d
play

d Inverse Trigonometric Functions i E 1 Lecture a l l u d b - PowerPoint PPT Presentation

Jan 14, 2024 •236 likes •449 views

Section 3.9 d Inverse Trigonometric Functions i E 1 Lecture a l l u d b Dr. Abdulla Eid A . College of Science r D MATHS 101: Calculus I Dr. Abdulla Eid (University of Bahrain) Inverse Trig 1 / 20 Inverse Trigonometric

  1. Section 3.9 d Inverse Trigonometric Functions i E 1 Lecture a l l u d b Dr. Abdulla Eid A . College of Science r D MATHS 101: Calculus I Dr. Abdulla Eid (University of Bahrain) Inverse Trig 1 / 20

  2. Inverse Trigonometric functions d i E 1 The derivative of the basic inverse trigonometric functions. a l 2 The derivative of functions that involve inverse trigonometric l u d functions. b A 3 Identities of inverse trigonometric functions using differential calculus. . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 2 / 20

  3. f ( x ) = sin x has an inverse if x ∈ [ − π 2 , π 2 ] . The inverse of the sine function is denoted by f − 1 ( x ) = sin − 1 x (= arcsin x ) with the following properties: d i 1 The graph of y = sin − 1 x is: E a l l 2 The domain of sin − 1 x is [ − 1, 1 ] u d 3 The range of sin − 1 x is [ − π b 2 ] 2 , π A � = x sin − 1 x . 4 sin � r D 5 sin − 1 ( sin x ) = x Exercise 1 Write a similar definition for the tan − 1 x and sec − 1 x and explore their properties. Dr. Abdulla Eid (University of Bahrain) Inverse Trig 3 / 20

  4. Derivative of sin − 1 x Example 2 sin − 1 x d � � Find . dx d Solution: i E sin y = x = x 1 = opp a y = sin − 1 x , find y ′ hypo l l u ( adj ) 2 + ( opp ) 2 = ( hypo ) 2 sin − 1 x sin y = sin � � d b ( adj ) 2 + x 2 = 1 sin y = x A cos y · y ′ = 1 ( adj ) 2 = 1 − x 2 . r D 1 y ′ = � adj = 1 − x 2 cos y = sec y sec y = hypo 1 adj = √ 1 − x 2 � = 1 y ′ = d sin − 1 x � √ dx 1 − x 2 Dr. Abdulla Eid (University of Bahrain) Inverse Trig 4 / 20

  5. Exercise 3 cos − 1 x d � � Derive a formula for dx d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 5 / 20

  6. Derivative of tan − 1 x Example 4 tan − 1 x d � � Find . dx d Solution: i E tan y = x = x 1 = opp a y = tan − 1 x , find y ′ l adj l u tan − 1 x ( adj ) 2 + ( opp ) 2 = ( hypo ) 2 � � tan y = tan d b 1 + x 2 = ( hypo ) 2 tan y = x A sec 2 y · y ′ = 1 ( hypo ) 2 = 1 + x 2 . r D 1 y ′ = sec 2 y = cos 2 y � hyp = 1 + x 2 cos y = adj 1 hypo = √ 1 + x 2 � = 1 y ′ = d tan − 1 x � 1 + x 2 dx Dr. Abdulla Eid (University of Bahrain) Inverse Trig 6 / 20

  7. Exercise 5 cot − 1 x d � � Derive a formula for dx d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 7 / 20

  8. Derivative of sec − 1 x Example 6 sec − 1 x d � � Find . dx d Solution: i E 1 = hypo sec y = x = x a y = sec − 1 x , find y ′ l adj l u sec − 1 x ( adj ) 2 + ( opp ) 2 = ( hypo ) 2 � � tan y = sec d b 1 + ( opp ) 2 = x 2 sec y = x A sec y tan y · y ′ = 1 ( opp ) 2 = x 2 − 1 . r D 1 y ′ = � x 2 − 1 sec y tan y = cos y cot y opp = cot y = adj 1 opp = √ x 2 − 1 � = 1 y ′ = d sec − 1 x � √ x 2 − 1 dx | x | Dr. Abdulla Eid (University of Bahrain) Inverse Trig 8 / 20

  9. Exercise 7 csc − 1 x d � � Derive a formula for dx d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 9 / 20

  10. Summary ( Derivative of the Inverse Trigonometic functions ) d i E a 1 − 1 ( sin − 1 x ) ′ = ( cos − 1 x ) ′ = √ l √ l u 1 − x 2 1 − x 2 d b A − 1 1 ( tan − 1 x ) ′ = ( cot − 1 x ) ′ = . 1 + x 2 1 + x 2 r D 1 − 1 ( sec − 1 x ) ′ = ( csc − 1 x ) ′ = √ √ x 2 − 1 x 2 − 1 | x | | x | Dr. Abdulla Eid (University of Bahrain) Inverse Trig 10 / 20

  11. 2 - Derivative of functions that involve the inverse trigonometric functions d i Example 8 E Find y ′ if y = tan − 1 ( √ x ) . a l l u Solution: We have y = tan − 1 ( √ x ) d b A � � 1 1 y ′ = 1 + ( √ x ) 2 · 2 √ x . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 11 / 20

  12. Exercise 9 √ d Find y ′ if y = tan − 1 x . i E √ a tan − 1 x Solution: We have y = l l u d 1 � 1 � y ′ = √ b · tan − 1 x 1 + x 2 A 2 . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 12 / 20

  13. Example 10 Find y ′ if y = cos − 1 ( e x ) . d i E Solution: We have y = cos − 1 ( e x ) a l l u − 1 d y ′ = 1 − ( e x ) 2 · ( e x ) b � A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 13 / 20

  14. Exercise 11 d Find y ′ if y = arcsin ( 3 − 2 x ) . i E a Solution: We have y = arcsin ( 3 − 2 x ) l l u d 1 y ′ = 1 − ( 3 − 2 x ) 2 · ( − 2 ) b � A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 14 / 20

  15. Example 12 Find y ′ if y = sin − 1 � √ � sin x . Solution: We have y = sin − 1 � √ d � sin x i E a � √ 1 � ′ l y ′ = l · sin x u � √ � d � 2 1 − sin x b A 1 1 = · √ · cos x . √ r � 2 sin x D 1 − ( sin x ) 2 1 1 = √ 1 − sin x · √ · cos x 2 sin x Dr. Abdulla Eid (University of Bahrain) Inverse Trig 15 / 20

  16. Exercise 13 √ Find y ′ if y = x sin − 1 x + 1 − x 2 and simplify your answer. d i E Solution: a l l 1 1 y ′ = ( 1 ) sin − 1 x + x · u √ √ 1 − x 2 + 1 − x 2 · ( − 2 x ) d 2 b x x y ′ = sin − 1 x + A √ 1 − x 2 − √ 1 − x 2 . r = sin − 1 x D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 16 / 20

  17. Identities of inverse trigonometric functions using differential calculus. Example 14 Find y ′ if y = sin − 1 + cos − 1 x . d i E Solution: a 1 − 1 y ′ = l √ 1 − x 2 + l √ u 1 − x 2 d y ′ = 0 b A The derivative is always zero, which means the function y is a constant. . r D y = sin − 1 + cos − 1 x = C → sin − 1 ( 0 ) + cos − 1 ( 0 ) = C → π = C 2 Hence the identity is sin − 1 + cos − 1 x = π 2 Dr. Abdulla Eid (University of Bahrain) Inverse Trig 17 / 20

  18. Exercise 15 Show using calculus that 1 cot − 1 x = π 2 − tan − 1 x 2 csc − 1 x = π 2 − sec − 1 x d i E a l l u d b A . r D Dr. Abdulla Eid (University of Bahrain) Inverse Trig 18 / 20

  19. Example 16 Find y ′ if y = tan − 1 x + tan − 1 � 1 � . x Solution: x 2 1 1 � 2 · − 1 d 1 1 + x 2 · − 1 y ′ = 1 + x 2 + x 2 → 1 + x 2 + � 1 i E x 2 1 + x a 1 1 y ′ = 1 + x 2 → y ′ = 0 l 1 + x 2 − l u d b The derivative is always zero, which means the function y is a constant. A � 1 � � 1 � . = C → π y = tan − 1 x + tan − 1 = C → tan − 1 ( 1 ) + tan − 1 r = C D 1 2 x Hence the identity is � 1 � tan − 1 x + tan − 1 = π x 2 Dr. Abdulla Eid (University of Bahrain) Inverse Trig 19 / 20

  20. Exercise 17 Show that sec − 1 ( − x ) + sec − 1 ( x ) = π . Exercise 18 d i (At home) Show that E � x + a a � l tan − 1 x + tan − 1 a = tan − 1 l u 1 − ax d b and use it to compute tan − 1 1 + tan − 1 2 + tan − 1 3. A . r D Exercise 19 (Challenging problem) What is wrong in � = d 1 1 sin − 1 x + sec − 1 x � √ 1 − x 2 + √ x 2 − 1 dx | x | Dr. Abdulla Eid (University of Bahrain) Inverse Trig 20 / 20

Recommend

More recommend