definition of inverse function lets you turn this new problem into the old problem of differentiating the tangent.
EXAMPLE 1 Differentiate: y = tan–1 x
Section 4-5: Derivatives of Inverse Trigonometric Functions © 2005 Key Curriculum Press 149
DEFINITIONS: Inverse Trigonometric Functions (Principal Branches)
y = sin–1 x if and only if sin y = x and y y = cos–1 x if and only if cos y = x and y [0, ] y = tan–1 x if and only if tan y = x and y y = cot–1 x if and only if cot y = x and y (0, )
y = sec–1 x if and only if sec y = x and y [0, ], but x ≠ y = csc–1 x if and only if csc y = x and y , but x≠0
Solution
Use algebra to solve for y′ = cos2y
Figure 4-5e
To find in terms of x, consider that y is an angle whose tangent and cosine are being found. Draw a right triangle with angle y in standard position (Figure 4-5e). By trigonometry,
Because tan y = x, which equals x/1, put x on the opposite leg and 1 on the adjacent leg. The hypotenuse is thus . Cosine equals (adjacent)/(hypotenuse), so you can write
In Example 1, both the left and right sides of the equation tan y = x are functions of x. As you learned in Section 3-9, the technique of differentiating both sides of such an equation with respect to x is called implicit
differentiation. You will study this technique more extensively in Section 4-8.
Derivative of the Inverse Secant Function
The derivative of the inverse secant is tricky. Example 2 shows what happens.
EXAMPLE 2 Differentiate: y = sec–1 x
Solution Transform the new problem into an old problem. y = sec–1 x sec y = x
sec y tan y · = 1 Remember the chain rule! That’s where y′ comes from. Use algebra to solve for y′.
Figure 4-5f
Consider y to be an angle in standard position as shown in Figure 4-5f. Secant equals (hypotenuse)/(adjacent) and sec y = x = x/1, so write x on the hypotenuse and 1 on the adjacent leg. By the Pythagorean theorem, the third side is . The range of sec–1 is
y [0, ] (excluding y = /2), so y can terminate in Quadrant I or II. Where sec y is negative, y terminates in Quadrant II. In this case, the hypotenuse, x, is negative, so you must draw it
in the negative direction, opposite the terminal side of y, as you do with negative radii in polar coordinates. (You’ll learn about polar coordinates in
150 © 2005 Key Curriculum Press Chapter 4: Products, Quotients, and Parametric Functions
=
=
=
y = tan–1 x tan y = x
sec2 y · = 1 The derivative of tan is sec2. Because y depends
on x, it is an inside function. The y is the derivative of this inside function (from the chain rule). Use the definition of tan–1 to write the equation in terms of tangent.
Chapter 8.) As shown in Figure 4-5f, you pick the negative square root for the
The derivatives of the six inverse trigonometric functions are shown here. In Problem Set 4-5 you will derive the four properties not yet discussed.
EXAMPLE 3 Differentiate: y = cos–1 e3x
Solution y = cos–1 e3x
= – · 3e3x= – Use the chain rule on the inside function.
Problem Set 4-5
Q1. x = —?— Q2. x = —?— Q3. x = —?— Q4. x = —?— Q5. x = —?— Q6. x = —?—Refer to Figure 4-5g for Problems Q7–Q10.
Q7. (1) = —?—
Q8. (3) = —?—
Q9. (4) = —?—
Q10. (6) = —?—
Figure 4-5g
Section 4-5: Derivatives of Inverse Trigonometric Functions © 2005 Key Curriculum Press 151
PROPERTIES: Derivatives of the Six Inverse Trigonometric Functions (sin–1x) = (cos–1x) = –
(tan–1x) = (cot–1x) = – (sec–1x) = (csc–1x) = – Note: Your grapher must be in radian mode.
Memory Aid: The derivative of each “co-” inverse function is the opposite of the derivative of the corresponding inverse function because each co-inverse function is decreasing at x = 0 or 1 (Figure 4-5d).
Quick Review
vertical coordinate. So the denominator of the derivative, sec y tan y, is if x > 0 or – if x < 0
In both cases, this quantity is positive. To avoid two different representations, use the notation |x| and write the derivative in this way.
For Problems 1–4, duplicate the graphs in Figure
than from – /2 to /2.
6. Explain why the principal branch of the inverse secant function cannot be continuous.
7. Evaluate: sin (sin–1 0.3) 8. Evaluate: cos–1 (cos 0.8)
For Problems 9–12, derive the formula. 9. (sin–1 x) =
10. (cos–1 x) = – 11. (csc–1 x) = – 12. (cot–1 x) = –
For Problems 13-24, find the derivative algebraically. 13. y = sin–1 4x 14. y = cos–1 10x 15. y = cot–1 e0.5x 16. y = tan–1 (ln x) 17. y = sec–1 18. y = csc–1 19. y = cos–1 5x2 20. f(x) = tan–1 x3 21. g(x) = (sin–1 x)2 22. u = (sec–1 x)2 23. v = x sin–1 x + (1 – x2)1/2 (Surprise!)
24. I(x) = cot–1 (cot x) (Surprise!)
25. Radar Problem: An officer in a patrol car sitting 100 ft from the highway observes a truck approaching (Figure 4-5h).
Figure 4-5h
a. At a particular instant, t, in seconds, the truck is at a distance x, in feet, down the highway. The officer’s line of sight to the truck makes an angle , in radians, to a perpendicular to the highway. Explain why
= tan–1 (x/100).
b. Find d / dx. Use the chain rule to write an equation for d / dt.
c. When the truck is at x = 500 ft, the officer notes that angle is changing at a rate d / dt = – 0.04 rad/s. How fast is the truck going? How many miles per hour is this? 26. Exit Sign Problem: The base of a 20-ft-tall exit
sign is 30 ft above the driver’s eye level (Figure 4-5i). When cars are far away, the sign is hard to read because of the distance. When they are very close, the sign is hard to read because drivers have to look up at a steep angle. The sign is easiest to read when the distance x is such that the angle at the driver’s eye is as large as possible.
Figure 4-5i
152 © 2005 Key Curriculum Press Chapter 4: Products, Quotients, and Parametric Functions 4-5d on your grapher. For Problems 3 and 4, recall
that csc y = 1/ sin y and cot y = 1/ tan y. 1. y = cos–1 x
2. y = sin–1 x 3. y = csc–1 x 4. y = cot–1 x
5. Explain why the principal branch of the inverse cotangent function goes from 0 to rather
27. Numerical Answer Check Problem: For f(x) = cos–1 x, make a table of values that show
(x) both numerically and by the formula. Start at x = –0.8 and go to x = 0.8, with
x = 0.2. Show that the formula and the numerical derivative give the same answers for each value of x.
28. Graphical Analysis Problem: Figure 4-5j shows the graph of y = sec–1 x.
Figure 4-5j
a. Calculate the derivative at x = 2. Based on the graph, why is the answer reasonable?
b. What does y equal when x is 2? What does (d / dy)(sec y) equal for this value of y? c. In what way is the derivative of the inverse
secant function related to the derivative of the secant function?
29. General Derivative of the Inverse of a Function: In this problem you will derive a general formula for the derivative of the inverse of a function.
a. Let y = sin–1 x. Show that .
b. By directly substituting sin–1 x for y in
part a, you get . Show that
this equation and the derivative of sin–1 x
given in this section give the same value when x = 0.6.
c. Show that this property is true for the derivative of the inverse of a function:
d. Suppose that f(x) = x3 + x. Let h be the
inverse function of f. Find x if f(x) = 10. Use the result and the property given above to calculate (10).
30. Quick! Which of the inverse trigonometric derivatives are preceded by a negative (–) sign?