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Tangent, Cotangent, Secant, and Cosecant

The Quotient Rule

In our last lecture, among other things, we discussed the function 1 x , its domain and its derivative. We also showed how to use the Chain Rule to find the domain and derivative of a function of the form k(x) =1 g(x);

whereg(x) is some function with a derivative. Today we go one step further: we discuss the domain and the

derivative of functions of the form h(x) =f(x) g(x);

wheref(x) andg(x) are functions with derivatives. The rule of differentiation we will derive is called the

quotient rule. We will then define the remaining trigonometric functions, and we will use the quotient rule

to find formulae for their derivatives.

The quotient rule has the following statement: letf(x) andg(x) be two functions with derivatives. Then

we can define a function h(x) =f(x) g(x) which has domain

Dom(h) =fx2R:g(x)6= 0g

and which is differentiable everywhere on its domain, with the formula h

0(x) =g(x)f0(x)¡f(x)g0(x)

(g(x))2:

We usually callf(x) the top function, andg(x) the bottom function, and so the formula for the derivative

of the quotienth(x) is given by "bottom times the derivative of the top minus top times the derivative of

the bottom all over bottom squared." Try to remember this phrase, or one like it. Let us study the quotient rule. First, we can rewrite the functionh(x) as h(x) =f(x) g(x)=f(x)¢1 g(x):

Thus, we can think of the quotienth(x) as being a product, with the first function beingf(x) and the second

the composition ofx¡1afterg(x).

To find the domain ofh(x), we first note that sincef(x) presumably has full domain (all real numbers),

the domain ofh(x) will be precisely the same as that of (g(x))¡1(since, wherever (g(x))¡1is defined,f(x)

will also be defined, and so their product will be defined there as well). We find the domain of (g(x))¡1,

a composition, by first looking at the domain of its outside function,x¡1: the domain ofx¡1is all real

numbers except 0. This tells us that, presumingg(x) is defined everywhere, (g(x))¡1is defined at all real

numbers except thosexfor whichg(x) = 0. Thus the domain ofh(x), which is the domain of (g(x))¡1, is

the set of all real numbersxsuch thatg(x)6= 0, which is precisely what the quotient rule tells us. Now, we can see thath(x) can be written as a product. This means that, whereverh(x) is defined, we can apply the product rule to find its derivative: h

0(x) =d

dxµ f(x)¢1 =f0(x)¢1 g(x)+f(x)¢d dxµ 1 To find the derivative of (g(x))¡1, we apply the Chain Rule: d dxµ 1 =¡(g(x))¡2¢g0(x) =¡g0(x) (g(x))2: 1

Combining the two fractions under a common denominator, we get the derivative formula for the quotient

rule: h

0(x) =f0(x)¢1

g(x)+f(x)¢d dxµ 1 =f0(x) g(x)¡f(x)g0(x) (g(x))2 g(x)f0(x) (g(x))2¡f(x)g0(x) (g(x))2=g(x)f0(x)¡f(x)g0(x) (g(x))2:

Let us do an example of the quotient rule. Let

h(x) =x+ 3 x

2¡4x+ 5:

The domain ofh(x) is all real numbersxsuch that the denominator,x2¡4x+5, is not 0. The denominator

is 0 precisely whenx= 1 orx= 5, so this gives us that

Dom(h) =fx2R:x6= 1; x6= 5g:

For anyxin the domain ofh, the functionh(x) has a derivative given by the quotient rule. That derivative

is h

0(x) =(x2¡4x+ 5)¢1¡(x+ 3)¢(2x¡4)

(x2¡4x+ 5)2=x2¡4x+ 5¡2x2¡2x+ 12 (x2¡4x+ 5)2=¡x2¡6x+ 17 (x2¡4x+ 5)2:

The functionh(x) is an example of a rational polynomial function. We will be studying rational polynomial

functions later in the course.

The Other Trigonometric Functions

So far in this course, the only trigonometric functions which we have studied are sine and cosine. Today

we discuss the four other trigonometric functions: tangent, cotangent, secant, and cosecant. Each of these

functions are derived in some way from sine and cosine. The tangent ofxis defined to be its sine divided

by its cosine: tanx=sinx cosx: The cotangent ofxis defined to be the cosine ofxdivided by the sine ofx: cotx=cosx sinx:

The secant ofxis 1 divided by the cosine ofx:

secx=1 cosx; and the cosecant ofxis defined to be 1 divided by the sine ofx: cscx=1 sinx:

If you are not in lecture today, you should use these formulae to make a numerical table for each of these

functions and sketch out their graphs. Below we list the major properties of these four functions, including

domain, range, period, oddness or evenness, and vertical asymptotes. None of these functions have horizontal

asymptotes. You should verify that your sketches reflect these properties: Tangent:The function tanxis defined for all real numbersxsuch that cosx6= 0, since tangent is the quotient of sine over cosine. Thus tanxis undefined for x=:::;¡3¼ 2 2 2 ;3¼ 2 2 Its range is all real numbers, that is, for any numbery, you can always find a numberxsuch that y= tanx. The period of tanxis¼. This is a departure from sinxand cosx, which have periods of

2¼. The reason is simple: opposite angles on the unit circle (like¼

4 and5¼ 4 ) have the same tangent because of the signs of their sines and cosines. For example: tan 4 =sin¼ 4 cos 4 =p 2 2 p 2 2 = 1 =¡p 2 2 p 2 2 =sin5¼ 4 cos 5¼ 4 = tan5¼ 4 The function tanxis an odd function, which you should be able to verify on your own. Finally, at the values ofxat which tanxis undefined, tanxhas both left and right vertical asymptotes. Specifically, ifais a value ofxoutside the domain of tanx, then lim x!a¡tanx= +1and limx!a+tanx=¡1: Cotangent:The function cotxis a lot like tanx. It is defined at all values ofxfor which sinx6= 0. In other words, the domain of cotxis all real numbersxexcept

Just as in the case of tanx, the range of cotxis all real numbers, and this should not be surprising,

since essentially cotxis 1 divided by the tangent ofx. For this same reason, the period of cotxis also

¼instead of 2¼. You can also verify that it is an odd function. Finally, like tanx, the function cotx

has left and right vertical asymptotes at each point at which it is undefined. Ifais a value ofxat which cotxis undefined, then we have that lim x!a¡tanx=¡1and limx!a+tanx= +1:

Secant:The function secxis defined to be the multiplicative inverse of cosine, so it is defined precisely

where cosxis not equal to 0. So the domain of secxis all real numbersxexcept x=:::;¡3¼ 2 2 2 ;3¼ 2 Thus, secxand tanxhave the same domains. The range of secxis a bit more complicated: remember

that the bounds oncosxare¡1·cosx·1. So, we see that if the secant ofxis positive, then it can

be no smaller than 1, and if it is negative, it can be no larger than¡1. Thus the range of secxis made

up of two intervals: secx¸1 or secx· ¡1: The period of secxis precisely the same as that of cosx, which means that the period of secxis

2¼. The function secxis an even function, and this is because cosxis an even function. Finally, at

every value ofxnot in the domain of secx, the function has both left and right vertical asymptotes.

Ifa=:::;¡3¼

2 2 ;5¼ 2 ;:::then the left vertical asymptote atagoes to positive infinity and the right vertical asymptote goes to negative infinity: lim x!a¡secx= +1and limx!a+secx=¡1:

If, on the other hand,b=:::;¡5¼

2 2 ;3¼ 2 ;:::, then the left vertical asymptote atbgoes to negative infinity and the right vertical asymptote goes to positive infinity: lim x!b¡secx=¡1and limx!b+secx= +1: Cosecant:Similar to the case of secx, the function cscxis defined precisely when sinxis not equal to 0. Thus the values ofxat which cscxis undefined are 3 The range of cscxis the same as that of secx, for the same reasons (except that now we are dealing with the multiplicative inverse of sine ofx, not cosine ofx). Therefore the range of cscxis cscx¸1 or cscx· ¡1: The period of cscxis the same as that of sinx, which is 2¼. Since sinxis an odd function, cscxis

also an odd function. Finally, at all of the points where cscxis undefined, the function has both left

and right vertical asymptotes, but just as in the case of secx, the behavior of the vertical asymptotes

depends on the point. Ifa=:::;¡2¼;0;2¼;:::, then the left vertical asymptote atagoes to negative

infinity, and the right vertical asymptote goes to positive infinity: lim x!a¡cscx=¡1and lim x!a+cscx= +1:

If, on the other hand,b=¡3¼;¡¼;¼;3¼;:::, then the left vertical asymptote atbgoes to positive

infinity, and the right vertical asymptote goes to negative infinity: lim x!b¡cscx= +1and limx!b+cscx=¡1:

Derivatives of Trigonometric Functions

Now that we know the properties of all of the trigonometric functions, we should take their derivatives. All

of the trigonometric functions are differentiable wherever they are defined. To find their derivatives, we use

the quotient rule and the Chain Rule: d dx(tanx) =d dxµ sinx cos

2x=cos2x+ sin2x

cos 2x=1 cos

2x= sec2x

d dx(cotx) =d dx³ cosx sinx´ sin

2x=¡sin2x+ cos2x

sin

2x=¡1

sin

2x=¡csc2x

d dx(secx) =d dxµ 1 =¡1 cos

2x¢(¡sinx) =sinx

cos 2x=1 cosx¢sinx cosx= secxtanx d dx(cscx) =d dxµ 1 =¡1 sin

2x¢(cosx) =¡cosx

sin

2x=¡1

sinx¢cosx sinx=¡cscxcotx:

Now that we have these formulae for the derivatives of trigonometric functions, let us do some examples

of using these formulae. Letf(x) = tan(1 +x2). Then, wheref(x) is defined, we have that f

0(x) = sec2(1 +x2)¢2x= 2xsec2(1 +x2):

As another example, letg(x) = esecxtanx. Then, by the Chain Rule, we get that whereg(x) is defined its derivative is g

0(x) = esecxtanx¢((secxtanx)¢tanx+ secx¢sec2x) = (secxtan2x+ sec3x)esecxtanx:

Finally, leth(x) =1

secx. Whereh(x) is defined, it should be equal to cosx. Let us verify that, in the domain ofh, the derivative ofh(x) is¡sinx: h

0(x) =¡1

sec

2x¢secxtanx=¡tanx

secx=¡sinx cosx 1 cosx=¡sinx:

Technically, cosxandh(x) are not the same function. Can you think of a reason why? (Hint: What are the

domains of these functions?) 4quotesdbs_dbs17.pdfusesText_23