1. 5.6 Inverse Trigonometric Functions: Differentiation Develop properties of the six inverse trig functions Differentiate an inverse trig function Review basic differentiation rules for elementary functions .

2. None of the six basic trigonometric functions have an inverse. Surprised? Since they are periodic, they do not have an inverse unless you restrict the domain.

3. Notice each makes 1 st quadrant part of its graph. What is the purpose of inverse trig functions?

4. Ex 1 p372 Evaluating Inverse Trig Functions By definition, y=arcsin ½ means sin y = ½ In the interval [- π /2, π /2], the angle with that sine value would be π /6 By definition, y=arccos 0 means cos y = 0 In the interval [0, π ], the angle with that cosine value would be π /2 By definition, means in the interval [- π /2, π /2], the angle with that tangent value would be - π /3 In radian mode, type in sin -1 0.6 and it produces approximately 0.6435. If you multiply by 180 and divide by pi, you could get it in degrees, or just set to degree mode. Most often you will want radians.

5. Remember that when applying these properties they have restricted domains. For x-values outside these domains the properties do not hold. For example arcsin(sin π )) = 0, not π .

6. Ex 2 p.373 Solving an equation arctan (2x – 5)= π /4 tan(arctan (2x – 5))= tan π /4 2x – 5 = 1 x = 3

7. Sometimes you will have to evaluate expressions like cos(arcsin x) Ex 3 p. 373 Using right triangles y x = cos y x 1 Therefore sin y= 2 y

8. y x 1 Looking at derivatives for inverse trig functions, even though they are transcendental, their derivatives are algebraic.

9. Ex 4 p. 374 Differentiating Inverse Trig Functions

10. Ex 5 p. 374 A Derivative that can be simplified Differentiate You can see one of the benefits of inverse trig functions – they can be used to integrate common algebraic functions.

11. Ex 6 p. 375 Analyzing an Inverse Trig Graph Analyze the graph of y = (arctan x) 2 Solution: From the derivative, The only critical number occurs when arctan x = 0, which happens at x = 0. By the first derivative test, this turns out to be a minimum. From the second derivative, Finally looking at there are horizontal asymptotes at y = The points of inflection occur when 2xarctan x = 1, and from Newton’s method, this occurs at x ≈  0.765

12. Here are the rules you should have memorized by test on Jan 27

13. The number pi is mighty fine 3.14159 The number e is very great 2.71828 Ted Mueller 2011 5.6 p. 377/ 1-65 EOO, 71-72, 94