinate. This is the same thing as saying that ln x = x. It is a goo exercise to think this through an unerstan why this is logically equivalent to the algebra with the chain rule I ve shown above. 4
5 3. Differentiating inverse trig functions In this subsection, I ll show how to compute the erivatives of sin x an tan x, using the same technique as I ve shown above. As before, the metho is: write own an equation satisfie by the function, ifferentiate both sies of this equation, an solve. First, consier tan x. It is an inverse function of tangent; this just means that it satisfies this equation. tan(tan x) = x Now, ifferentiate both sies of the equation. The right sie is easy. The left sie requires the chain rule. It also requires the fact that the erivative of tan x is sec 2 x, which we foun in a previous lecture using the quotient rule. tan(tan x) = x sec 2 (tan x) tan x = (Chain rule on the left; easy erivative on the right) tan x = sec 2 (tan x) Now, we at least have a formula for the erivative of tan x, an in principle this means our work is one. But it is a goo iea to simplify this expression slightly, because it will become something much simpler. To o this, remember exactly what tan x is. tan x is precisely that angle in ( π/2, π/2) whose tangent is x. So if we raw the following right triangle, the marke angle will have measure tan x raians. x tan x Remember: tan x is an angle. Since tan x takes angles to ratios, tan takes ratios back to angles. In the triangle above, it is the angle marke with the little arc in the corner. Where i this triangle come from? This was the most pressing question when we talke about this picture in class. The answer is: I mae it up. It is a prop. But I ve mae it up with a specific purpose: it s purpose is to have the angle tan x in it. The whole purpose of the triangle is to exhibit this angle so that I can stuy it. Back to the man iscussion, remember what we re trying to fin here. We want to know what sec 2 (tan x) is. I ve rawn a triangle with the angle tan x in it. Since the secant of an angle is hypotenuse, we just ajacent nee to fin the hypotenuse of this triangle. We can get this from the Pythagorean theorem. 5
6 + x 2 x tan x So the secant of the angle tan x is precisely + x 2 / = + x 2. So sec 2 (tan x) = + x 2. Using this, we can put the erivative of tan in its simplest form. tan x = + x 2 Alternative metho. Recall that there is an ientity (one of the many equivalent forms of the Pythagorean theorem) sec 2 x = tan 2 x +. We coul also have use this ientity to o this algebra, as follows: sec 2 (tan x) = + tan 2 (tan x) = + ( tan(tan x) ) 2 = + x 2 Asie. Notice that the only fact we ve use about tan is that it is an inverse function of tan x. So actually, +x is the erivative of any inverse function of tan x, not just the stanar one that we call 2 tan. If you look at the picture in the last section, you ll see why this makes sense: if you flip the graph of y = tan x, there are many arcs, of which y = tan x is only one, but all of them are just vertical translates of each other. So they all have the same erivative function. Now let s move on to sin x. The technique is the same. Write an equation escribing sin x implicitly, an ifferentiate both sies of it. The right sie is easy, an the left sie uses the chain rule. sin(sin x) = x (because it s an inverse function) sin(sin x) = x (ifferentiate both sies) cos(sin x) sin x = (chain rule on left; easy erivative on right) sin x = cos(sin (ivie on both sies) x) Like before, in principle we re one here: we have a totally wellefine expression for the erivative of sin, which you can plug into a computer an everything. But again, it s worth reexpressing it in a simpler way. Like before, there are two stanar ways to o this, which both just boil own to the Pythagorean theorem. The quickest one is to use the ientity cos 2 x + sin 2 x =. Letting x be the angle sin x, this says: 6
7 So this gives the simplification that we want. cos 2 ( sin x ) + sin 2 ( sin x ) = [ cos(sin x) ] 2 + x 2 = [ cos(sin x) ] 2 = x 2 cos(sin x) = x 2 sin x = x 2 Technical point I ve swept uner the rug: I took the square root of both sies of the equation [ cos(sin x) ] 2 = x 2 above, to get cos(sin x) = x 2. But technically, all I coul conclue here is cos(sin x) = ± x 2 (two numbers whose squares are equal are not necessarily equal: they are either equal or opposite). So how o we know that the plus sign is correct? Here is where we must use the specific efinition of sin x: it oesn t return just any angle whose sine is x: it return the angle in [ π 2, π 2 ] whose sine is x. An the cosine of any angle in [ π/2, π/2] is not negative. If we chose a ifferent inverse function of sin x, the erivative coul be x instea. Look at the picture in the previous section to see that this makes 2 sense. If this confuses you, on t worry too much about it; I mention it for completeness. Differentiating cos x with this metho is completely analogous. It is left to you as a homework problem. 3.2 Differentiating any inverse function The metho shown above works in general to ifferentiate any inverse function you like. Here s how it works: suppose that f(x) is any function, an f (x) is an inverse function. Follow the same steps as above. f(f (x)) = x (inverse function) f(f (x)) = x (ifferentiate both sies) f (f (x)) f (x) = (chain rule) f (x) = f (f (x)) This last line is a vali formula for inverse function. Here are some examples. Example 3.. Suppose f(x) = e x. Then f (x) = e x an f (x) = ln x. So ln x = = e ln x x. Example 3.2. Suppose f(x) = sin x an f (x) = sin (x). Then sin x =, as in the last cos(sin (x)) section. Example 3.3. Suppose f(x) = x 2 an f (x) = x. Then f (x) = 2x, so x = 2f (x) = 2 x. 4 Implicit an explicit functions In the previous sections, we ve seen that a useful way to ifferentiate inverse functions like ln x of sin x is not to attack them irectly, but instea write some equation that escribes them implicitly, ifferentiate both sies of the equation, an solve. We re now going to iscuss this technique in more generality. As a first step, I want to elaborate on what it means to write an implicit equation. An implicit equation is just an equation in two variables x an y that is not necessarily of the form y = f(x). Here are some 7
8 examples. Explicit equations: y = x y = x 2 y = x x Implicit equations: y 2 = x x 2 + y 2 = xy = x + y One major istinction between explicit an implicit equations is that implicit equations on t necessarily escribe graphs of functions. Instea, they escribe graphs of curves. Perhaps the simplest example is this implicit equation, which efines a circle. x 2 + y 2 = escribes this curve: Now there are two graphs that lie on the circle above: y = x 2 an y = x 2 (one is the upper semicircle, an one is the lower semicircle). So the implicit equation x 2 + y 2 = escribes two ifferent explicit equations. This is often the case with implicit equations. Here are two more examples of implicit equations, which we ll revisit in the last section. Example 4.. (Descartes leaf) Consier the following equation. x 3 + y 3 = 6xy This is an implicit equation. It escribes the curve shown below (mae with Wolfram alpha). 8
9 This curve is traitionally calle the folium of Descartes ( folium is Latin for leaf ). This equation first occurre in a letter from Descartes to Fermat (Fermat was a lawyer by profession, but i mathematics as a hobby). Fermat claime to have a metho to fin tangent lines to any curve, an Descartes invente this curve as a challenge to Fermat. Fermat was successful in fining tangent lines to the curve. Toay, however, the problem of fining tangent lines is very easy, ue to the invention of calculus some time later. This event was notable enough in the history of calculus that it is memorialize on the following Albanian postage stamp. Example 4.2. Consier the following implicit equation. tan(x + y) = sin(xy) If you graph its solution curve, it is the following very complex picture (shown at two ifferent levels of zoom). As you can see, there are many ifferent functions that obey this implicit equation, because there are many ifferent values of y for a given value of x. I am not aware of any relation between Descartes an Albania, but I only spent a couple minutes googling for it. 9
10 5 Implicit ifferentiation Now we ll look at how you can use an implicit equation to fin erivatives. The basic technique will always be as follows. Differentiate both sies of the implicit equation. Remember that y is a function of x. Use algebra to solve for y. The result will be an expression in x an y. Let s begin with a simple example. Example 5.. Consier the implicit equation of a circle. x 2 + y 2 = Now imagine that y is some function of x that obeys this implicit equation. We can ifferentiate both sies of the equation with respect to x. (x2 + y 2 ) = x2 + y2 = 0 2x + 2y y = 0 (chain rule) 2y y = 2x y = 2y 2x y = y x This shows that the slope of the tangent line to the circle escribe by x 2 + y 2 = is always given by y/x at a point (x, y). Note. When you solve for y, it will almost always be an expression in terms of both x an y. In some cases, you can reexpress it as something purely in terms of x, but not always. The next example shows one case where you can reexpress it. Example 5.2. Suppose that y = 3 cos x + 7. Fin y using implicit ifferentiation. Note. You can ifferentiate this using the chain rule as well. Of course you will get the same answer as we get below, but you may fin one technique or the other easier. Solution. Cube both sies to obtain y 3 = cos x + 7. Differentiate both sies: y3 = (cos x + 7) (ifferentiate both sies) 3y 2 y = sin x (chain rule on the left, known erivatives on the right) y = sin x 3y 2 (ivie both sies by 3y 2 ) In this case, we have an explicitly equation for y, namely y = 3 cos x + 7, so we can substitute that back into the answer here to get the erivative purely in terms of x. 0
11 y = sin x 3 ( 3 cos x + 7 ) 2 sin x = 3 (cos x + 7) 3/2 To summarize, there are two main reasons to ifferentiate implicit equations.. Because you have no explicit equation for y in terms of x (you can still obtain y in terms of x an y). 2. Because you have an explicit equation, but an implicit equation is much simpler (in this case, you can substitute your explicit equation back at the en to get y purely in terms of x). 6 Examples As examples of the technique of implicit ifferentiation, we will fin some tangent lines to the two implicit curves given in section 4. Example 6.. Consier the equation for Descartes leaf. x 3 + y 3 = 6xy. Fin the tangent line to this curve at the point (3, 3). 2. Fin the tangent line to this curve at the point ( 4 3, 8 3 ). Solution. Begin by ifferentiating both sies of the equation, as functions of x. ( x 3 + y 3) = (6xy) 3x 2 + 3y 2 y = 6 y y + 6x 3x 2 + 3y 2 y y = 6y + 6x 3y 2 y y 6x = 6y 3x2 (move all the y (3y 2 6x) y = 6y 3x2 (group like terms) y y = 6y 3x2 3y 2 6x = 2y x2 y 2 2x (prouct rule use on the right sie) (ivie on both sies) to one sie) (cancel the factor of 3 on top an bottom) Now, we can use this expression to compute the two esire tangent lines.. At the point (3, 3), the slope of the tangent line is y = = 3 3 =. So the tangent line is the line with slope through the point (3, 3). Therefore this line is given by the equation (y 3) = ( )(x 3), or alternatively y = x At the point ( 4 3, ), the slope of the tangent line is given by 3 ( 4 3 )2 ( 8 3 ) This simplifies to 6/3 6/9 64/9 8/3 = 48/9 6/9 64/9 24/9 = 32/9 50/9 = = 4 5. So the equation of the tangent line is (y 8 3 ) = 4 5 (x 4 3 ), or if you prefer, y = 4 5 x
12 The curve, with these two tangent lines, is shown below. y = 4 5 x ( 4 3, 8 3 ) (3, 3) y = x + 6 Example 6.2. Consier the curve given by the following implicit equation. tan(x + y) = sin(xy) Fin the tangent line to this curve at the point ( π, π). Solution. Begin by ifferentiating both sies of the equation. As usual, regar y as a function of x. tan(x + y) = sin(xy) tan(x + y) = sin(xy) sec 2 (x + y) (x + y) = cos(xy) (xy) (chain rule on both sies) ( sec 2 (x + y) + y ) ( = cos(xy) y + x y ) (prouct rule on the right) ( sec 2 (x + y) + y ) ( = cos(xy) y + x y ) (because = ) So far so goo. Now solve this whole monster for y. 2
13 sec 2 (x + y) + sec 2 (x + y) y y = y cos(xy) + x cos(xy) (istributing terms) ( sec 2 (x + y) x cos(xy) ) y = y cos(xy) sec2 (x + y) (moving all the y to one sie) y = y cos(xy) sec2 (x + y) sec 2 (ivie on both sies) (x + y) x cos(xy) To fin the slope at the specific point ( π, π), just substitute the values of x an y into this expression. Notice first of all that xy = π an x + y = 0; these occur in multiple places in the expression. slope = π cos( π) sec 2 (0) sec 2 (0) π cos( π) = π ( ) π ( ) π = π Therefore the equation of the tangent line is (y + π) = this is y = π π + (x π). In slopeintercept form, π x 2π. If you use a calculator to approximate all these numbers, you can also π + π + write this equation as (y +.77) = 0.28 (x.77) or y = 0.28x 2.27 (any of these four boxe answers woul be acceptable on homework). This line is shown on the plot below (Wolfram alpha has marke all the points of intersection; the point ( π, π) is the secon from the right, where the line is tangent to the curve). 3
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