UNIT I-Module 6 Differentiation of trigonometric and Inverse trigonometric functions
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UNIT I - THE DERIVATIVES Module 6: Differentiation of Trigonometric and Inverse Trigonometric Functions Introduction In this topic, we will discuss differentiation of new class of functions called the transcendental functions. The trigonometric functions and their inverses, together with the logarithmic and exponential functions are the simplest transcendental functions. This topic covers differentiation of trigonometric functions and their inverses. Learning Objectives At the end of this topic, you should be able to demonstrate the following: a. Determine the derivative of trigonometric functions; and b. Find the derivative of inverse trigonometric functions. Presentation of Contents Differentiation of Trigonometric Functions Let u be a differentiable function of x. That is u = g(x). The derivative of the six trigonometric functions with respect to x are as follows: 1. š (sin š¢) šš„ = cos š¢ . šš„ šš¢ 2. š (cos š¢) šš„ = −sin š¢ . šš„ 3. š (tan š¢) šš„ = š šš 2 š¢ . šš„ š šš¢ 4. šš„ (cot š¢) = −šš š 2 š¢ . šš„ šš¢ 5. šš„ (sec š¢) = sec š¢ tan š¢ . šš„ š šš¢ šš¢ 6. šš„ (csc š¢) = −csc š¢ cot š¢. šš„ š šš¢ The following examples illustrate the use of the formulas in differentiating functions involving trigonometric expressions. EXAMPLE 1 a. Find šš¦ šš„ if y = csc(4š„ + 5). Solution šš¦ y = csc(4š„ + 5) takes the form of y = csc u with u = 4x + 5. Hence to find šš„, we use the formula šš¦ šš„ = š (csc š¢) šš„ šš¢ = −csc š¢ cot š¢. šš„ u = 4x + 5; šš¦ šš„ šš¦ šš„ b. šš¢ šš„ =4 = −csc(4š„ + 5) cot(4š„ + 5). (4) = −4csc(4š„ + 5) cot(4š„ + 5) Given f(x) = š šš2 4š„ 2 , determine f ‘(x). Solution f (x) = š šš2 4š„ 2 can be express as f (x) = (sin 4š„ 2 )2. This function takes the form of f (x) = š¢š with u = sin 4š„ 2 and n = 2. This suggests the use of the GPR, that is šš¢ f ‘(x) = nš¢š−1. šš„ u = sin 4š„ 2 and n = 2, gives f ‘(x) = 2 (sin 4š„ 2 )2−1. š šš„ (sin 4š„ 2 ); we further differentiate š šš 4š„ 2 using š (sin š¢) šš„ šš¢ with u = 4š„ 2 , we have = cos š¢ . šš„ š (sin 4š„ 2 ) šš„ š = cos 4š„ 2 . šš„ (4š„ 2 ) š (sin 4š„ 2 ) šš„ = (cos 4š„ 2 ) (8x) š (sin 4š„ 2 ) šš„ = 8xcos 4š„ 2 f ‘(x) = 2( sin 4š„ 2 ) (8xcos 4š„ 2 ) f ‘(x) = 16x sin 4š„ 2 cos 4š„ 2 šš¦ 1 Find šš„ if y = 2 x + 3tan 2š„ c. Solution 1 šš¦ y = 2 x + 3tan 2š„ is a sum of two functions, so to determine šš„, we use the sum rule. 1 d. Find y = 2 x + 3tan 2š„ differentiate both sides with respect to x šš¦ šš„ = šš„ (2 x) + šš„ (3 tan 2š„) using constant multiple rule šš¦ šš„ =2 šš¦ šš„ = 2 (1) + 3 šš„ (tan 2š„) šš¦ šš„ = 1 2 + 3 [š šš2 2š„ . šš¦ šš„ = 1 2 + 3[(š šš2 2š„) (2) ] šš¦ šš„ = 2 + 6š šš 2 2š„ šš¦ šš„ = šš¦ šš„ if sin(š„ + š¦) = x + y. š 1 1 š (x) šš„ 1 š š + 3 šš„ (tan 2š„) š š šš„ (2š„) ] š (š„) šš„ =1 differentiate š”šš 2š„ using š šš„ (š”šš š¢) = š šš 2 š¢ . šš¢ šš„ š but šš„ (2š„) = 2, so we have simplifying further 1 1+12 š šš2 2š„ 2 šš¦ The function is implicit, so we use implicit differentiation to obtain šš„. sin(š„ + š¦) = x + y differentiate both sides with respect to x gives with u = 2x š [sin(š„ šš„ cos(š„ + š (x šš„ + š¦)] = š š¦) šš„ (x + y) = š cos(š„ + š¦) [šš„ (š„) + cos(š„ + š¦) [ 1 + + y) šš¦ ] šš„ to differentiate š šš(š„ + š¦), use š š (š„) šš„ š (y)] šš„ š (š¦) šš„ + šš„ šš¦ šš¦ šš„ šš„ ;u=x+y šš¦ šš¦ = 1 + šš„ Using DPMA on the left side of the equation šš¦ šš¦ šš¢ = 1 + šš„ šš¦ cos(š„ + š¦) + cos(š„ + š¦) šš„ = 1 + šš„ cos(š„ + š¦) šš„ - šš„ (š šš š¢) = ššš š¢ . = 1 - cos(š„ + š¦) [cos(š„ + š¦) − 1] = 1 - cos(š„ + š¦) šš¦ šš„ = 1 − cos(š„+š¦) cos(š„+š¦)−1 šš¦ šš„ = −1[cos(š„+š¦)−1] cos(š„+š¦)−1 šš¦ šš„ = -1 šš¦ collect terms with šš„ on one side of the equation šš¦ factor out šš„ on one side of the equation by MPE simplify further by factoring out -1 in the numerator Application Activity 1 šš¦ Find šš„ and simplify your answer whenever possible. 1. y = sin š„ + cos š„ + tan š„ 2. y = š šš 4 4š„ 3. y = š„ 2 cot 2x 1−š ššš„ 4. y = 1+cos š„ 5. cos(š„š¦) = x – y Differentiation of Inverse Trigonometric Functions Let u be a differentiable function of x; u = g(x). The derivative of the six inverse trigonometric functions with respect to x are as follows: 1. 2. 3. 4. 5. š šš„ š šš„ š šš„ š šš„ š šš„ (Arcsin š¢) = (Arccos š¢) = (Arctan š¢) = (Arccot š¢) = (Arcsec š¢) = 1 šš¢ √1− š¢2 šš„ −1 šš¢ √1− š¢2 šš„ 1 šš¢ 1+ š¢2 šš„ −1 šš¢ 1+ š¢2 šš„ 1 šš¢ š¢√š¢2 −1 šš„ š 6. šš„ (Arccsc š¢) = −1 šš¢ š¢√š¢2 −1 šš„ The notation š šš−1 š„ used in trigonometry is considered inconvenient for this particular topic since it might be read as “sin x with an exponent of -1” because -1 is not an exponent but it is the symbol for 1 inverse of sin. Therefore š šš−1 š„ does not mean (š šš š„)−1 or š šš š„. EXAMPLE 2 šš¦ a. Find šš„ if y = arccot š„ 2 . y = arccot š„ 2 takes the form of y = arccot u with u = š„ 2 . Hence to find š (Arccot š¢) šš„ −1 we use the formula šš¢ = 1+ š¢2 šš„ y = arccot š„ 2 ; u = = š„ 2 −1 šš¦ , šš„ šš¦ šš„ = 1+ (š„ 2 )2 šš„ (š„ 2 ) šš¦ šš„ = 1+š„4 (2x) šš¦ šš„ = 1+š„4 by substitution š −1 −2š„ šš¦ b. Find šš„ if y = Arctan x + Arcsec √1 + š„ 2 šš¦ y = Arctan x + Arcsec √1 + š„ 2 is a sum of two functions, so to determine šš„, we use the sum rule. y = Arctan x + Arcsec √1 + š„ 2 šš¦ šš„ šš¦ šš„ š š = šš„ (Arctan x) + šš„ (Arcsec √1 + š„2 ) 1 š 1 = 1+ š„ 2 + šš„ šš¦ šš¦ = 1 1+ š„ 2 + 1 = 1+ š„ 2 + šš„ šš¦ šš„ šš¦ šš„ šš¦ šš„ šš¦ 1 = 1+ š„ 2 + √1+ š„2 √(√1+ š„2 ) −1 1 1 š„√1+ š„2 1 š„√1+ š„2 1 2 . . š„ 1 (1+ š„2 )2 š„ 1 (1+ š„2 )2 š„ š„√1+ š„2 √1+ š„2 1 1 š . [ (1 + š„2 )−2 . 2š„ ] 2 1 1 1 . [2 (1 + š„2 )−2 . šš„ (1 + š„2 ) ] 1 š„√1+ š„ 1 1 . šš„ (1 + š„2 )2 1 √1+ š„2 √š„2 . šš„ (√1 + š„2 ) š √1+ š„2 √(1−š„2 ) −1 1 = 1+ š„ 2 + š 2 1 1 = 1+ š„ 2 + šš„ šš„ 1 = 1+ š„ 2 . šš„ (š„) + šš¦ šš¦ differentiate both sides of the equation with respect to x = 1+ š„ 2 + 1+ š„2 = 1+ š„ 2 + 1+ š„2 = šš„ 2 1+ š„ 2 šš¦ c. Find šš„ if arcsin x + arcos y = y šš¦ arcsin x + arcos y = y is implicit, so to obtain šš„, we use implicit differentiation. arcsin x + arcos y = y differentiate both sides of the equation with respect to x š šš¦ šš„ (arcsin x) + 1 š šš„ š √1− š„2 1 √1− š„2 1 1 šš¦ 1 = šš„ + 1 šš¦ . šš„ (š¦) = šš„ šš¦ 1 šš¦ √1− š¦2 šš„ = šš„ (1 + 1 √1− š¦ 2 ) √1− š¦2 +1 šš¦ = šš„ ( 2 √1− š„ šš„ = šš„ šš¦ √1− š„2 š √1− š¦ 2 √1− š¦ 2 šš„ šš¦ √1− š„2 −1 . šš„ (š„) + - (arcos y) = √1− š¦ 2 ) 1 šš¦ šš„ = √1− š„2 √1− š¦2 +1 √1− š¦2 šš¦ 1 = √1− šš„ šš¦ = šš„ š„2 . √1− š¦2 √1− š¦ 2 +1 √1− š¦ 2 √1− š„ 2 (√1− š¦ 2 +1) Application Activity 2 šš¦ Find šš„ and simplify your answer whenever possible. 1. y = Arccot (tan 2x) 2. y = 3arcsin x – 2arctanx tan š„ 3. f(x) = arctan š„ 4. f(x) = sin š„ arccosx 5. arccot x – tan y = 1 + y
