[PDF] Lecture 17: Implicit di erentiation - Nathan Pflueger

INVERSE TRIGONOMETRIC FUNCTIONS

arcsin(x), a set typically with an infinite number of angle values, and Arcsin(x), a specific representative angle from that set They use the “small a” notation, arcsin(x), to mean the one principal value Similarly for Arccos(x)and Arctan(x)

ChapitreVFonctions arcsin arccos arctan 1 Définitions

cours du mercredi 1/3/17 ChapitreVFonctionsarcsin; arccos; arctan 1 Définitions 1 1 arcsin Proposition1 1 La fonction sin : [ ˇ=2;ˇ=2] [ 1;1] est une bijection

Inverse functions

lnbdoes not exist if b 0; arcsin(x) does not exist if jxj>1: Similar facts hold for arctan, arccos and so on By the end of your rst calculus course, you should be able to compute the derivative of an inverse

Lecture 17: Implicit di erentiation - Nathan Pflueger

on calculators by sin 1;cos ;tan 1, and they are often called in other places by the names arcsin;arccos;arctan (there are also, of course, inverse functions of sec;csc, and cot, but we won’t discuss these as much) In these cases, ipping the graph of the original functions give plots that have many yvalues of each xvalue, so there

Exo7 - Cours de mathématiques

ter à notre catalogue de nouvelles fonctions : ch,sh,th,arccos,arcsin,arctan,argch,argsh,argth Ces fonctions apparaissent naturellement dans la résolution de problèmes simples, en particulier issus de la physique

Conseils de travail pour les vacances de Toussaint

i) Tracer sur un m^eme gure les graphes de Arcsin et Arccos Que dire? ii) Donner une relation simple entre Arccos(−x) et Arccos(x) pour tout x∈[−1;1] 3) Etude de la fonction Arctan a) Etude seulement a partir de la d e nition comme \fonction r ecip "

Planche no 13 Fonctions circulaires réciproques : corrigé

Arctan a+b 1−ab si ab < 1 Arctan a+b 1−ab +π si ab > 1 et a > 0 Arctan a+b 1−ab −π si ab > 1 et a < 0 Exercice no 3 Pour x réel, on pose f(x)= Z sin2x 0 Arcsin √ t dt+ Z cos2 x 0 Arccos √ t dt La fonction t 7→ Arcsin √ t est continue sur [0,1] Donc, la fonction y 7→ Z y 0 Arcsin √ t dt est définie et dérivable sur

˘ ˇ - melusineeuorg

Title (Microsoft Word - 12 Fonctions circulaires r\351ciproques doc) Author: Ismael Created Date: 4/8/2006 7:31:40

Cours de mathématiques MPSI - AlloSchool

La fonction f: x 7arccos(cos(x)) n’est pas l’identité, elle est 2 - périodique et paire, il suffit donc l’étudier sur [0; ] intervalle sur lequel f ( x ) ˘ x Attention

Fonctionsusuelles - GitHub Pages

©LaurentGarcin MPSILycéeJean-BaptisteCorot Fonctionsusuelles 1 Fonctionslogarithme,exponentielleetpuissances 1 1 Fonctionlogarithmeetexponentielle

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Lecture 17: Implicit dierentiation

Nathan P

ueger

18 October 2013

1 Introduction

Today we discuss a technique called implicit dierentiation, which provides a quicker and easier way to

compute many derivatives we could already do, and also can be used to evaluate some new derivatives. The

main example we will see of new derivatives are the derivatives of the inverse trigonometric functions.

Implicit dierentiation is not a new dierentiation rule; instead, it is a technique that can be applied with

the rules we've already learned. The idea is this: instead of trying to tackle the desired function explicitly,

instead just nd a simple equation that the function satises (called animplicit equation.If you dierentiate

both sides of this equation, then you can usually recover the derivative of the function you actually cared

about with just a little algebra.

The reference for today is Stewart sectionx3:5 (for implicit dierentiation generally) andx3:6 (for inverse

trig functions specically).

2 Review: Inverse functions

Functions like

px;lnx;tan1x(which is also written arctan(x)) and so forth are examples of what are called inverse functions.The idea is that they invert to the eect for some other function. In these cases: (px)2=x, sopxis the inverse ofx2. elnx=x, so lnxis the inverse ofex. tan(tan1x) =x, so tan1xis the inverse of tanx. The general denition is:g(x) is an inverse function off(x) iff(g(x)) =xfor allxin the domain ofg. Note.Many functions have more than one inverse function. For example,pxis also an inverse function ofx2.

Graphically, you obtain inverse functions by re

ecting the graph of the original function across the line y=x; that is, switch the roles ofxandy, as in the following pictures. 1 y=exy= lnxy=x2y=px y=px Note in the second picture that the two possible inverse functionstogetherform the re ection of the original graph, but neither does individually.

The main inverse functions we are interested in are the inverse trigonometric functions. These are labeled

on calculators by sin

1;cos1;tan1, and they are often called in other places by the names arcsin;arccos;arctan

(there are also, of course, inverse functions of sec;csc, and cot, but we won't discuss these as much). In these

cases, ipping the graph of the original functions give plots that havemanyyvalues of eachxvalue, 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

1;cos1;tan1. The following pictures show the arc chosen in solid red, and

the rest of the \ ipped graph" in dashed red.

Note in particular that the conventional \inverse trigonometric functions" have the following ranges.

It is confusing that the inverse cosine has a dierent range from the other two, but you can always visualize these pictures to remember which range should be which.

Function Range

sin

1(x) [=2;=2]

cos

1(x) [0;]

tan

1(x) (=2;=2)

2 y= sin(x)y= sin1(x)y= cos(x)y= cos1(x)y= tanxy= tan1(x)3

3 Derivatives of inverse functions

If you can dierentiate a function, you can always write an expression for the derivative of its inverse. The

technique is the rst example of what we'll later callimplicit dierentiation.I'll illustrate the idea rst by

nding the derivative of the natural log function.

First, write the following equation.

e lnx=x This equation can be regrading as basically just thedenitionof the natural logarithm. Except it

doesn't dene itexplicitly; it dened itimplicitly, but telling an equation it must satisfy. Now the technique

is: dierentiate both sides of this equation. Since the two sides are equal, their derivatives must be equal

also. ddx elnx=ddx x Now consider the two sides of this equation separately.

Left side:ddx

elnx. This is a composition of two functions: lnxandex. So you can dierentiate it with the chain rule, using the fact that ddx ex=x. In Leibniz notation, it looks like this. ddx elnx=dd(lnx)elnxddx lnx(chain rule) =elnxddx lnx(derivative ofex) =xddx lnx(sinceelnx=x)

Right side:

ddx x. This is easy: the derivative ofxis just 1. ddx x= 1 Putting these together:We can now just set these two expressions equal to each other, and thensolve the equation. ddx elnx=ddx x(starting point) x ddx lnx= 1 (analysis of the two sides, above) ddx lnx=1x (divide both sides byx) The result of this work is that, as if by magic, we were able tosolvefor the derivative of lnx.

Aside.Here's a quick and easy way to remember the derivative of natural log, if you have a good visual

imagination. If this doesn't make any sense to you, don't worry about; it isn't essential. The derivative of

e

xis justex. That means that if (a;b) is any point on the graphy=ex, then the slope of the tangent line is

b. Now, \ ip" this picture be switchingxandy. That means that (b;a) is a point on the graph of natural log. The slope of the tangent line ips becomes 1b (because rise and run have traded placed). So: the slope

of the tangent line to the graphy= lnxis just the reciprocal of thex-coordinate. This is the same thing

as saying that ddx lnx=1x . It is a good exercise to think this through and understand why this is logically equivalent to the algebra with the chain rule I've shown above. 4

3.1 Dierentiating inverse trig functions

In this subsection, I'll show how to compute the derivatives of sin

1xand tan1x, using the same technique

as I've shown above. As before, the method is: write down an equation satised by the function, dierentiate

both sides of this equation, and solve.

First, consider tan

1x. It is an inverse function of tangent; this just means that it satises this equation.

tan(tan

1x) =x

Now, dierentiate both sides of the equation. The right side is easy. The left side requires the chain rule.

It also requires the fact that the derivative of tanxis sec2x, which we found in a previous lecture using the

quotient rule. ddx tan(tan1x) =ddx x sec

2(tan1x)ddx

tan1x= 1 (Chain rule on the left; easy derivative on the right) ddx tan1x=1sec

2(tan1x)

Now, we at least have a formula for the derivative of tan

1x, and in principle this means our work is

done. But it is a good idea to simplify this expression slightly, because it will become something much

simpler. To do this, remember exactly what tan 1xis. tan

1xis precisely that angle in (=2;=2) whose tangent isx. So if we draw the following right

triangle, the marked angle will have measure tan

1xradians.1x

tan

1xRemember:tan1xis anangle. Since tanxtakes angles to ratios, tan1takes ratios back to angles.

In the triangle above, it is the angle marked with the little arc in the corner. Where did this triangle come from?This was the most pressing question when we talked about

this picture in class. The answer is: I made it up. It is a prop. But I've made it up with a specic purpose:

it's purpose is to have the angle tan

1xin it. The whole purpose of the triangle is to exhibit this angle so

that I can study it. Back to the man discussion, remember what we're trying to nd here. We want to know what sec

2(tan1x)

is. I've drawn a triangle with the angle tan

1xin it. Since the secant of an angle ishypotenuseadjacent

, we just need to nd the hypotenuse of this triangle. We can get this from the Pythagorean theorem. 5

1xp1 +x2tan

1xSo the secant of the angle tan

1xis preciselyp1 +x2=1 =p1 +x2. So sec2(tan1x) = 1 +x2. Using

this, we can put the derivative of tan

1in its simplest form.

ddx

tan1x=11 +x2Alternative method.Recall that there is an identity (one of the many equivalent forms of the Pythagorean

theorem) sec

2x= tan2x+ 1. We could also have used this identity to do this algebra, as follows:

sec

2(tan1x) = 1 + tan2(tan1x)

= 1 +tan(tan1x)2 = 1 +x2 Aside.Notice thatthe only fact we've used abouttan1is that it is an inverse function oftanx.So actually,

11+x2is the derivative ofanyinverse function of tanx, not just the \standard" one that we call

tan

1. If you look at the picture in the last section, you'll see why this makes sense: if you

ip the graph of

y= tanx, there are many arcs, of whichy= tan1xis only one, but all of them are just vertical translates

of each other. So they all have the same derivative function.

Now let's move on to sin

1x. The technique is the same. Write an equation describing sin1ximplicitly,

and dierentiate both sides of it. The right side is easy, and the left side uses the chain rule. sin(sin

1x) =x(because it's an inverse function)

ddx sin(sin1x) =ddx x(dierentiate both sides) cos(sin

1x)ddx

sin1x= 1 (chain rule on left; easy derivative on right) ddx sin1x=1cos(sin

1x)(divide on both sides)

Like before, in principle we're done here: we have a totally well-dened expression for the derivative of

sin

1, which you can plug into a computer and everything. But again, it's worth re-expressing it in a simpler

way. Like before, there are two standard ways to do this, which both just boil down to the Pythagorean

theorem. The quickest one is to use the identity cos

2x+sin2x= 1. Lettingxbe the angle sin1x, this says:

6 cos

2sin1x+ sin2sin1x= 1

cos(sin1x)2+x2= 1 cos(sin1x)2= 1x2 cos(sin

1x) =p1x2

So this gives the simplication that we want.

ddx

sin1x=1p1x2Technical point I've swept under the rug:I took the square root of both sides of the equationcos(sin1x)2=

1x2above, to get cos(sin1x) =p1x2. But technically, all I could conclude here is cos(sin1x) =

p1x2(two numbers whose squares are equal are not necessarily equal: they are either equal or oppo-

site). So how do we know that the plus sign is correct? Here is where we must use the specic denition

of sin

1x: it doesn't return just any angle whose sine isx: it return the angle in [2

;2 ] whose sine isx.

And the cosine of any angle in [=2;=2] is not negative. If we chose a dierent inverse function of sinx,

the derivative could be1p1x2instead. Look at the picture in the previous section to see that this makes

sense. If this confuses you, don't worry too much about it; I mention it for completeness.

Dierentiating cos

1xwith this method is completely analogous. It is left to you as a homework problem.

3.2 Dierentiating any inverse function

The method shown above works in general to dierentiate any inverse function you like. Here's how it works:

suppose thatf(x) is any function, andf1(x) is an inverse function. Follow the same steps as above. f(f1(x)) =x(inverse function) ddx f(f1(x)) =ddx x(dierentiate both sides) f

0(f1(x))ddx

f1(x) = 1 (chain rule) ddx f1(x) =1f

0(f1(x))

This last line is a valid formula for inverse function. Here are some examples. Example3.1.Supposef(x) =ex. Thenf0(x) =exandf1(x) = lnx. Soddx lnx=1e lnx=1x Example3.2.Supposef(x) = sinxandf1(x) = sin1(x). Thenddx sin1x=1cos(sin

1(x)), as in the last

section. Example3.3.Supposef(x) =x2andf1(x) =px. Thenf0(x) = 2x, soddx px=12f1(x)=12 px

4 Implicit and explicit functions

In the previous sections, we've seen that a useful way to dierentiate inverse functions like lnxof sin1x

is not to attack them directly, but instead write some equation that describes themimplicitly, dierentiate

both sides of the equation, and solve. We're now going to discuss this technique in more generality.

As a rst step, I want to elaborate on what it means to write an implicit equation. An implicit equation

is just an equation in two variablesxandythat is not necessarily of the formy=f(x). Here are some 7 examples.

Explicit equations:

y=px y=p1x2 y=xx1

Implicit equations:

y2=x x2+y2= 1 xy=x+y

One major distinction between explicit and implicit equations is that implicit equationsdon't necessarily

describe graphs of functions.Instead, they describe graphs of curves. Perhaps the simplest example is this

implicit equation, which denes a circle.x

2+y2= 1describes this curve:

Now there are two graphs that lie on the circle above:y=p1x2andy=p1x2(one is the upper semicircle, and one is the lower semicircle). So theimplicitequationx2+y2= 1 describes two dierent explicitequations. This is often the case with implicit equations. Here are two more examples of implicit equations, which we'll revisit in the last section. Example4.1.(Descartes' leaf) Consider the following equation. x

3+y3= 6xy

This is an implicit equation. It describes the curve shown below (made with Wolfram alpha).8

This curve is traditionally called the \folium of Descartes" (\folium" is Latin for \leaf"). This equation

rst occurred in a letter from Descartes to Fermat (Fermat was a lawyer by profession, but did mathematics

as a hobby). Fermat claimed to have a method to nd tangent lines to any curve, and Descartes invented

this curve as a challenge to Fermat. Fermat was successful in nding tangent lines to the curve. Today,

however, the problem of nding tangent lines is very easy, due to the invention of calculus some time later.

This event was notable enough in the history of calculus that it is memorialized on the following Albanian

postage stamp

1.Example4.2.Consider 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 dierent levels of

zoom).As you can see, there aremanydierent functions that obey this implicit equation, because there are

many dierent values ofyfor a given value ofx.1

I am not aware of any relation between Descartes and Albania, but I only spent a couple minutes googling for it.

9

5 Implicit dierentiation

Now we'll look at how you can use an implicit equation to nd derivatives. The basic technique will always

be as follows. Dierentiate both sides of the implicit equation. Remember thatyis a function of x.

Use algebra to solve fordydx

. The result will be an expression inxandy.

Let's begin with a simple example.

Example5.1.Consider the implicit equation of a circle. x

2+y2= 1

Now imagine thatyis some function ofxthat obeys this implicit equation. We can dierentiate both sides of the equation with respect tox. ddx (x2+y2) =ddx 1 ddx x2+ddx y2= 0

2x+ 2ydydx

= 0 (chain rule)

2ydydx

=2x dydx =2y2x dydx =yx This shows that the slope of the tangent line to the circle described byx2+y2= 1 is always given by y=xat a point (x;y).

Note.When you solve fordydx

, it will almost always be an expression in terms of bothxandy. In some

cases, you can re-express it as something purely in terms ofx, but not always. The next example shows one

case where you can re-express it.

Example5.2.Suppose thaty=3pcosx+ 7. Finddydx

using implicit dierentiation.

Note.You can dierentiate this using the chain rule as well. Of course you will get the same answer as we

get below, but you may nd one technique or the other easier. Solution.Cube both sides to obtainy3= cosx+ 7. Dierentiate both sides: ddx y3=ddx (cosx+ 7) (dierentiate both sides)

3y2dydx

=sinx(chain rule on the left, known derivatives on the right) dydx =sinx3y2(divide both sides by 3y2) In this case, we have an explicitly equation fory, namelyy=3pcosx+ 7, so we can substitute that back into the answer here to get the derivative purely in terms ofx. 10 dy dx =sinx3

3pcosx+ 72

=sinx3(cosx+ 7)3=2 To summarize, there are two main reasons to dierentiate implicit equations. 1. Because y ouha veno explicit equation for yin terms ofx(you can still obtaindydx in terms ofxandy). 2.

Because y ouha vean expl icitequation, b utan implicit equation is m uchsimpler (in this case, y oucan

substitute your explicit equation back at the end to get dydx purely in terms ofx).

6 Examples

As examples of the technique of implicit dierentiation, we will nd some tangent lines to the two implicit

curves given in section 4. Example6.1.Consider the equation for Descartes' leaf. x

3+y3= 6xy

1. Find the tangen tline to this curv eat the p oint(3 ;3). 2. Find the tangen tline to this curv eat the p oint( 43
;83 Solution.Begin by dierentiating both sides of the equation, as functions ofx. ddx x3+y3=ddx (6xy)

3x2+ 3y2dydx

= 6dxdx y+ 6xdydx (product rule used on the right side)

3x2+ 3y2dydx

= 6y+ 6xdydx

3y2dydx

6xdydx

= 6y3x2(move all thedydx to one side) (3y26x)dydx = 6y3x2(group like terms) dydx =6y3x23y26x(divide on both sides) dydx =2yx2y

22x(cancel the factor of 3 on top and bottom)

Now, we can use this expression to compute the two desired tangent lines. 1. A tthe p oint(3 ;3), the slope of the tangent line isdydx =23323

223=33

=1. So the tangent line is the line with slope1 through the point (3;3). Therefore this line is given by the equation (y3) = (1)(x3), or alternativelyy=x+ 6.2.A tthe p oint( 43
;83 ), the slope of the tangent line is given by283 (43 )2( 83
)2243 . This simplies to16=316=964=98=3=

48=916=964=924=9=32=950=9=3240

=45 . So the equation of the tangent line is (y83 ) =45 (x43 ), or if you prefer,y=45 x+85 .11 The curve, with these two tangent lines, is shown below. y=x+ 6y=45 x+85 (3;3)( 43
;83 )Example6.2.Consider the curve given by the following implicit equation. tan(x+y) = sin(xy) Find the tangent line to this curve at the point ( p;p). Solution.Begin by dierentiating both sides of the equation. As usual, regardyas a function ofx. tan(x+y) = sin(xy) ddx tan(x+y) =ddx sin(xy) sec

2(x+y)ddx

(x+y) = cos(xy)ddx (xy) (chain rule on both sides) sec

2(x+y)dxdx

+dydx = cos(xy)dxdx y+xdydx (product rule on the right) sec

2(x+y)

1 +dydx

= cos(xy)quotesdbs_dbs8.pdfusesText_14