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Tangent, Cotangent, Secant, and Cosecant
The Quotient Rule
In our last lecture, among other things, we discussed the function (^) x^1 , 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) =
g(x)
where g(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)
where f (x) and g(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: let f (x) and g(x) be two functions with derivatives. Then we can define a function
h(x) =
f (x) g(x)
which has domain Dom(h) = {x ∈ R : g(x) 6 = 0}
and which is differentiable everywhere on its domain, with the formula
h′(x) =
g(x)f ′(x) − f (x)g′(x) (g(x))^2
We usually call f (x) the top function, and g(x) the bottom function, and so the formula for the derivative of the quotient h(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 function h(x) as
h(x) = f (x) g(x)
= f (x) ·
g(x)
Thus, we can think of the quotient h(x) as being a product, with the first function being f (x) and the second the composition of x−^1 after g(x). To find the domain of h(x), we first note that since f (x) presumably has full domain (all real numbers), the domain of h(x) will be precisely the same as that of (g(x))−^1 (since, wherever (g(x))−^1 is 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 of x−^1 is all real numbers except 0. This tells us that, presuming g(x) is defined everywhere, (g(x))−^1 is defined at all real numbers except those x for which g(x) = 0. Thus the domain of h(x), which is the domain of (g(x))−^1 , is the set of all real numbers x such that g(x) 6 = 0, which is precisely what the quotient rule tells us. Now, we can see that h(x) can be written as a product. This means that, wherever h(x) is defined, we can apply the product rule to find its derivative:
h′(x) =
d dx
f (x) ·
g(x)
= f ′(x) ·
g(x)
d dx
g(x)
To find the derivative of (g(x))−^1 , we apply the Chain Rule:
d dx
g(x)
= −(g(x))−^2 · g′(x) = −
g′(x) (g(x))^2
Combining the two fractions under a common denominator, we get the derivative formula for the quotient rule:
h′(x) = f ′(x) ·
g(x)
g(x)
f ′(x) g(x)
f (x)g′(x) (g(x))^2
=
g(x)f ′(x) (g(x))^2
f (x)g′(x) (g(x))^2
g(x)f ′(x) − f (x)g′(x) (g(x))^2
Let us do an example of the quotient rule. Let
h(x) = x + 3 x^2 − 4 x + 5
The domain of h(x) is all real numbers x such that the denominator, x^2 − 4 x + 5, is not 0. The denominator is 0 precisely when x = 1 or x = 5, so this gives us that
Dom(h) = {x ∈ R : x 6 = 1, x 6 = 5}.
For any x in the domain of h, the function h(x) has a derivative given by the quotient rule. That derivative is
h′(x) =
(x^2 − 4 x + 5) · 1 − (x + 3) · (2x − 4) (x^2 − 4 x + 5)^2
x^2 − 4 x + 5 − 2 x^2 − 2 x + 12 (x^2 − 4 x + 5)^2
−x^2 − 6 x + 17 (x^2 − 4 x + 5)^2
The function h(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 of x is defined to be its sine divided by its cosine:
tan x =
sin x cos x
The cotangent of x is defined to be the cosine of x divided by the sine of x:
cot x = cos x sin x
The secant of x is 1 divided by the cosine of x:
sec x =
cos x
and the cosecant of x is defined to be 1 divided by the sine of x:
csc x =
sin x
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 tan x is defined for all real numbers x such that cos x 6 = 0, since tangent is the quotient of sine over cosine. Thus tan x is undefined for
x =... , − 3 π 2
π 2
π 2
3 π 2
The range of csc x is the same as that of sec x, for the same reasons (except that now we are dealing with the multiplicative inverse of sine of x, not cosine of x). Therefore the range of csc x is
csc x ≥ 1 or csc x ≤ − 1.
The period of csc x is the same as that of sin x, which is 2π. Since sin x is an odd function, csc x is also an odd function. Finally, at all of the points where csc x is undefined, the function has both left and right vertical asymptotes, but just as in the case of sec x, the behavior of the vertical asymptotes depends on the point. If a =... , − 2 π, 0 , 2 π,.. ., then the left vertical asymptote at a goes to negative infinity, and the right vertical asymptote goes to positive infinity:
lim x→a−^
csc x = −∞ and lim x→a+^
csc x = +∞.
If, on the other hand, b = − 3 π, −π, π, 3 π,.. ., then the left vertical asymptote at b goes to positive infinity, and the right vertical asymptote goes to negative infinity:
lim x→b−^
csc x = +∞ and lim x→b+^
csc x = −∞.
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
(tan x) =
d dx
sin x cos x
(cos x) · (cos x) − (sin x) · (− sin x) cos^2 x
cos^2 x + sin^2 x cos^2 x
cos^2 x
= sec^2 x
d dx
(cot x) =
d dx
( (^) cos x sin x
(sin x) · (− sin x) − (cos x) · (cos x) sin^2 x
sin^2 x + cos^2 x sin^2 x
sin^2 x
= − csc^2 x
d dx
(sec x) =
d dx
cos x
cos^2 x
· (− sin x) =
sin x cos^2 x
cos x
sin x cos x
= sec x tan x
d dx
(csc x) =
d dx
sin x
sin^2 x
· (cos x) = −
cos x sin^2 x
sin x
cos x sin x
= − csc x cot x.
Now that we have these formulae for the derivatives of trigonometric functions, let us do some examples of using these formulae. Let f (x) = tan(1 + x^2 ). Then, where f (x) is defined, we have that
f ′(x) = sec^2 (1 + x^2 ) · 2 x = 2x sec^2 (1 + x^2 ).
As another example, let g(x) = esec^ x^ tan^ x. Then, by the Chain Rule, we get that where g(x) is defined its derivative is
g′(x) = esec^ x^ tan^ x^ · ((sec x tan x) · tan x + sec x · sec^2 x) = (sec x tan^2 x + sec^3 x)esec^ x^ tan^ x.
Finally, let h(x) = (^) sec^1 x. Where h(x) is defined, it should be equal to cos x. Let us verify that, in the domain of h, the derivative of h(x) is − sin x:
h′(x) = −
sec^2 x
· sec x tan x = −
tan x sec x
sin x cos x 1 cos x
= − sin x.
Technically, cos x and h(x) are not the same function. Can you think of a reason why? (Hint: What are the domains of these functions?)
