Chapter 1 Differentiation PDF

Chapter 1: Differentiation 1. The Derivative and the Equation of Tangent Line The derivative is so important in all parts of pure and applied mathematics that we must devote a great deal of effort to find formulas for the derivatives of various kinds of functions. Definition: The derivative of the...

Chapter 1: Differentiation 1. The Derivative and the Equation of Tangent Line The derivative is so important in all parts of pure and applied mathematics that we must devote a great deal of effort to find formulas for the derivatives of various kinds of functions. Definition: The derivative of the function 𝑓 at 𝑥 = 𝑎, denoted by 𝑓 ′ (𝑎) and read 𝑓 prime of 𝑎 is 𝑓(𝑥) − 𝑓(𝑎) 𝑓 ′ (𝑎) = lim , 𝑥→𝑎 𝑥−𝑎 provided the limit exists. In this case we say that 𝑓 is differentiable at 𝑎. Geometrically, this is the slope of the tangent line through the point (𝑎, 𝑓(𝑎)). A more computationally useful formula for the derivative is obtained by letting 𝑥 = 𝑎 + ℎ and looking at the limit as ℎ approaches 0 (equivalent to letting 𝑥 approach 𝑎). Definition: The derivative of the function 𝑓 at 𝑥 = 𝑎, denoted by 𝑓 ′ (𝑎) and read 𝑓 prime of 𝑎 is 𝑓(𝑎 + ℎ) − 𝑓(𝑎) 𝑓 ′ (𝑎) = lim , ℎ→0 ℎ provided the limit exists. In this case we say that 𝑓 is differentiable at 𝑎. The derivative can also be thought of as a function of 𝑥. Its domain consists of all values in the domain of 𝑓 for which 𝑓 is differentiable. The function 𝑓 ′ can be defined by adapting the second definition above. Definition: If 𝑦 = 𝑓(𝑥), then the derivative of the function 𝑓 with respect to 𝑥, is the function 𝑓 ′ whose value at 𝑥 is 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim , ℎ→0 ℎ for all values of 𝑥 where the limit exists. 1 Remark: Let 𝑦 = 𝑓(𝑥). To denote the derivative of a function 𝑓, we have the following notations; 𝑑𝑦 𝑑𝑓(𝑥) 𝑦′, 𝑓 ′ (𝑥), , , 𝐷𝑓(𝑥), 𝐷𝑦, 𝐷𝑥 𝑓(𝑥), 𝐷𝑥 𝑦. 𝑑𝑥 𝑑𝑥 To denote the derivative of a function 𝑓 at 𝑥0 , we have the following notations; 𝑑𝑦 𝑑𝑓(𝑥0 ) 𝑦 ′ |𝑥=𝑥0 , 𝑓 ′ (𝑥0 ), | , , 𝐷𝑓(𝑥0 ), 𝐷𝑦|𝑥=𝑥0 , 𝐷𝑥 𝑓(𝑥0 ),. 𝑑𝑥 𝑥=𝑥0 𝑑𝑥 2. Rules for Differentiation: Rule 1: Derivative of Constant : The derivative of a constant function is 0 (the zero function), that is 𝑑 𝑐 = 0 , where 𝑐 𝑖s a constant. 𝑑𝑥 Example: 1 𝑑 1 1. If 𝑓(𝑥) = , then 𝑓 ′ (𝑥) = = 0. 2 𝑑𝑥 2 𝑑 2. If 𝑓(𝑥) = −6, then 𝑓 ′ (𝑥) = (−6) = 0. 𝑑𝑥 𝑑 3. If 𝑓(𝑥) = √3, then 𝑓 ′ (𝑥) = √3 = 0. 𝑑𝑥 𝑑 4. If 𝑓(𝑥) = 𝜋, then 𝑓 ′ (𝑥) = 𝜋 = 0. 𝑑𝑥 𝑑 5. If 𝑓(𝑥) = 𝑒 5 , then 𝑓 ′ (𝑥) = 𝑒 5 = 0. 𝑑𝑥 𝑑 6. If 𝑓(𝑥) = ln7, then 𝑓 ′ (𝑥) = ln7 = 0. 𝑑𝑥 Rule 2: Derivative of Identity Function : The derivative of the identity function, 𝑓(𝑥) = 𝑥, is the constant function 1, that is 𝑑 𝑥 = 1. 𝑑𝑥 Rule 3: Power Rule for Positive Integers : Let 𝑛 be a positive integer. Then the power function 𝑥 𝑛 is differentiable on ℝ and we have 𝑑 𝑛 𝑥 = 𝑛𝑥 𝑛−1. 𝑑𝑥 2 Remark: If 𝑛 is any real number and 𝑓(𝑥) = 𝑥 𝑛 , then 𝑓 ′ (𝑥) = 𝑛𝑥 𝑛−1. Example: 𝑑𝑦 𝑑 1. If 𝑦 = 𝑥 7 , then = 𝑥 7 = 7𝑥 6. 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑 2. If 𝑦 = 𝑥 −10 , then = 𝑥 −10 = −10𝑥 −11. 𝑑𝑥 𝑑𝑥 1 𝑑𝑦 𝑑 1 𝑑 2 3. If 𝑦 = , then = = 𝑥 −2 = −2𝑥 −3 = − 3. 𝑥2 𝑑𝑥 𝑑𝑥 𝑥 2 𝑑𝑥 𝑥 𝑑𝑦 𝑑 𝜋 4. If 𝑦 = 𝑥 𝜋 , then = 𝑥 = 𝜋𝑥 𝜋−1. 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑 5 5. If 𝑦 = 𝑥 5/6 , then = 𝑥 5/6 = 𝑥 −1/6. 𝑑𝑥 𝑑𝑥 6 4 𝑑𝑦 𝑑 4 𝑑 3 3 1 6. If If 𝑦 = √𝑥 3 , then = √𝑥 3 = 𝑥 3/4 = 𝑥 −1/4 =. 𝑑𝑥 𝑑𝑥 𝑑𝑥 4 4 4√𝑥 1 𝑑𝑦 𝑑 1 𝑑 2 2 1 2 1 7. If 𝑦 = 3 , then = 3 = 𝑥 −2/3 = − 𝑥 −5/3 = − 3 =− 3. √𝑥 2 𝑑𝑥 𝑑𝑥 √𝑥 2 𝑑𝑥 3 3 √𝑥 5 3 𝑥 √𝑥 2 Remark: The equation of the tangent line to the curve 𝑦 = 𝑓(𝑥) at the point (𝑥0 , 𝑦0 ) has the form 𝑦 − 𝑦0 = 𝑚(𝑥 − 𝑥0 ), where 𝑚 is the slope of the tangent line at the point (𝑥0 , 𝑦0 ) and is given by 𝑑𝑦 𝑚= |. 𝑑𝑥 𝑥=𝑥0 Example: Show that the equation of tangent line to curve 𝑦 = 𝑥 4 at the point (1,1) is 4𝑥 − 𝑦 − 3 = 0. Solution: The equation of tangent line to curve 𝑦 = 𝑥 4 at the point (1,1) is 𝑑𝑦 𝑦 − 1 = 𝑚(𝑥 − 1), 𝑤ℎ𝑒𝑟𝑒 𝑚 = |. 𝑑𝑥 𝑥=1 𝑑𝑦 Since = 4𝑥 3 , then 𝑚 = 4 × 13 = 4. Then the equation of tangent line to curve 𝑦 = 𝑥 4 at 𝑑𝑥 the point (1,1) is 𝑦 − 1 = 4(𝑥 − 1) ⇒ 𝑦 − 1 = 4𝑥 − 4 ⇒ 4𝑥 − 𝑦 − 3 = 0. 3 Rule 4: Constant Multiple Rule for Differentiation : Let 𝑓 be a function and let 𝑐 be a constant. Suppose that 𝑓 is differentiable. Then the function 𝑐𝑓 is also differentiable. Moreover, we have 𝑑 𝑑 (𝑐𝑓)(𝑥) = 𝑐 𝑓(𝑥). 𝑑𝑥 𝑑𝑥 Example: 𝑑 𝑑 1. If 𝑦 = √3𝑥 4 , then 𝑦 ′ = √3𝑥 4 = √3 𝑑𝑥 𝑥 4 = 4√3𝑥 3. 𝑑𝑥 𝑥 10 𝑑 𝑥 10 1 𝑑 10 2. If 𝑦 = , then 𝑦 ′ = = 𝑥 10 = 𝑥 9 = 2𝑥 9. 5 𝑑𝑥 5 5 𝑑𝑥 5 4 𝑑 4 𝑑 1 𝑑 28 3. If 𝑦 = , then 𝑦 ′ = =4 = 4 𝑥 −7 = −28𝑥 −8 = − 8. 𝑥7 𝑑𝑥 𝑥7 𝑑𝑥 𝑥7 𝑑𝑥 𝑥 Rule 5: Sum Rule for Differentiation (Term by Term Differentiation) : Let 𝑓 and 𝑔 be two differentiable functions. Then the function 𝑓 + 𝑔 is also differentiable. Moreover, we have 𝑑 𝑑 𝑑 (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥), 𝑥 ∈ Dom(𝑓) ∩ Dom(𝑔). 𝑑𝑥 𝑑𝑥 𝑑𝑥 Remark: If 𝑓1 , 𝑓2 , … , 𝑓𝑛 are differentiable functions at 𝑥, then 𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 is also differentiable function and 𝑑 𝑑 𝑑 𝑑 (𝑓1 + 𝑓2 + ⋯ + 𝑓𝑛 )(𝑥) = 𝑓1 (𝑥) + 𝑓2 (𝑥) + ⋯ + 𝑓 (𝑥). 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑛 Example: In the following find 𝑦 ′ : 3 1 7 18 1. 𝑦 = 𝑥 20 + √2 2. 𝑦 = 2𝑥 3 + 3√𝑥 + +5 3. 𝑦 = 𝑥 2 + 6√𝑥 6 + 9 + 9. 5𝑥 4 2 √𝑥 Solution: 1. Since 𝑦 = 𝑥 20 + √2, then 𝑑 𝑑 𝑦 = 𝑥 20 + √2 ⟹ 𝑦′ = 𝑥 20 + √2 ⟹ 𝑦 ′ = 20𝑥 19 + 0 ⟹ 𝑦 ′ = 20𝑥 19. 𝑑𝑥 𝑑𝑥 3 2. The equation 𝑦 = 2𝑥 3 + 3√𝑥 + + 5 can be rewrite as 5𝑥 4 3 𝑦 = 2𝑥 3 + 3𝑥 1/2 + 𝑥 −4 + 5. 5 4 Term by term differentiation, we get 3 𝑑 𝑑 3 𝑑 𝑑 𝑦 = 2𝑥 3 + 3𝑥 1/2 + 𝑥 −4 + 5 ⟹ 𝑦′ = 2 𝑥3 + 3 𝑥 1/2 + 𝑥 −4 + 5. 5 𝑑𝑥 𝑑𝑥 5 𝑑𝑥 𝑑𝑥 Then 3 12 3 1 12 1 𝑦 ′ = 6𝑥 2 + 𝑥 −1/2 − 𝑥 −5 + 0 ⟹ 𝑦 ′ = 6𝑥 2 + −. 2 5 2 √𝑥 5 𝑥5 1 7 18 3. The equation 𝑦 = 𝑥 2 + 6√𝑥 6 + 9 + 9 can be rewrite as 2 √𝑥 1 𝑦 = 𝑥 2 + 6𝑥 6/7 + 18𝑥 −1/9 + 9. 2 Term by term differentiation, we obtain 1 1 𝑑 𝑑 𝑑 𝑑 𝑦 = 𝑥 2 + 6𝑥 6/7 + 18𝑥 −1/9 + 9 ⟹ 𝑦′ = 𝑥2 + 6 𝑥 6/7 + 18 𝑥 −1/9 + 9. 2 2 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 Then 36 18 36 1 1 𝑦′ = 𝑥 + 𝑥 −1/7 + 𝑥 −10/9 + 0 ⟹ 𝑦′ = 𝑥 + 7 −29. 7 −9 7 √𝑥 √𝑥 10 Rule 6: Difference Rule for Differentiation : Let 𝑓 and 𝑔 be two differentiable functions. Then the function 𝑓 − 𝑔 is also differentiable. Moreover, we have 𝑑 𝑑 𝑑 (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥), 𝑥 ∈ Dom(𝑓) ∩ Dom(𝑔). 𝑑𝑥 𝑑𝑥 𝑑𝑥 Remark: If 𝑓1 , 𝑓2 , … , 𝑓𝑛 are differentiable functions at 𝑥, then 𝑓1 − 𝑓2 − ⋯ − 𝑓𝑛 is also differentiable function and 𝑑 𝑑 𝑑 𝑑 (𝑓1 − 𝑓2 − ⋯ − 𝑓𝑛 )(𝑥) = 𝑓1 (𝑥) − 𝑓2 (𝑥) − ⋯ − 𝑓 (𝑥). 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑛 Example: 1 Show that the equation of tangent line to curve 𝑦 = 3𝑥 − at the point (1,2) is 𝑥 4𝑥 − 𝑦 − 2 = 0. Solution: 1 The equation of tangent line to curve 𝑦 = 3𝑥 − at the point (1,2) is 𝑥 5  𝑑𝑦 𝑦 − 2 = 𝑚(𝑥 − 1), 𝑤ℎ𝑒𝑟𝑒 𝑚 = |. 𝑑𝑥 𝑥=1 𝑑𝑦 1 1 Since = 3 + 𝑥 −2 = 3 + , then 𝑚 = 3 + = 3 + 1 = 4. 𝑑𝑥 𝑥2 12 1 Then equation of tangent line to curve 𝑦 = 3𝑥 − at the point (1,2) is 𝑥 𝑦 − 2 = 4(𝑥 − 1) ⇒ 𝑦 − 2 = 4𝑥 − 4 ⇒ 4𝑥 − 𝑦 − 2 = 0. Rule 7: Product Rule for Differentiation : Let 𝑓 and 𝑔 be two differentiable functions. Then the function 𝑓 ⋅ 𝑔 is also differentiable. Moreover, we have 𝑑 𝑑 𝑑 (𝑓 ⋅ 𝑔)(𝑥) = 𝑓(𝑥) ⋅ 𝑔(𝑥) + 𝑔(𝑥) ⋅ 𝑓(𝑥), 𝑥 ∈ Dom(𝑓) ∩ Dom(𝑔). 𝑑𝑥 𝑑𝑥 𝑑𝑥 Example: In the following find the derivative of 𝑦: 1. 𝑦 = (𝑥 4 + 3)(6 − 𝑥 5 ) 2. 𝑦 = (√𝑥 + 1)(𝑥 3 + 4). Solution: 1. Since 𝑦 = (𝑥 4 + 3)(6 − 𝑥 5 ), then 𝑑 𝑑 𝑑 𝑦′ = [(𝑥 4 + 3)(6 − 𝑥 5 )] ⟹ 𝑦 ′ = (𝑥 4 + 3) (6 − 𝑥 5 ) + (6 − 𝑥 5 ) (𝑥 4 + 3) 𝑑𝑥 𝑑𝑥 𝑑𝑥 ⟹ 𝑦 ′ = (𝑥 4 + 3)(−5𝑥 4 ) + (6 − 𝑥 5 )(4𝑥 ) 3 ⟹ 𝑦 ′ = −5𝑥 8 − 15𝑥 4 + 24𝑥 3 − 4𝑥 8 ⟹ 𝑦 ′ = −9𝑥 8 − 15𝑥 4 + 24𝑥 3. 2. Since 𝑦 = (√𝑥 + 1)(𝑥 3 + 4), then 𝑑 𝑑 𝑑 𝑦′ = [(√𝑥 + 1)(𝑥 3 + 4)] ⟹ 𝑦 ′ = (√𝑥 + 1) (𝑥 3 + 4) + (𝑥 3 + 4) (√𝑥 + 1) 𝑑𝑥 𝑑𝑥 𝑑𝑥 1 ⟹ 𝑦 ′ = (√𝑥 + 1)(3𝑥 2 ) + (𝑥 3 + 4) 2√ 𝑥 3 𝑥 4 ⟹ 𝑦 ′ = 3√𝑥 × 𝑥 2 + 3𝑥 2 + + 2 √𝑥 2 √𝑥 1 2 ⟹ 𝑦 ′ = 3𝑥 5/2 + 3𝑥 2 + 𝑥 5/2 + 2 √𝑥 7 2 ⟹ 𝑦 ′ = 𝑥 5/2 + 3𝑥 2 +. 2 √𝑥 6 Rule 8: Quotient Rule for Differentiation : Let 𝑓 and 𝑔 be two differentiable functions. Then 𝑓 the function is also differentiable. Moreover, we have 𝑔 𝑑 𝑑 𝑑 𝑓 𝑔(𝑥) ⋅ 𝑓(𝑥) − 𝑓(𝑥) ⋅ 𝑔(𝑥) ( ) (𝑥) = 𝑑𝑥 𝑑𝑥 , 𝑥 ∈ Dom(𝑓) ∩ Dom(𝑔) − {𝑥: 𝑔(𝑥) = 0}. 𝑑𝑥 𝑔 (𝑔(𝑥))2 Example: In the following find the derivative of 𝑦: 3𝑥 2 √𝑥 1. 𝑦 = 2. 𝑦 =. 𝑥 5 +4 √𝑥+1 Solution: 3𝑥 2 1. Since 𝑦 = , then 𝑥 5 +4 𝑑 𝑑 5 𝑑 3𝑥 2 (𝑥 5 + 4) (3𝑥 2 ) − 3𝑥 2 (𝑥 + 4) 𝑦′ = [ 5 ] ⟹ 𝑦′ = 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥 + 4 (𝑥 5 + 4)2 (𝑥 5 + 4)(6𝑥) − 3𝑥 2 (5𝑥 4 ) ⟹ 𝑦′ = (𝑥 5 + 4)2 6𝑥 6 + 24𝑥 − 15𝑥 6 ⟹ 𝑦′ = (𝑥 5 + 4)2 −9𝑥 6 + 24𝑥 ⟹ 𝑦′ =. (𝑥 5 + 4)2 √𝑥 2. Since 𝑦 = , then √𝑥+1 𝑑 𝑑 (√𝑥 + 1) (√𝑥) − √𝑥 ( 𝑥 + 1) 𝑦′ = 𝑑 [ √𝑥 ] ⟹ 𝑦′ = 𝑑𝑥 𝑑𝑥 √ 𝑑𝑥 √𝑥 + 1 (√𝑥 + 1)2 1 1 (√𝑥 + 1) × − √𝑥 × 2 √𝑥 2 √𝑥 ⟹ 𝑦′ = 2 (√𝑥 + 1) 1 1 + −2 2 2 √𝑥 ⟹ 𝑦′ = (√𝑥 + 1)2 1 3 − 2 √𝑥 2 1 𝑥 −1/2 − 3 ⟹ 𝑦′ = ⟹𝑦 = ′. (√𝑥 + 1)2 2 (√𝑥 + 1)2 7  Exercises (1) If y = x + 3x + x − 9 then y  = 4 3 2 Ⓐ 4x 3 + 9x 2 + 2x Ⓑ 4 x 3 + x 2 + 2x Ⓒ x 3 + 9x 2 + 2x Ⓓ 4x 3 + 9x 2 + x (2) If y = 2x − 5x + 4x + 7 then y  = 3 2 Ⓐ 3x 2 − 10x + 4 Ⓑ 6 x 2 − 5x + 4 Ⓒ 6x 2 − 10x + 4 Ⓓ 6x 2 + 10x + 4 3 (3) If y = 5 x + − 18 then y  = x 5 3 5 3 5 3 5 3 Ⓐ + 2 Ⓑ − 2 Ⓒ − 2 − 18 Ⓓ − 2 2 x x x x 2 x x 2 x x (4) If y = x + x + 12 then y  = 2 1 1 1 1 Ⓐ 2x + + 12 Ⓑ 2x + + 12 Ⓒ 2x + Ⓓ 2x + x 2 x 2 x x 1 1 (5) If y = x 2 + + + 1 then y  = x x3 1 3 1 3 1 3 1 3 Ⓐ 2x + − Ⓑ 2x − − Ⓒ 2x − + Ⓓ 2x + + x2 x4 x2 x4 x2 x4 x2 x4 1− x 2 (6) If f (x ) = then f (x ) = 1+ x 2 x 4x 4x x Ⓐ− Ⓑ Ⓒ− Ⓓ ( ) 2 2 ( ) 2 2 ( ) 2 2 ( ) 2 1+ x 1+ x 1+ x 1+ x 2 8  1 − 2s du (7) If u = then = 1 + 2s ds 4 2 4 2 Ⓐ Ⓑ Ⓒ− Ⓓ− (1 + 2s ) (1 + 2s ) (1 + 2s ) (1 + 2s ) 2 2 2 2 2 + 3t du (8) If u = then = 2 − 3t dt 12 3 3 12 Ⓐ− Ⓑ Ⓒ− Ⓓ ( 2 − 3t ) ( 2 − 3t ) ( 2 − 3t ) ( 2 − 3t ) 2 2 2 2 1 + 2t du (9) If u = then = 1 − 2t dt 4 4 2 2 Ⓐ− Ⓑ Ⓒ− Ⓓ (1 − 2t ) (1 − 2t ) (1 − 2t ) (1 − 2t ) 2 2 2 2 (10) If y = (x + 1)(x + 4) then y  = 2 3 Ⓐ 5x 4 + 3x 2 + 8x Ⓑ x 4 + 3x 2 + 8x Ⓒ 5x 4 + x 2 + 8x Ⓓ 5x 4 + 3x 2 + x (11) If y = (1 + 2x )(1 − 3x ) then y  = Ⓐ −1 − 12x Ⓑ1 + 12x Ⓒ −1 + 12x Ⓓ1 − 12x (12) If f (t) = (1 − 2t )(1 + 3t ) then f (t ) = Ⓐ −1 − 12t Ⓑ 1 + 12t Ⓒ −1 + 12t Ⓓ 1 − 12t dy (13) If y = (1 + 3s )(1 − 4s ) then = ds Ⓐ 1 + 24s Ⓑ −1 − 24s Ⓒ 1 − 24s Ⓓ −1 + 24s 9 (14) The equation of the tangent line to the curve y = x 3 + 1 at the point ( −1,0) is Ⓐ 3 y = −x + 1 Ⓑ y = 3x + 3 Ⓒ 3 y = −x − 1 Ⓓ y = 3x − 3 (15) The equation of the tangent line to the curve y = x 3 − 1 at the point (1,0) is Ⓐ y = 3x + 3 Ⓑ 3 y = −x + 1 Ⓒ y = 3x − 3 Ⓓ 3 y = −x − 1 (16) The equation of the tangent line to the curve y = x 3 + 2 at the point (−1,1) is Ⓐ y = 3x + 4 Ⓑ 3 y = −x + 2 Ⓒ 3 y = −x − 2 Ⓓ y = 3x − 4 (17) The equation of the tangent line to the curve y = x 3 − 2 at the point (1, − 1) is Ⓐ y = 3x + 4 Ⓑ 3 y = −x + 2 Ⓒ 3 y = −x − 2 Ⓓ y = 3x − 4 10 3. The Chain Rule This section discuss one of the most powerful differentiation rules the Chain Rule. In calculus, the chain rule is a formula for computing the derivative of the composition of two or more functions. That is, if 𝑓 and 𝑔 are two functions, then the chain rule expresses the derivative of the composite function 𝑓 ∘ 𝑔 in terms of the derivatives of 𝑓 and 𝑔. The Chain Rule: Let 𝑦 = 𝑓(𝑢) and 𝑢 = 𝑔(𝑥) be functions such that 𝑓 is compatible for composition with 𝑔. Suppose 𝑔 is differentiable at 𝑥 and 𝑓 is differentiable at 𝑢𝑢 = 𝑔(𝑥), then the composite function 𝑓 ∘ 𝑔 is differentiable at 𝑥, and (𝑓 ∘ 𝑔)′ (𝑥) = 𝑓 ′ (𝑔(𝑥)) ⋅ 𝑔′ (𝑥). In Leibniz notation, if 𝑦 = 𝑓(𝑢) and 𝑢 = 𝑔(𝑥) are both differentiable functions, then 𝑑𝑦 𝑑𝑦 𝑑𝑢 = ⋅ , 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑦 where is evaluated at 𝑢 = 𝑔(𝑥). 𝑑𝑢 Remark: The reason for the name Chain Rule becomes clear when we make a longer chain by adding another link. If 𝑦 = 𝑓(𝑢), 𝑢 = 𝑔(𝑥) and 𝑥 = ℎ(𝑡) are both differentiable functions, then 𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑥 = ⋅ ⋅. 𝑑𝑡 𝑑𝑢 𝑑𝑥 𝑑𝑡 The Power Rule Combined with the Chain Rule: If 𝑛 is any real number and 𝑓 is differentiable function of 𝑥, then 𝑑 𝑑 [𝑓(𝑥)]𝑛 = 𝑛[𝑓(𝑥)]𝑛−1 ⋅ 𝑓(𝑥). 𝑑𝑥 𝑑𝑥 1 In particular, if 𝑛 = , then 2 𝑑 1 𝑑 √𝑓(𝑥) = ⋅ 𝑓(𝑥). 𝑑𝑥 2√𝑓(𝑥) 𝑑𝑥 Example: Find the derivative of the following functions 1 1. 𝑓(𝑥) = (𝑥 4 + 6)30 2. 𝑔(𝑡) = √𝑡 2 + 4 3. ℎ(𝑥) = 4. √𝑥 3 +4𝑥 2 +5𝑥+1 11 Solution: 𝑑 1. 𝑓(𝑥) = (𝑥 4 + 6)30 ⟹ 𝑓 ′ (𝑥) = 30(𝑥 4 + 6)29 (𝑥 4 + 6) ⟹ 𝑓 ′ (𝑥) = 120𝑥 3 (𝑥 4 + 6)29. 𝑑𝑥 1 𝑑 𝑡 2. 𝑔(𝑡) = √𝑡 2 + 4 ⟹ 𝑔′ (𝑡) = (𝑡 2 + 4) ⟹ 𝑔′ (𝑡) =. 2√𝑡 2 +4 𝑑𝑡 √𝑡 2 +4 3. The function ℎ rewrite as ℎ(𝑥) = (𝑥 3 + 4𝑥 2 + 5𝑥 + 1)−1/4. Then 1 𝑑 3 ℎ′ (𝑥) = − (𝑥 3 + 4𝑥 2 + 5𝑥 + 1)−5/4 (𝑥 + 4𝑥 2 + 5𝑥 + 1) 4 𝑑𝑥 1 = − (3𝑥 2 + 8𝑥 + 5)(𝑥 3 + 4𝑥 2 + 5𝑥 + 1)−5/4. 4 Example: 𝑑𝑦 If 𝑦 = (𝑥 3 + 4)4 (𝑥 4 + 3)3 , find. 𝑑𝑥 Solution: 𝑑𝑦 𝑑 4 𝑑 3 = (𝑥 3 + 4)4 (𝑥 + 3)3 + (𝑥 4 + 3)3 (𝑥 + 4)4 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 = (𝑥 3 + 4)4 (3(𝑥 4 + 3)2 ) (𝑥 4 + 3) + (𝑥 4 + 3)3 (4(𝑥 3 + 4)3 ) (𝑥 3 + 4) 𝑑𝑥 𝑑𝑥 3 3 4 4 2 2 4 3 3 3 = 12𝑥 (𝑥 + 4) (𝑥 + 3) + 12𝑥 (𝑥 + 3) (𝑥 + 4) = 12𝑥 2 (𝑥 3 + 4)3 (𝑥 4 + 3)2 [𝑥(𝑥 3 + 4) + (𝑥 4 + 3)] = 12𝑥 2 (𝑥 3 + 4)3 (𝑥 4 + 3)2 (2𝑥 4 + 4𝑥 + 3). Example: 1+𝑠 8 If 𝑓(𝑠) = ( ) , find 𝐷𝑠 𝑓(𝑠). 1+𝑠 2 Solution: 1+𝑠 7 𝑑 1+𝑠 𝐷𝑠 𝑓(𝑠) = 8( ) ( ) 1 + 𝑠 2 𝑑𝑠 1 + 𝑠 2 2 𝑑 𝑑 2 1 + 𝑠 7 (1 + 𝑠 ) 𝑑𝑠 (1 + 𝑠) − (1 + 𝑠) 𝑑𝑠 (1 + 𝑠 ) = 8( ) ( ) 1 + 𝑠2 (1 + 𝑠 2 )2 1 + 𝑠 7 (1 + 𝑠 2 ) × 1 − (1 + 𝑠)(2𝑠) = 8( ) ( ) 1 + 𝑠2 (1 + 𝑠 2 )2 1 + 𝑠 7 1 − 2𝑠 − 𝑠 2 8(1 + 𝑠)7 (1 − 2𝑠 − 𝑠 2 ) = 8( ) ( )=. 1 + 𝑠2 (1 + 𝑠 2 )2 (1 + 𝑠 2 )9 12  Exercises 1 (1) If f (x ) = then f (x ) = x +1 2 2x 2x x x Ⓐ− Ⓑ Ⓒ Ⓓ− ( ) ( ) (x ) (x ) 2 2 2 2 x 2 +1 x 2 +1 2 +1 2 +1 1 dy (2) If y = − then = 4 −t 4 dt 4t 3 4t 3 4t 3 4t 3 Ⓐ− Ⓑ Ⓒ Ⓓ− ( 4 3 ) ( ) (4 − t ) (4 − t ) 3 4 2 4 2 4 −t 4 −t 4 (3) If f (x ) = x 2 + 1 , then f (x ) = x 2x x 1 Ⓐ Ⓑ Ⓒ Ⓓ 2 x 2 +1 x 2 +1 x 2 +1 2 x 2 +1 (4) If f (x ) = (7x − 3) , then f (x ) = 10 Ⓐ10(7 x − 3) Ⓑ 70(7 x − 3) Ⓒ 7(7 x − 3) Ⓓ (7x − 3) 9 9 9 9 (5) If y = (2x + 3) , then y  = 6 Ⓐ 6(2x + 3) Ⓑ 2(2x + 3) Ⓒ (2x + 3) Ⓓ12(2x + 3) 5 5 5 5 (6) If y = (4 − x ) , then y  = 2 10 Ⓐ −20(4 − x ) Ⓑ 20x (4 − x ) Ⓒ −20x (4 − x ) Ⓓ −10x (4 − x ) 2 9 2 9 2 9 2 9 13  ( ) 10 (7) If f (x ) = x 2 − 5 , then f (x ) = ( ) ( ) ( ) ( ) 9 9 9 9 Ⓐ10x x 2 − 5 Ⓑ 20 x 2 − 5 Ⓒ 20x x 2 − 5 Ⓓ10 x 2 − 5 (8) If f (x ) = 1 − 3x 2 , then f (x ) = −6x −3x −3x 3x Ⓐ Ⓑ Ⓒ Ⓓ 1 − 3x 2 1 − 3x 2 2 1 − 3x 2 1 − 3x 2 1 (9) If f (x ) = , then f (x ) = x +4 2 x x 2x 2x Ⓐ 3 Ⓑ− 3 Ⓒ 3 Ⓓ− 3 (x 2 +4 ) 2 (x 2 +4 ) 2 (x 2 +4 ) 2 (x 2 +4 ) 2 10  x +1 (10) If y =   , then y  =  x −1 10( x + 1)9 10( x + 1)9 20(x + 1)9 20( x + 1)9 Ⓐ Ⓑ− Ⓒ Ⓓ− (x − 1) (x − 1) (x − 1) (x − 1) 11 11 11 11 14 4. Differentiation of Trigonometric Functions We write the six trigonometric functions as: sin𝑥 𝑦 = 𝑓(𝑥) = sin𝑥 𝑦 = 𝑓(𝑥) = cos𝑥 𝑦 = 𝑓(𝑥) = tan𝑥 = cos𝑥 cos𝑥 1 1 𝑦 = 𝑓(𝑥) = cot𝑥 = 𝑦 = 𝑓(𝑥) = sec𝑥 = 𝑦 = 𝑓(𝑥) = csc𝑥 =. sin𝑥 cos𝑥 sin𝑥 Trigonometric formulas: Pythagorean identities: 1. sin2 𝜃 + cos 2 𝜃 = 1. 2. 1 + tan2 𝜃 = sec 2 𝜃. 3. 1 + cot 2 𝜃 = csc 2 𝜃. Sum and difference formulas: 1. sin(𝛼 ± 𝛽) = sin𝛼 cos𝛽 ± sin𝛽 cos𝛼. 2. cos(𝛼 ± 𝛽) = cos𝛼 cos𝛽 ∓ sin𝛼 sin𝛽. tan𝛼±tan𝛽 3. tan(𝛼 ± 𝛽) =. 1∓tan𝛼 tan𝛽 Double angle formulas: 1. sin2𝜃 = 2sin𝜃 cos𝜃. 2. cos2𝜃 = cos 2 𝜃 − sin2 𝜃. 3. cos2𝜃 = 1 − 2sin2 𝜃. 4. cos2𝜃 = 2cos 2 𝜃 − 1. 2tan𝜃 5. tan2𝜃 =. 1−tan2 𝜃 15 In the following we state the derivative of six basic trigonometric functions. Rule 1: The derivative of sine function : The sine function is differentiable on ℝ and its derivative is the cosine function, that is, 𝑑 sin𝑥 = cos𝑥, 𝑥 ∈ ℝ. 𝑑𝑥 Rule 2: The derivative of cosine function : The cosine function is differentiable on ℝ and its derivative is the negative of sine function, that is, 𝑑 cos𝑥 = −sin𝑥, 𝑥 ∈ ℝ. 𝑑𝑥 Rule 3: The derivative of tangent function : The tangent function is differentiable on its domain and its derivative is the square of the secant function, that is, 𝑑 𝜋 3𝜋 5𝜋 tan𝑥 = sec 2 𝑥, 𝑥 ∈ ℝ − {± , ± , ± , … }. 𝑑𝑥 2 2 2 Rule 4: The derivative of cotangent function : The cotangent function is differentiable on its domain and its derivative is the negative square of the cosecant function, that is, 𝑑 cot𝑥 = −csc 2 𝑥, 𝑥 ∈ ℝ − {0, ±𝜋, ±2𝜋, ±3𝜋, … }. 𝑑𝑥 Rule 5: The derivative of secant function : The secant function is differentiable on its domain and its derivative is 𝑑 𝜋 3𝜋 5𝜋 sec𝑥 = sec𝑥 tan𝑥, 𝑥 ∈ ℝ − {± , ± , ± , … }. 𝑑𝑥 2 2 2 Rule 6: The derivative of cosecant function : The cosecant function is differentiable on its domain and its derivative is 𝑑 csc𝑥 = −csc𝑥 cot𝑥, 𝑥 ∈ ℝ − {0, ±𝜋, ±2𝜋, ±3𝜋, … }. 𝑑𝑥 Example: Find 𝑦 ′ , if 1. 𝑦 = 5cos𝑥 + 3sin𝑥 2. 𝑦 = sin2 𝑥 cos2 𝑥. 16 Solution: 𝑑 𝑑 1. 𝑦 = 5cos𝑥 + 3sin𝑥 ⇒ 𝑦 ′ = 5 cos𝑥 + 3 sin𝑥 ⇒ 𝑦 ′ = −5sin𝑥 + 3cos𝑥. 𝑑𝑥 𝑑𝑥 2. Since 𝑦 = sin2 𝑥 cos2 𝑥 ⇒ 𝑦 = (sin𝑥)2 (cos𝑥)2 , then 𝑑 𝑑 𝑦′ = (sin𝑥)2 (cos𝑥)2 + (cos𝑥)2 (sin𝑥)2 𝑑𝑥 𝑑𝑥 (sin𝑥)2 (2cos𝑥(−sin𝑥)) (cos𝑥) 2 (2sin𝑥 = + cos𝑥) 2 2 = 2sin𝑥 cos𝑥[(cos𝑥) − (sin𝑥) ] = sin2𝑥 cos2𝑥. Example: sin𝑥+cos𝑥 2 If 𝑓(𝑥) = , show that 𝑓 ′ (𝑥) = −. sin𝑥−cos𝑥 (sin𝑥−cos𝑥)2 Solution: 𝑑 sin𝑥 + cos𝑥 𝑓 ′ (𝑥) = [ ] 𝑑𝑥 sin𝑥 − cos𝑥 𝑑 𝑑 (sin𝑥 − cos𝑥) (sin𝑥 + cos𝑥) − (sin𝑥 + cos𝑥) (sin𝑥 − cos𝑥) = 𝑑𝑥 𝑑𝑥 (sin𝑥 − cos𝑥)2 (sin𝑥 − cos𝑥)(cos𝑥 − sin𝑥) − (sin𝑥 + cos𝑥)(cos𝑥 + sin𝑥) = (sin𝑥 − cos𝑥)2 (sin𝑥 − cos𝑥)2 + (sin𝑥 + cos𝑥)2 =− (sin𝑥 − cos𝑥)2 sin2 𝑥 − 2sin𝑥 cos𝑥 + cos 2 𝑥 + sin2 𝑥 + 2sin𝑥 cos𝑥 + cos 2 𝑥 =− (sin𝑥 − cos𝑥)2 2(sin2 𝑥 + cos 2 𝑥) 2 =− = −. (sin𝑥 − cos𝑥)2 (sin𝑥 − cos𝑥)2 Example: tan𝑥 If 𝑓(𝑥) = ,find 𝑓 ′ (𝑥). 1+cot𝑥 17 Solution: 𝑑 𝑑 𝑑 tan𝑥 (1 + cot𝑥) tan𝑥 − tan𝑥 (1 + cot𝑥) 𝑓 ′ (𝑥) = [ ] ⇒ 𝑓 ′ (𝑥) = 𝑑𝑥 𝑑𝑥 𝑑𝑥 1 + cot𝑥 (1 + cot𝑥)2 ′ (1 + cot𝑥)sec 2 𝑥 − tan𝑥(−csc 2 𝑥) ⇒ 𝑓 (𝑥) = (1 + cot𝑥)2 ′ sec 2 𝑥 + sec𝑥 csc𝑥 + sec𝑥 csc𝑥 ⇒ 𝑓 (𝑥) = (1 + cot𝑥)2 sec𝑥(sec𝑥 + 2csc𝑥) ⇒ 𝑓 ′ (𝑥) =. (1 + cot𝑥)2 Example: Find 𝐷𝑥 𝑦, if 1. 𝑦 = √7 + sec𝑥 2. 𝑦 = sec 2 𝑥 csc 2 𝑥. Solution: 𝑑 1 𝑑 sec𝑥 tan𝑥 1. 𝐷𝑥 𝑦 = [√7 + sec𝑥] ⇒ 𝐷𝑥 𝑦 = 2 (7 + sec𝑥) ⇒ 𝐷𝑥 𝑦 =. 𝑑𝑥 √7+sec𝑥 𝑑𝑥 2√7+sec𝑥 2. Since 𝑦 = sec 2 𝑥 csc 2 𝑥 = (sec𝑥)2 (csc𝑥)2 , then 𝑑 𝑑 𝐷𝑥 𝑦 = (sec𝑥)2 (csc𝑥)2 + (csc𝑥)2 (sec𝑥)2 𝑑𝑥 𝑑𝑥 = (sec𝑥)2 (2csc𝑥(−csc𝑥 cot𝑥)) + (csc𝑥)2 (2sec𝑥(sec𝑥 tan𝑥)) = −2sec𝑥 csc 3 𝑥 + 2csc𝑥 sec 3 𝑥 = 2sec𝑥 csc𝑥(sec 2 𝑥 − csc 2 𝑥). The Differentiation of Trigonometric Functions Combined with the Chain Rule: If 𝑓 is differentiable function of 𝑥, then 𝑑 𝑑 𝑑 𝑑 sin(𝑓(𝑥)) = cos(𝑓(𝑥)) 𝑓(𝑥) cos(𝑓(𝑥)) = −sin(𝑓(𝑥)) 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑 𝑑 tan(𝑓(𝑥)) = sec 2 (𝑓(𝑥)) 𝑓(𝑥) cot(𝑓(𝑥)) = −csc 2 (𝑓(𝑥)) 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 sec(𝑓(𝑥)) = sec(𝑓(𝑥)) tan(𝑓(𝑥)) 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 𝑑 𝑑 csc(𝑓(𝑥)) = −csc(𝑓(𝑥)) cot(𝑓(𝑥)) 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 18 Example: Find 𝑦 ′ , if 1. 𝑦 = sin(𝑥 2 + 4) 2. 𝑦 = tan2 (cos𝑥) 3. 𝑦 = cot√𝑥 2 − 9 4. 𝑦 = sec(1 + sin𝑥) Solution: 𝑑 1. 𝑦 = sin(𝑥 2 + 4) ⇒ 𝑦 ′ = cos(𝑥 2 + 4) (𝑥 2 + 4) ⇒ 𝑦 ′ = 2𝑥 cos(𝑥 2 + 4). 𝑑𝑥 2 2. Since 𝑦 = tan2 (cos𝑥) = (tan(cos𝑥)) , then 𝑑 2 𝑦′ = (tan(cos𝑥)) 𝑑𝑥 𝑑 = 2tan(cos𝑥) tan(cos𝑥) 𝑑𝑥 𝑑 = 2tan(cos𝑥)sec 2 (cos𝑥) cos𝑥 𝑑𝑥 = 2tan(cos𝑥)sec 2 (cos𝑥)(−sin𝑥) = −2sin𝑥 tan(cos𝑥) sec 2 (cos𝑥). 𝑑 𝑑 3. 𝑦 ′ = cot√𝑥 2 − 9 ⇒ 𝑦 ′ = −csc 2 √𝑥 2 − 9 √𝑥 2 − 9 𝑑𝑥 𝑑𝑥 1 𝑑 ⇒ 𝑦 ′ = −csc 2 √𝑥 2 − 9 (𝑥 2 − 9) 2√𝑥 2 −9 𝑑𝑥 𝑥 ⇒ 𝑦′ = − csc 2 √𝑥 2 − 9. √𝑥 2 −9 𝑑 𝑑 4. 𝑦 ′ = sec(1 + sin𝑥) ⇒ 𝑦 ′ = sec(1 + sin𝑥)tan(1 + sin𝑥) (1 + sin𝑥) 𝑑𝑥 𝑑𝑥 ⇒ 𝑦 ′ = cos𝑥 sec(1 + sin𝑥)tan(1 + sin𝑥). Example: Find 𝑓 ′ if 1. 𝑓(𝑥) = sin(tan2 (3𝑥 3 )) 2. 𝑓(𝑥) = cot (cos(sin(2𝑥 2 + 5))). Solution: 1. Since 𝑓(𝑥) = sin(tan2 (3𝑥 3 )) = sin((tan(3𝑥 3 ))2 ), then 19  𝑑 𝑑 𝑓 ′ (𝑥) = sin((tan(3𝑥 3 ))2 ) ⇒ 𝑓 ′ (𝑥) = cos((tan(3𝑥 3 ))2 ) (tan(3𝑥 3 ))2 𝑑𝑥 𝑑𝑥 𝑑 ⇒ 𝑓 ′ (𝑥) = cos((tan(3𝑥 3 ))2 )(2tan(3𝑥 3 )) tan(3𝑥 3 ) 𝑑𝑥 𝑑 ⇒ 𝑓 ′ (𝑥) = 2cos((tan(3𝑥 3 ))2 )tan(3𝑥 3 )sec 2 (3𝑥 3 ) (3𝑥 3 ) 𝑑𝑥 ′ 2 3 2 )tan(3𝑥 3 ⇒ 𝑓 (𝑥) = 18𝑥 cos((tan(3𝑥 )) )sec (3𝑥 3 ). 2 2. 𝑑 𝑓 ′ (𝑥) = cot (cos(sin(2𝑥 2 + 5))) 𝑑𝑥 𝑑 = −csc 2 (cos(sin(2𝑥 2 + 5))) cos(sin(2𝑥 2 + 5)) 𝑑𝑥 𝑑 = csc 2 (cos(sin(2𝑥 2 + 5))) sin(sin(2𝑥 2 + 5)) sin(2𝑥 2 + 5) 𝑑𝑥 𝑑 = csc 2 (cos(sin(2𝑥 2 + 5))) sin(sin(2𝑥 2 + 5))cos(2𝑥 2 + 5) (2𝑥 2 + 5) 𝑑𝑥 = 4𝑥csc 2 (cos(sin(2𝑥 2 + 5))) sin(sin(2𝑥 2 + 5))cos(2𝑥 2 + 5). 20  Exercises (1) If f (x ) = cos ( 3x ) , then f (x ) = Ⓐ −3sin ( 3x ) Ⓑ − sin ( 3x ) Ⓒ 3sin ( 3x ) Ⓓ −3cos ( 3x ) ( ) (2) If f (x ) = tan x 3 − 4 , then f (x ) = 2 3 ( Ⓐ sec x − 4 ) Ⓑ −3x sec x − 4 2 2 3 ( ) 2 2 3 ( Ⓒ 3x sec x − 4 ) ( Ⓓ − sec x − 4 2 3 ) (3) If f (x ) = sin ( x 2 ) , then f (x ) = Ⓐ 2 cos ( x 2 ) Ⓑ cos ( x 2 ) Ⓒ −2 x cos ( x 2 ) Ⓓ 2 x cos ( x 2 ) (4) If y = sin ( x ) + cos ( 3x ) , then y  = Ⓐ cos ( x ) − sin ( 3x ) Ⓒ  cos ( x ) + 3sin ( 3x ) Ⓑ  cos ( x ) − 3sin ( 3x ) Ⓓ cos ( x ) + sin ( 3x ) (5) If f (x ) = sin x cos x , then f (x ) = Ⓐ − sin x cos x Ⓑ cos 2 x + sin 2 x Ⓒ sin 2 x − cos 2 x Ⓓ cos 2 x − sin 2 x cos x dy (6) If y = , then = 1 − sin x dx 1 1 1 1 Ⓐ− Ⓑ Ⓒ Ⓓ− 1 − sin x 1 − sin x (1 − sin x ) (1 − sin x ) 2 2 (7) If y = 3x + cot  x  , then dy = 2 dx Ⓐ 3 − csc2  x  Ⓑ 3 + 1 csc2  x  Ⓒ 3 − 1 csc2  x  Ⓓ 3 + csc2  x  2 2 2 2 2 2 21  dy (8) If y = sin x , then 3 = dx Ⓐ cos3 x Ⓑ −3sin 2 x cos x Ⓒ 3sin 2 x Ⓓ 3sin 2 x cos x dy (9) If y = cos x , then = dx sin x sin x sin x Ⓓ − sin x Ⓐ− Ⓑ Ⓒ− 2 x 2 x x dy (10) If y = sin ( 2cos x ) , then = dx Ⓐ 2cos ( 2cos x ) sin x Ⓒ −2cos ( 2cos x ) sin x Ⓑ cos ( 2cos x ) Ⓓ 2cos ( 2cos x ) cos x (11) If f (x ) = x 2 cos ( 3x ) , then f (x ) = Ⓐ −3x 2 sin ( 3x ) + 2x cos ( 3x ) Ⓒ −3x 2 sin ( 3x ) − 2x cos ( 3x ) Ⓑ 3x 2 sin ( 3x ) + 2x cos ( 3x ) Ⓓ 3x 2 sin ( 3x ) − 2x cos ( 3x ) (12) If y = csc(x + 2) , then y  = 2 Ⓐ − csc ( x 2 + 2 ) cot ( x 2 + 2 ) Ⓒ 2x csc ( x 2 + 2 ) cot ( x 2 + 2 ) Ⓑ csc ( x 2 + 2 ) cot ( x 2 + 2 ) Ⓓ −2x csc ( x 2 + 2 ) cot ( x 2 + 2 ) (13) If f (x ) = x 2 + 2cos x , then f (x ) = 2x 1 2sin x x − sin x Ⓐ Ⓑ Ⓒ− Ⓓ x 2 + 2cos x 2 x 2 + 2cos x x 2 + 2cos x x 2 + 2cos x 22 (14) If f (x ) = ( x + sin x ) , then f (x ) = 9 Ⓐ 9 ( x + sin x ) Ⓒ 9 ( x + sin x ) (1 + cos x ) 8 8 Ⓑ 9 ( x + sin x ) cos x Ⓓ ( x + sin x ) (1 + cos x ) 8 8 sin x + cos x (15) If f (x ) = , then f (x ) = sin x − cos x sin x + cos x 2 Ⓐ Ⓒ ( sin x − cos x ) ( sin x − cos x ) 2 2 sin x + cos x 2 Ⓑ− Ⓓ− ( sin x − cos x ) ( sin x − cos x ) 2 2 dy (16) If y = 8 + sec x , then = dx sec x tan x 1 sec x tan x sec x tan x Ⓐ Ⓑ Ⓒ− Ⓓ 8 + sec x 2 8 + sec x 2 8 + sec x 2 8 + sec x 23 5. Differentiation of Inverse Trigonometric Functions In the following we state the derivative of six basic inverse trigonometric functions. The derivative of inverse sine function : 𝑑 1 sin−1 𝑥 = , − 1 < 𝑥 < 1. 𝑑𝑥 √1−𝑥 2 The derivative of inverse cosine function : 𝑑 1 cos −1 𝑥 = − , − 1 < 𝑥 < 1. 𝑑𝑥 √1 − 𝑥 2 The derivative of inverse tangent function : 𝑑 1 tan−1 𝑥 = , − ∞ < 𝑥 < ∞. 𝑑𝑥 1 + 𝑥2 The derivative of inverse cotangent function : 𝑑 1 cot −1 𝑥 = − , − ∞ < 𝑥 < ∞. 𝑑𝑥 1 + 𝑥2 The derivative of inverse secant function : 𝑑 1 sec −1 𝑥 = , 𝑥 ∈ (−∞, −1) ∪ (1, ∞). 𝑑𝑥 |𝑥|√𝑥 2 − 1 The derivative of inverse cosecant function : 𝑑 1 csc −1 𝑥 = − , 𝑥 ∈ (−∞, −1) ∪ (1, ∞). 𝑑𝑥 |𝑥|√𝑥 2 − 1 Example: Find 𝑦 ′ 1. 𝑦 = √sin−1 𝑥 2. 𝑦 = 𝑥 3 sin−1 𝑥. Solution: 𝑑 1 𝑑 1 1 1. 𝑦 ′ = √sin−1 𝑥 ⇒ 𝑦′ = sin−1 𝑥 ⇒ 𝑦′ = ×. 𝑑𝑥 2√sin−1 𝑥 𝑑𝑥 2√sin−1 𝑥 √1−𝑥 2 𝑑 𝑑 𝑑 𝑥3 2. 𝑦 ′ = [𝑥 3 sin−1 𝑥] ⇒ 𝑦 ′ = 𝑥 3 sin−1 𝑥 + sin−1 𝑥 𝑥3 ⇒ 𝑦′ = + 3𝑥 2 sin−1 𝑥. 𝑑𝑥 𝑑𝑥 𝑑𝑥 √1−𝑥 2 24 Example: 1 If 𝑓(𝑥) = , find 𝑓 ′ (𝑥). cos−1 𝑥 Solution: 1 −1 −1 ′ −1 −2 𝑑 𝑓(𝑥) = −1 ⇒ 𝑓(𝑥) = (cos 𝑥) ⇒ 𝑓 (𝑥) = −(cos 𝑥) cos −1 𝑥 cos 𝑥 𝑑𝑥 1 1 1 ⇒ 𝑓 ′ (𝑥) = −(cos −1 𝑥)−2 (− ) ⇒ 𝑓 ′ (𝑥) = ×. √1−𝑥 2 √1−𝑥 2 (cos−1 𝑥)2 Example: If 𝑦 = tan−1 (𝑥 2 + 1), find 𝐷𝑥 𝑦. Solution: 𝑑 1 𝑑 2 2𝑥 𝐷𝑥 𝑦 = tan−1 (𝑥 2 + 1) ⇒ 𝐷𝑥 𝑦 = (𝑥 + 1) = 𝑑𝑥 1 + 𝑥 2 𝑑𝑥 1 + 𝑥2 Example: 𝑑𝑦 If 𝑦 = sin(√16 + cot −1 𝑡), find. 𝑑𝑡 Solution: 𝑑𝑦 𝑑 𝑑𝑦 𝑑 = sin (√16 + cot −1 𝑡) ⇒ = cos (√16 + cot −1 𝑡) × √16 + cot −1 𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑦 cos(√16 + cot −1 𝑡) 𝑑 ⇒ = × (16 + cot −1 𝑡) 𝑑𝑡 2√16 + cot −1 𝑡 𝑑𝑡 𝑑𝑦 cos(√16 + cot −1 𝑡) 1 ⇒ = × (− ) 𝑑𝑡 2√16 + cot −1 𝑡 1 + 𝑡2 𝑑𝑦 1 cos(√16 + cot −1 𝑡) ⇒ =−. 𝑑𝑡 2 (1 + 𝑡 2 )√16 + cot −1 𝑡 25 Example: Find 𝑦 ′ 3 1. 𝑦 = √(sec −1 𝑥)2 2. 𝑦 = (𝑥 3 + 1)(csc −1 𝑥)2. Solution: 3 1. Since 𝑦 = √(sec −1 𝑥)2 = (sec −1 𝑥)2/3 , then 𝑑 2 𝑑 𝑦′ = (sec −1 𝑥)2/3 ⇒ 𝑦 ′ = (sec −1 𝑥)−1/3 sec −1 𝑥 𝑑𝑥 3 𝑑𝑥 2 1 ⇒ 𝑦 ′ = (sec −1 𝑥)−1/3 3 |𝑥|√𝑥 2 − 1 2 1 1 ⇒ 𝑦′ = 3 ×. 3 √sec −1 𝑥 |𝑥|√𝑥 2 − 1 𝑑 2. 𝑦 ′ = [(𝑥 3 + 1)(csc −1 𝑥)2 ] 𝑑𝑥 𝑑 𝑑 ⇒ 𝑦 ′ = (𝑥 3 + 1) (csc −1 𝑥)2 + (csc −1 𝑥)2 (𝑥 3 + 1) 𝑑𝑥 𝑑𝑥 𝑑 ⇒ 𝑦 ′ = (𝑥 3 + 1) (2csc −1 𝑥 × csc −1 𝑥) + (csc −1 𝑥)2 (3𝑥 2 ) 𝑑𝑥 −1 ⇒ 𝑦 ′ = 2(𝑥 3 + 1)csc −1 𝑥 × + 3𝑥 2 (csc −1 𝑥)2 |𝑥|√𝑥 2 −1 −2(𝑥 3 +1) ⇒ 𝑦 ′ = csc −1 𝑥 ( + 3𝑥 2 csc −1 𝑥). |𝑥|√𝑥 2 −1 The Differentiation of Inverse Trigonometric Functions Combined with the Chain Rule: If 𝑓 is differentiable function of 𝑥, then 𝑑 1 𝑑 𝑑 −1 𝑑 sin−1 (𝑓(𝑥)) = 𝑓(𝑥) cos−1 (𝑓(𝑥)) = 𝑓(𝑥) 𝑑𝑥 √1−(𝑓(𝑥))2 𝑑𝑥 𝑑𝑥 √1−(𝑓(𝑥))2 𝑑𝑥 𝑑 1 𝑑 𝑑 −1 𝑑 tan−1 (𝑓(𝑥)) = 𝑓(𝑥) cot−1 (𝑓(𝑥)) = 𝑓(𝑥) 𝑑𝑥 1+(𝑓(𝑥))2 𝑑𝑥 𝑑𝑥 1+(𝑓(𝑥))2 𝑑𝑥 𝑑 1 𝑑 sec −1 (𝑓(𝑥)) = 𝑓(𝑥) 𝑑𝑥 |𝑓(𝑥)|√(𝑓(𝑥))2 −1 𝑑𝑥 𝑑 −1 𝑑 csc −1 (𝑓(𝑥)) = 𝑓(𝑥) 𝑑𝑥 |𝑓(𝑥)|√(𝑓(𝑥))2 − 1 𝑑𝑥 26 Example: Find 𝑓 ′ (𝑥) if 1 1. 𝑓(𝑥) = cos −1 ( ) 2. 𝑓(𝑥) = tan−1 (√𝑥 2 + 5) 𝑥 3. 𝑓(𝑥) = sec −1 (𝑥 2 ) + (sec −1 𝑥)2. Solution: 𝑑 1 −1 𝑑 1 −1 1 1. 𝑓 ′ (𝑥) = cos −1 ( ) ⇒ 𝑓 ′ (𝑥) = × ( ) ⇒ 𝑓 ′ (𝑥) = (− 𝑥 2) 𝑑𝑥 𝑥 1 𝑑𝑥 𝑥 1 √1− 2 √1− 2 𝑥 𝑥 1 1 ⇒ 𝑓 ′ (𝑥) = ⇒ 𝑓 ′ (𝑥) =. 1 𝑥√𝑥 2 −1 𝑥 2 √1− 2 𝑥 𝑑 1 𝑑 2. 𝑓 ′ (𝑥) = tan−1 (√𝑥 2 + 5) ⇒ 𝑓 ′ (𝑥) = 2 × √𝑥 2 + 5 𝑑𝑥 1+(√𝑥 2 +5) 𝑑𝑥 1 1 𝑑 1 1 ⇒ 𝑓 ′ (𝑥) = × × (𝑥 2 + 5) ⇒ 𝑓 ′ (𝑥) = × (2𝑥) 6+𝑥 2 2√𝑥 2 +5 𝑑𝑥 6+𝑥 2 2√𝑥 2 +5 1 𝑥 ⇒ 𝑓 ′ (𝑥) = 2 ×. 6+𝑥 √𝑥 2 +5 𝑑 𝑑 𝑑 3. 𝑓 ′ (𝑥) = (sec −1 (𝑥 2 ) + (sec −1 𝑥)2 ) ⇒ 𝑓 ′ (𝑥) = sec −1 (𝑥 2 ) + (sec −1 𝑥)2 𝑑𝑥 𝑑𝑥 𝑑𝑥 1 𝑑 𝑑 ⇒ 𝑓 ′ (𝑥) = × 𝑥 2 + 2sec −1 𝑥 × sec −1 𝑥 |𝑥 2 |√(𝑥 2 )2 −1 𝑑𝑥 𝑑𝑥 1 1 ⇒ 𝑓 ′ (𝑥) = × (2𝑥) + 2sec −1 𝑥 × 𝑥 2 √𝑥 4 −1 |𝑥|√𝑥 2 −1 2 2sec−1 𝑥 ⇒ 𝑓 ′ (𝑥) = +. 𝑥√𝑥 4 −1 |𝑥|√𝑥 2 −1 27  Exercises (1) If y = sin −1 ( x 2 ) , then y  = x 2x x 2x Ⓐ Ⓑ− Ⓒ− Ⓓ 1− x 4 1− x 4 1− x 4 1− x 4 (2) If y = cos −1 ( x 2 ) , then y  = x 2x x 2x Ⓐ Ⓑ− Ⓒ− Ⓓ 1− x 4 1− x 4 1− x 4 1− x 4 (3) If y = sin −1 ( cos x ) , then y  = cos x sin x sin x cos x Ⓐ− Ⓑ Ⓒ− Ⓓ 1 − cos 2 x 1 − cos 2 x 1 − cos 2 x 1 − cos 2 x (4) If y = cos −1 ( sin x ) , then y  = cos x sin x sin x cos x Ⓐ Ⓑ Ⓒ− Ⓓ− 1 − sin 2 x 1 − sin 2 x 1 − sin 2 x 1 − sin 2 x (5) If y = tan −1 ( cos x ) , then y  = sin x sin x cos x cos x Ⓐ Ⓑ− Ⓒ− Ⓓ 1 + cos 2 x 1 + cos 2 x 1 + cos 2 x 1 + cos 2 x (6) If y = cot −1 ( sin x ) , then y  = cos x cos x sin x sin x Ⓐ Ⓑ− Ⓒ− Ⓓ 1 + sin 2 x 1 + sin 2 x 1 + sin 2 x 1 + sin 2 x 28 (7) If y = cos −1 ( x ) , then y= 1 1 1 1 Ⓐ Ⓑ− Ⓒ− Ⓓ 2 x −x2 x −x 2 2 x −x 2 x −x 2 (8) If f (x ) = sin −1 ( x ) , then f (x ) = 1 1 1 1 Ⓐ Ⓑ Ⓒ − Ⓓ− 2 x 1− x 2 2 x −x2 2 x 1− x 2 2 x −x 2 (9) If y = sec−1 ( x 2 ) , then y  = 1 1 2 2 Ⓐ− Ⓑ Ⓒ− Ⓓ x x 4 −1 x x 4 −1 x x 4 −1 x x 4 −1 (10) If y = csc−1 ( x 2 ) , then y  = 1 1 2 2 Ⓐ− Ⓑ Ⓒ− Ⓓ x x 4 −1 x x 4 −1 x x 4 −1 x x 4 −1 29 6. Differentiation of logarithmic and exponential functions 6.1 Differentiation of logarithmic functions Let 𝑎 be a positive number with 𝑎 ≠ 1. The logarithmic function with base 𝑎, is denoted by log 𝑎 𝑥 and is defined by 𝒇(𝒙) = 𝐥𝐨𝐠 𝒂 𝒙 (𝒂 > 𝟎), 𝒂 ≠ 𝟏. The domain is the set of positive real numbers or (0, ∞). The range is the set of all real numbers or (−∞, ∞). Properties of the logarithmic function: Let 𝑎, 𝑥 and 𝑦 be positive real numbers, 𝑎 ≠ 1. 1. log 𝑎 (𝑥𝑦) = log 𝑎 𝑥 + log 𝑎 𝑦. 𝑥 2. log 𝑎 ( ) = log 𝑎 𝑥 − log 𝑎 𝑦. 𝑦 3. log 𝑎 𝑥 𝑟 = 𝑟log 𝑎 𝑥, 𝑟 ∈ ℝ. 4. log 𝑎 𝑎 = 1, log 𝑎 1 = 0. Remark: 1. The logarithm with base 10 is called the common logarithm and is denoted by omitting the base: log𝑥 = log10 𝑥. 2. The logarithm with base 𝑒 is called the natural logarithm and is denoted by 𝑙𝑛: log 𝑒 𝑥 = ln𝑥. ln𝑥 3. log 𝑎 𝑥 = ln𝑎 The derivative of general logarithmic function : If 𝑓(𝑥) = log 𝑎 𝑥, 𝑎 ≠ 1, 𝑎 > 0, then 1 1 ln𝑒 1 𝑓 ′ (𝑥) = log 𝑎 𝑒 = , where log 𝑎 𝑒 = = 𝑥 𝑥 ln𝑎 ln𝑎 ln𝑎 30 The derivative of natural logarithmic function: The derivative of natural logarithmic function is given by 𝑑 1 ln𝑥 =. 𝑑𝑥 𝑥 The differentiation of logarithmic functions combined with the Chain Rule: If 𝑓 is differentiable function of 𝑥, then 𝑑 1 𝑑 𝑑 1 𝑑 log 𝑎 (𝑓(𝑥)) = 𝑓(𝑥) ln(𝑓(𝑥)) = 𝑓(𝑥). 𝑑𝑥 𝑓(𝑥) ln𝑎 𝑑𝑥 𝑑𝑥 𝑓(𝑥) 𝑑𝑥 Example: Find 𝑓 ′ (𝑥) if 1. 𝑓(𝑥) = log 2 (𝑥 4 + 9) 2. 𝑓(𝑥) = tan−1 (ln𝑥) + ln(tan−1 𝑥) 𝑥 3. 𝑓(𝑥) = log ( ). 1+𝑥 2 Solution: 𝑑 1 𝑑 4 𝑥3 1. 𝑓 ′ (𝑥) = log 2 (𝑥 4 + 9) ⇒ 𝑓 ′ (𝑥) = × (𝑥 4 + 9) ⇒ 𝑓 ′ (𝑥) =. 𝑑𝑥 (𝑥 4 +9)ln2 𝑑𝑥 ln2 𝑥 4 +9 𝑑 2. 𝑓 ′ (𝑥) = [tan−1 (ln𝑥) + ln(tan−1 𝑥)] 𝑑𝑥 𝑑 𝑑 ⇒ 𝑓 ′ (𝑥) = tan−1 (ln𝑥) + ln(tan−1 𝑥) 𝑑𝑥 𝑑𝑥 1 𝑑 1 𝑑 ⇒ 𝑓 ′ (𝑥) = × ln𝑥 + × tan−1 𝑥 1+(ln𝑥)2 𝑑𝑥 tan−1 𝑥 𝑑𝑥 1 1 1 1 ⇒ 𝑓 ′ (𝑥) = 2 × + ×. 1+(ln𝑥) 𝑥 tan−1 𝑥 1+𝑥 2 𝑥 3. Since 𝑓(𝑥) = log ( 1+𝑥 2 ) = log𝑥 − log(1 + 𝑥 2 ), then 𝑑 𝑓 ′ (𝑥) = [log𝑥 − log(1 + 𝑥 2 )] 𝑑𝑥 𝑑 𝑑 ⇒ 𝑓 ′ (𝑥) = log𝑥 − log(1 + 𝑥 2 ) 𝑑𝑥 𝑑𝑥 1 1 𝑑 ⇒ 𝑓 ′ (𝑥) = − (1+𝑥 2)ln10 (1 + 𝑥 2 ) 𝑥ln10 𝑑𝑥 1 1 ⇒ 𝑓 ′ (𝑥) = − (2𝑥) 𝑥ln10 (1+𝑥 2 )ln10 1 1+𝑥 2 −2𝑥 2 1 1−𝑥 2 ⇒ 𝑓 ′ (𝑥) = =. ln10 𝑥(1+𝑥 2 ) ln10 𝑥(1+𝑥 2 ) 31 6.2 Differentiation of exponential functions An exponential function with base 𝑎 is defined for all real numbers 𝑥 by 𝑓(𝑥) = 𝑎 𝑥 , where where 𝑎 is a positive real number (𝑎 > 0), 𝑎 ≠ 1. The domain is the set of all real numbers or (−∞, ∞). The range is the set of positive real numbers or (0, ∞). Rules of the exponential functions: Let 𝑎 and 𝑏 be two positive real numbers. Then for every real numbers 𝑥 and 𝑦, we have 1. 𝑎 𝑥+𝑦 = 𝑎 𝑥 × 𝑎 𝑦. 𝑎𝑥 2. 𝑎 𝑥−𝑦 =. 𝑎𝑦 3. (𝑎 𝑥 )𝑦 = (𝑎 𝑦 )𝑥 =. 4. (𝑎 × 𝑏)𝑥 = 𝑎 𝑥 × 𝑏 𝑥. 𝑎 𝑥 𝑎𝑥 5. ( ) = 𝑥. 𝑏 𝑏 1 6. 𝑎−𝑥 = , 𝑎0 = 1. 𝑎𝑥 Remark: If 𝑎 = 𝑒, the function 𝑓(𝑥) = 𝑒 𝑥 is called the natural exponential function, where the irrational number 𝑒 ≃ 2.71828182845904523536 is called the natural base. The derivative of general exponential function : 𝑑 𝑥 𝑎 = 𝑎 𝑥 ln𝑎, 𝑎 ≠ 1, 𝑎 > 0. 𝑑𝑥 The derivative of natural exponential function: The derivative of natural exponential function is given by 𝑑 𝑥 𝑒 = 𝑒 𝑥. 𝑑𝑥 32 The differentiation of exponential functions combined with the Chain Rule: If 𝑓 is differentiable function of 𝑥, then 𝑑 𝑑 𝑑 𝑑 𝑎 𝑓(𝑥) = 𝑎 𝑓(𝑥) ln𝑎 𝑓(𝑥) 𝑒 𝑓(𝑥) = 𝑒 𝑓(𝑥) 𝑓(𝑥). 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 Example: Find 𝑓 ′ (𝑥) if 3𝑥 1. 𝑓(𝑥) = 2csc 2. 𝑓(𝑥) = 𝑒 𝑥 sin𝑥 3. 𝑓(𝑥) = 𝑥𝑒 √1−𝑥. Solution: 𝑑 3𝑥 3 𝑑 1. 𝑓 ′ (𝑥) = 2csc ⇒ 𝑓 ′ (𝑥) = 2csc 𝑥 ln2 × (csc𝑥)3 𝑑𝑥 𝑑𝑥 3 𝑑 ⇒ 𝑓 ′ (𝑥) = 2csc 𝑥 ln2(3(csc𝑥)2 ) × csc𝑥 𝑑𝑥 3 ⇒ 𝑓 ′ (𝑥) = 2csc 𝑥 ln2(3(csc𝑥)2 ) × (−csc𝑥cot𝑥) 3 ⇒ 𝑓 ′ (𝑥) = −3ln2 × 2csc 𝑥 csc 3 𝑥cot𝑥. 𝑑 𝑑 2. 𝑓 ′ (𝑥) = 𝑒 𝑥sin𝑥 ⇒ 𝑓 ′ (𝑥) = 𝑒 𝑥sin𝑥 [𝑥sin𝑥] 𝑑𝑥 𝑑𝑥 𝑑 𝑑 ⇒ 𝑓 ′ (𝑥) = 𝑒 𝑥sin𝑥 [𝑥 sin𝑥 + sin𝑥 𝑥] ⇒ 𝑓 ′ (𝑥) = 𝑒 𝑥sin𝑥 [𝑥cos𝑥 + sin𝑥]. 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑 3. 𝑓 ′ (𝑥) = [𝑥𝑒 √1−𝑥 ] ⇒ 𝑓 ′ (𝑥) = 𝑥 𝑒 √1−𝑥 + 𝑒 √1−𝑥 𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑 ⇒ 𝑓 ′ (𝑥) = 𝑥𝑒 √1−𝑥 × √1 − 𝑥 + 𝑒 √1−𝑥 × 1 𝑑𝑥 1 𝑑 ⇒ 𝑓 ′ (𝑥) = 𝑥𝑒 √1−𝑥 × (1 − 𝑥) + 𝑒 √1−𝑥 2√1−𝑥 𝑑𝑥 −𝑥𝑒 √1−𝑥 ⇒ 𝑓 ′ (𝑥) = + 𝑒 √1−𝑥. 2√1−𝑥 33  Exercises 2 −3 x (1) If y = e x , then y  = −3 x Ⓑ ( 2x + 3 ) e x −3x Ⓒ ( 2x − 3 ) e x −3x −3x 2 2 2 2 Ⓐ ex Ⓓ 2xe x (2) If y = 1 + e 2 x , then y  = e 2x e 2x 1 2e 2 x Ⓐ Ⓑ Ⓒ Ⓓ 2 1 + e 2x 1 + e 2x 2 1 + e 2x 1 + e 2x (3) If f (x ) = 2 , then f (x ) = sec x Ⓐ − sec ( x ) tan ( x ) 2sec x ln 2 Ⓒ sec ( x ) tan ( x ) 2sec x Ⓑ − sec ( x ) tan ( x ) 2sec x Ⓓ sec ( x ) tan ( x ) 2sec x ln 2 (4) If f (x ) = csc (e x ) , then f (x ) = Ⓐ −e x csc (e x ) cot (e x ) Ⓒ − csc (e x ) cot (e x ) Ⓑ e x csc (e x ) cot (e x ) Ⓓ csc (e x ) cot (e x ) (5) If g (x ) = 3 + ln x , then g ( x ) = sin x ( Ⓐ − cos x 3sin x ln 3 + x ) 1 ( ) Ⓒ cos x 3sin x ln3 + 1 x ( Ⓑ cos x 3sin x + x 1 ) ( Ⓓ − cos x 3sin x + ) 1 x ex dy (6) If y = , then = 1+e x dx 2e x ex Ⓒ 1 Ⓓ − 1 Ⓐ Ⓑ (1 + e x ) 2 (1 + e x ) 2 (1 + e x ) 2 (1 + e x ) 2 34 (7) If f (x ) = e x sin x , then f (x ) = Ⓐ e x sin x Ⓒ ( x sin x + cos x )e x sin x Ⓑ ( x cos x − sin x )e x sin x Ⓓ ( x cos x + sin x )e x sin x dy (8) If y = xe , then 4x = dx Ⓐ 4(x + 1)e Ⓑ (x + 4)e Ⓒ (x + 1)e Ⓓ (4x + 1)e 4x 4x 4x 4x −1 dy (9) If y = e sin x , then = dx −1 −1 −1 −1 e sin x e sin x e sin x e sin x Ⓐ Ⓑ − Ⓒ − Ⓓ 1− x 2 1− x 2 x 2 −1 x 2 −1 (10) If f (x ) = ln(1 + csc x ) , then f (x ) = csc x cot x 1 1 csc x cot x Ⓐ− Ⓑ− Ⓒ Ⓓ 1 + csc x 1 + csc x 1 + csc x 1 + csc x dy (11) If y = log 3 (sin x ) , then = dx Ⓐ − cos x Ⓑ sin x Ⓒ cos x Ⓓ − sin x sin x ln 3 cos x ln 3 sin x ln 3 cos x ln 3 dy (12) If y = log 3 (cos x ) , then = dx Ⓐ − cos x Ⓑ sin x Ⓒ cos x Ⓓ − sin x sin x ln 3 cos x ln 3 sin x ln 3 cos x ln 3 df (13) If f (x ) = ln(sin x ) , then = dx Ⓐ tan x Ⓑ cot x Ⓒ − cot x Ⓓ − tan x 35  df (14) If f (x ) = ln(cos x ) , then = dx Ⓐ tan x Ⓑ cot x Ⓒ − cot x Ⓓ − tan x (15) If f (x ) = ln(1 + x 2 ) , then f (x ) = 1 x 2x x2 Ⓐ Ⓑ Ⓒ Ⓓ 1+ x 2 1+ x 2 1+ x 2 1+ x 2 (16) If f (x ) = ln 1 + x 2 , then f (x ) = x x 2x 2x Ⓐ Ⓑ Ⓒ Ⓓ 1+ x 2 1+ x 2 1+ x 2 1+ x 2 (17) If f (x ) = cot −1 ( ln x ) , then f (x ) = x x 1 1 Ⓐ Ⓑ− Ⓒ Ⓓ− 1 + ( ln x ) 2 1 + ( ln x ) 2 ( x 1 + ( ln x ) 2 ) ( x 1 + ( ln x ) 2 ) 36 7. Differentiation of Hyperbolic Functions and Inverse Hyperbolic Functions 7.1 Differentiation of Hyperbolic Functions The hyperbolic functions have similar names to the trigonmetric functions, but they are defined in terms of the exponential function. Hyperbolic cosine of x: 𝑒 𝑥 + 𝑒 −𝑥 𝑓(𝑥) = cosh𝑥 = 2 Hyperbolic sine of x: 𝑒 𝑥 −𝑒 −𝑥 𝑓(𝑥) = sinh𝑥 = 2 Hyperbolic tangent of x: sinh𝑥 𝑒 𝑥 −𝑒 −𝑥 𝑓(𝑥) = tanh𝑥 = = cosh𝑥 𝑒 𝑥 +𝑒 −𝑥 Hyperbolic cotangent of x: cosh𝑥 𝑒 𝑥 +𝑒 −𝑥 𝑓(𝑥) = coth𝑥 = = sinh𝑥 𝑒 𝑥 −𝑒 −𝑥 Hyperbolic secant of x: 1 2 𝑓(𝑥) = sech𝑥 = = cosh𝑥 𝑒 𝑥 +𝑒 −𝑥 Hyperbolic cosecant of x: 1 2 𝑓(𝑥) = csch𝑥 = = sinh𝑥 𝑒 𝑥 −𝑒 −𝑥 Basic Identities: 1. sinh𝑥 + cosh𝑥 = 𝑒 𝑥 2. cosh𝑥 − sinh𝑥 = 𝑒 −𝑥 3. cosh2 𝑥 − sinh2 𝑥 = 1 4. 1 − tanh2 𝑥 = sech2 𝑥 5. coth2 𝑥 − 1 = csch2 𝑥 37  6. sinh2𝑥 = 2sinh𝑥 cosh𝑥 7. cosh2𝑥 = sinh2 𝑥 + cosh2 𝑥 8. cosh2𝑥 = 2sinh2 𝑥 + 1 9. cosh2𝑥 = 2cosh2 𝑥 − 1 10. sinh(𝑥 ± 𝑦) = sinh𝑥 cosh𝑦 ± cosh𝑥 sinh𝑦 11. cosh(𝑥 ± 𝑦) = cosh𝑥 cosh𝑦 ± sinh𝑥 sinh𝑦 In the following we state the derivative of hyperbolic functions 𝑑 𝑑 (1) sinh𝑥 = cosh𝑥 (2) cosh𝑥 = sinh𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 (3) tanh𝑥 = sech2 𝑥 (4) coth𝑥 = −csch2 𝑥 𝑑𝑥 𝑑𝑥 𝑑 𝑑 (5) sech𝑥 = −sech𝑥 tanh𝑥 (6) csch 𝑥 = −csch𝑥 coth𝑥 𝑑𝑥 𝑑𝑥 The Differentiation of Hyperbolic Functions Combined with the Chain Rule: If 𝑓 is differentiable function of 𝑥, then 𝑑 (1) sinh [𝑓(𝑥)] = cosh [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 𝑑 (2) cosh [𝑓(𝑥)] = sinh [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 𝑑 (3) tanh [𝑓(𝑥)] = sech2 [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 𝑑 (4) coth [𝑓(𝑥)] = −csch2 [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 𝑑 (5) sech [𝑓(𝑥)] = −sech [𝑓(𝑥)] tanh [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 𝑑 (6) csch [𝑓(𝑥)] = −csch [𝑓(𝑥)] coth [𝑓(𝑥)] ⋅ 𝑓′(𝑥) 𝑑𝑥 Example: Find 𝑓 ′ (𝑥) if (1) 𝑓(𝑥) = sinh(𝑥 2 + 3) (2) 𝑓(𝑥) = cosh4 (√𝑥) 2 (3) 𝑓(𝑥) = tanh(𝑒 2𝑥 ) (4) 𝑓(𝑥) = sech(4𝑥 ) (5) 𝑓(𝑥) = ln(sinh𝑥) + sinh(ln𝑥) 38 Solution: (1) 𝑓(𝑥) = sinh(𝑥 2 + 3) ⇒ 𝑓′(𝑥) = 2𝑥cosh(𝑥 2 + 3) 1 (2) 𝑓(𝑥) = cosh4 (√𝑥) ⇒ 𝑓′(𝑥) = 4sinh(√𝑥)cosh3 (√𝑥) × 2√𝑥 2 ⇒ 𝑓′(𝑥) = sinh(√𝑥)cosh3 (√𝑥) √𝑥 2𝑥 ) (3) 𝑓(𝑥) = tanh(𝑒 ⇒ 𝑓′(𝑥) = 2𝑒 2𝑥 sech2 (𝑒 2𝑥 ) 2 2 2 2 (4) 𝑓(𝑥) = sech(4𝑥 ) ⇒ 𝑓′(𝑥) = −sech(4𝑥 )tanh(4𝑥 ) × 4𝑥 ln4 × 2𝑥 2 2 2 ⇒ 𝑓′(𝑥) = −2𝑥 sech(4𝑥 )tanh(4𝑥 ) × 4𝑥 ln4 (5) 𝑓(𝑥) = ln(sinh𝑥) + sinh(ln𝑥) cosh𝑥 1 ⇒ 𝑓′(𝑥) = + cosh(ln𝑥) × sinh𝑥 𝑥 7.2 Differentiation of Inverse Hyperbolic Functions 𝑑 1 (1) sinh−1 𝑥 = 𝑑𝑥 √1+𝑥 2 𝑑 1 (2) cosh−1 𝑥 = , 𝑥>1 𝑑𝑥 √𝑥 2 −1 𝑑 1 (3) tanh−1 𝑥 = , |𝑥| < 1 𝑑𝑥 1−𝑥 2 𝑑 1 (4) coth−1 𝑥 = , |𝑥| > 1 𝑑𝑥 1−𝑥 2 𝑑 −1 (5) sech−1 𝑥 = , 0