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Lecture 32
Derivatives of inverse
trigonometric functions
Today we’re going to learn some new derivatives! Let’s begin with some review. This is an exercise we’ve already seen in class:
Exercise 32.0.1. Using the fact that e ln( x )^ = x , find the derivative of ln( x ). (I know you know what the derivative of ln is; I want you to be able to prove your answer.) Hint: You’ll need to use the chain rule.
The reason that I want you to prove your answer: Now you can find new deriva- tives!
Exercise 32.0.2. Remember that sin 2 ◊ + cos 2 ◊ = 1. So, for example
Ô 1 ≠ cos 2 ◊ = sin ◊ and
Ò 1 ≠ sin 2 ◊ = cos ◊
whenever sin ◊ Ø 0 and cos ◊ Ø 0.
(a) Using the fact that sin(arcsin( x )) = x , find the derivative of arcsin( x ).
(b) Using the fact that cos(arccos( x )) = x , find the derivative of arccos( x ).
(c) Using the fact that tan(arctan( x )) = x , find the derivative of arctan( x ).
Here are solutions to the exercises.
2 LECTURE 32. DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
32.1 Derivative of arcsin
Note that arcsin( x ) is defined when x œ [≠ 1 , 1], with an output/range of [≠ fi/ 2 , fi/ 2]. An important thing to note is that cos ◊ Ø 0 whenever ◊ œ [≠ fi/ 2 , fi/ 2]. As a result:
sin(arcsin( x )) = x (32.1) cos(arcsin( x )) · (arcsin( x ))Õ^ = 1 (32.2)
(arcsin( x ))Õ^ =
cos(arcsin( x ))
Ò 1 ≠ sin 2 arcsin( x )
Ò 1 ≠ (sin arcsin( x ))^2
Ô
1 ≠ x^2
Note that even though arcsin( x ) is continuous along [≠ 1 , 1] , its derivative only exists along (≠ 1 , 1). (That is, arcsin( x ) is not di erentiable at -1 and at 1.) Can you explain why using the graph of arcsin? Also note that the derivative of arcsin has two vertical asymptotes, at x = 1 and x = ≠1. You can see this by taking the limit of Ô 11 ≠ x 2 at x = 1 (approaching from
the left) and the limit at x = ≠1 (approaching from the right):
lim x æ 1 ≠
Ô
1 ≠ x^2
= Œ and lim x æ≠ 1 +
Ô
1 ≠ x^2
4 LECTURE 32. DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
32.2 Derivative of arccos
Now let’s try arccos. This has a domain of [≠ 1 , 1] and a range of [0 , fi ]. On this interval, sin takes on values between 0 and 1.
cos(arccos( x )) = x (32.7) ≠ sin(arccos( x )) · (arccos( x ))Õ^ = 1 (32.8)
(arccos( x ))Õ^ =
≠ sin(arccos( x ))
Ò 1 ≠ cos 2 arccos( x )
Ò 1 ≠ (cos arccos( x ))^2
Ô
1 ≠ x^2
Ô
1 ≠ x^2
arccos( x ) fi/^2
fi
Ô^ ≠^1 1 ≠ x^2
Figure 32.2: The graph of arccos( x ) on the left. The graph of its derivative on the right.
32.3. DERIVATIVE OF ARCTAN 5
32.3 Derivative of arctan
Finally, let’s compute the derivative of arctan. This has a domain of (≠Œ , Œ) and a range (≠ fi/ 2 , fi/ 2). We will make use of the following:
sec 2 ( ◊ ) =
cos 2 ( ◊ )
sin 2 ( ◊ ) + cos 2 ( ◊ ) cos 2 ( ◊ )
sin 2 ( ◊ ) cos 2 ( ◊ )
cos 2 ( ◊ ) cos 2 ( ◊ )
= (tan ◊ )^2 + 1_._ (32.17)
Here’s the computation of the derivative:
tan(arctan( x )) = x (32.18) sec 2 (arctan( x )) · (arctan( x ))Õ^ = 1_._ (32.19)
(arctan( x ))Õ^ =
sec 2 (arctan( x ))
(tan(arctan( x )))^2 + 1
x^2 + 1
32.4. PREPARATION FOR NEXT TIME 7
32.4 Preparation for next time
(We did this in class, but make sure you could do this on a test!) Using the fact that sin(arcsin( x )) = x and using a trig identity, prove that
(arcsin( x ))Õ^ =
Ô
1 ≠ x^2
(We did this in class, but make sure you could do this on a test!) Using the fact that tan(arctan( x )) = x and using a trig identity, prove that
(arctan( x ))Õ^ =
x^2 + 1
Compute the following indefinite integrals:
(a) (^) ⁄ 1 9 x^2 + 1
dx
(b) (^) ⁄ 1 Ô 1 ≠ 4 x^2
dx
(c) (^) ⁄ 1 Ô 3 ≠ 4 x^2
dx
32.4.4 (Plus One)
Let f be a function, and x (^) 0 a real number. Find the equation of a line that goes through the point ( x (^) 0 , f ( x (^) 0 )), and is tangent to f at this point.
