1. Derivatives of Inverse Functions
Trigonometric Functions
Functions in General

2. Introduction and Importance
The inverse trigonometric functions we
introduced at the start of the semester are
major players in Calculus. In this section we:
Develop the derivatives of the six trigonometric functions and begin an exploration of their many applications.
The method for differentiating the inverses of more general functions is also presented.

3. Inverse Sine and Its Derivative
Recall that y = sin-1 x is the value of y such that x = sin y, where –π/2 ≤ y ≤ π/2. The domain of sin-1 x is {x: -1 ≤ x ≤ 1}.

5. The derivative of y = sin-1 x follows differentiating both sides of x = sin y with respect to x, simplifying, and solving for dy/dx:
y = sin-1 x iff x = sin y
Differentiate with respect to x.
Chain Rule on the right side.
Solve for dy/dx.

6. The identity sin2 y + cos2 y = 1 is used to express this derivative in terms of x. Solving for cos y yields…
x = sin y implies x2 = sin2 y
Since y is restricted to the interval –π/2 ≤ y ≤ π/2, we have cos y ≥ 0. Therefore, we choose the positive branch of the square root, and it follows that…

7. This result is consistent with the graph of f(x) = sin-1 x.

9. Example 1: Compute the following derivatives.
We apply the Chain Rule for both derivatives.

10. This result is valid for -1 < x < 1, where sin(sin-1 x) = x.

11. Derivatives of Inverse Tangent
Recall that y = tan-1 x is the value of y such that x = tan y, where –π/2 < y < π/2. The domain of y = tan-1 x is {x: -∞ < x < ∞}.

13. To find dy/dx, we differentiate both sides of x = tan y with respect to x and simplify…
y = tan-1 x iff x = tan y
Differentiate with respect to x.
Chain Rule
Solve for dy/dx.

14. To express this derivative in terms of x, we combine the trigonometric identity sec2 y = 1 + tan2 y with x = tan y to obtain sec2 y = 1 + x2. Substituting this result into the expression for dy/dx, it follows that…

17. Evaluate f ‘(2√3), where f(x) = x tan-1 (x/2).
Example 2:
Evaluate f ‘(2√3), where f(x) = x tan-1 (x/2).
Find an equation of the line tangent to the graph of g(x) = sec-1 (2x) at the point (1, π/3).
Product Rule and Chain Rule

18. We evaluate f ‘ at x = 2√3 and note that tan-1 √3 = π/3…

19. b. The slope of the tangent line at (1, π/3) is g ‘(1)
b. The slope of the tangent line at (1, π/3) is g ‘(1). Using the Chain Rule, we have…
It follows that g ‘(1) = 1/√3. An equation of the tangent line is…

20. Derivatives of Inverse Functions in General
We found the derivatives of the inverse trigonometric functions using implicit differentiation. However, this approach does not always work. For example, suppose we know only f and its derivative f ‘ and want to evaluate the derivative of f -1. The key to finding the derivative of the inverse function lies in the symmetry of the graphs of f and f -1.

21. Example 3: Consider the general linear function y = f(x) = mx + b, where m ≠ 0 and b are constants. a. Write the inverse of f in the form y = f -1(x).
Solving y = mx + b for x, we find that mx = y – b, or…
Writing this function in the form y = f -1(x) (by reversing the roles of x and y), we have…
which describes a line with slope of 1/m.

22. b. Find the derivative of the inverse
The derivative of f -1 is…
Notice that f ‘(x) = m, so the derivative of f -1 is the reciprocal of f ‘.

23. c. Consider the specific case f(x) = 2x – 6
c. Consider the specific case f(x) = 2x – 6. Graph f and f ‘, and find the slope of each line.
f -1(x) = x/2 + 3

24. The reciprocal property obeyed by f ‘ and (f -1) ‘ in Example 4 holds for all function with inverses. The figure below shows the graphs of a typical one-to-one function and its inverse. It also shows a pair of symmetric points – (x0, y0) on the graph of f and (y0, x0) on the graph of f -1 – along with the tangent lines at these points.
Notice that as the lines tangent to the graph of f get steeper (as x increases), the corresponding lines tangent to the graph of f -1 get less steep.

25. To understand this theorem, suppose that (x0, y0) is a point on the graph of f, which means that (y0, x0) is the corresponding point on the graph of f -1. Then the slope of the line tangent to the graph of f -1 at the point (y0, x0) is the reciprocal of the slope of the line tangent to the graph of f at the point (x0, y0). Importantly, the theorem says that we can evaluate the derivative of the inverse function without finding the inverse function itself.
The result of Theorem 7.3 is also written in the form…

26. Example 4: The function f(x) = √x + x2 + 1 is one-to-one, for x ≥ 0, and has an inverse on the interval. Find the slope of the curve y = f -1(x) at the point (3, 1).
The point (1, 3) is on the graph of f; therefore, (3, 1) is on the graph of f -1. In this case, the slope of the curve y = f -1(x) at the point (3, 1) is the reciprocal of the slope of the curve y = f(x) at (1, 3).

27. Note that…
which means that…
Therefore…
Observe that it is NOT necessary to find a formula for f -1 to evaluate its derivative at a point.

29. Example 5: If f(x) = 2x + cos x, find (f -1)’(1).
Notice that f is one-to-one since f ‘(x) = 2 – sin x > 0.
So, f is increasing.
To use Theorem 7.3, we need to know f -1(1) and we can find it by inspection:
Therefore…

30. Example 6: Use the values of a one-to-one differentiable function f in the table to complete the indicated derivatives or state that the derivative cannot be determined. x -1 1 2 3
f(x) 5 6 7
f ‘(x)
1/2
3/2
2/3

31. a. (f -1) ‘(5)
In this case, y0 = f(x0) = 5. Using the table, we see that x0 = 1 and f ‘(1) = 3/2. Therefore…

32. b. (f -1) ‘(2)
In this case, y0 = f(x0) = 2, which implies that x0 = -1 and f ‘(-1) = 1/2. Therefore…

33. c. (f -1) ‘(1)
With y0 = f(x0) = 1, the table does not supply a value of x0. Therefore, neither f ‘(x0) nor f ‘(1) can be determined.