Questions and Answers #37 Derivatives

Questions and Answers Sheet 37 Derivatives Question #1 Find the indicate d partial derivative. 1 𝑓(𝑥, 𝑦) = 𝑦 sin−1 (𝑥𝑦) ; 𝑓𝑦 (1, ) 2 Answer: 𝑓(𝑥, 𝑦) = 𝑦 sin−1 𝑥 𝑦 Differentiate with respect to y, treat x like a constants 𝑓𝑦 (𝑥, 𝑦) = sin−1 𝑥 𝑦 + 𝑦 ∗ 𝑓𝑦 (𝑥, 𝑦) = sin−1 𝑥 𝑦 + 1 √1 − 𝑥 2 𝑦 2 ∗𝑥 𝑥𝑦 √1 − 𝑥 2 𝑦 2 Therefore 1 1 𝑓𝑦 (1, ) = sin−1 + 2 2 1 2 √1 − 1 4 1 1 𝜋 𝑓𝑦 (1, ) = + 2 2 6 √3 2 1 𝜋 1 𝑓𝑦 (1, ) = + 2 6 √3 Result: 1 𝜋 1 𝑓𝑦 (1, ) = + 2 6 √3 Question #2 Suppose that 𝑓(5) = 2, 𝑓 ′ (5) = 3, 𝑔(5) = −6, and 𝑔′ (5) = 1. Find the following values: (𝑓𝑔)′ (5) Answer: (1) Rewrite the function as a product. (𝑓𝑔)(𝑥) = 𝑓(𝑥)𝑔(𝑥) (2) Find the derivative using the product rule. (𝑓𝑔)′ (𝑥) = 𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′ (𝑥) (3) Substitute x=5. (𝑓𝑔)′ (5) = 𝑓 ′ (5)𝑔(5) + 𝑓(5)𝑔′ (5) (4) Substitute 𝑓 ′ (5) = 3, 𝑔(5) = −6, 𝑓(5) = 2, and 𝑔′ (5) = 1. = 3(−6) + 2(1) (5) Simplify = −16 Question #3 Suppose that 𝑓(5) = 2, 𝑓 ′ (5) = 3, 𝑔(5) = −6, and 𝑔′ (5) = 1. Find the following values: (𝑓/𝑔)′ (5) Answer: (1)Rewrite the function as a quotient. (𝑓/𝑔)(𝑥) = 𝑓(𝑥) 𝑔(𝑥) (2)Find the derivative using the quotient rule. (𝑓/𝑔)′ (𝑥) = 𝑓 ′ (𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′ (𝑥) [𝑔(𝑥)]2 (3)Substitute x=5. (𝑓/𝑔)′ (5) = 𝑓 ′ (5)𝑔(5) − 𝑓(5)𝑔′ (5) [𝑔(5)]2 Substitute 𝑓 ′ (5) = 3, 𝑔(5) = −6, 𝑓(5) = 2, and 𝑔′ (5) = 1. = 3(−6) − 2(1) (−6)2 (5)Simplify = −20 36 (6)Reduce the fraction. =− 5 9 Question #4 Suppose that 𝑓(5) = 2, 𝑓 ′ (5) = 3, 𝑔(5) = −6, and 𝑔′ (5) = 1. Find the following values: (𝑔/𝑓)′ (5) Answer: (1)Rewrite the function as a quotient. (𝑔/𝑓)(𝑥) = 𝑔(𝑥) 𝑓(𝑥) (2)Find the derivative using the quotient rule. (𝑔/𝑓)′ (𝑥) = 𝑔′ (𝑥)𝑓(𝑥) − 𝑔(𝑥)𝑓 ′ (𝑥) [𝑓(𝑥)]2 (3)Substitute 𝑥 = 5. (𝑔/𝑓)′ (5) 𝑔′ (5)𝑓(5) − 𝑔(5)𝑓 ′ (5) = [𝑓(5)] 2 (4)Substitute 𝑓 ′ (5) = 3, 𝑔(5) = −6, 𝑓(5) = 2, and 𝑔′ (5) = 1. = 1(2) − (−6)(3) 22 (5)Simplify. = 20 4 (6)Reduce the fraction. =5 Question #5 Find the 50th derivative of 𝑦 = cos 2 𝑥. Answer: Trigonometric derivatives cycle, as shown. 𝑦 ′ = −2 sin(2𝑥) 𝑦 ′′ = −4 cos(2𝑥) 𝑦 ′′′ = 8 sin(2𝑥) 𝑦 ′′′′ = 16 cos(2𝑥) So, for the 50th derivative, the co-efficient will be 250 . The remainder of 50 divided by 4 is 2, so we know it corresponds with y''. 50th derivative of 𝑦 = −250 cos(2𝑥) Quadratic Formula. 𝑟= − − 4 ± √(−4)2 − 4 ∗ 1 ∗ 1 2∗1 𝑟= 4 ± √16 − 4 2 𝑟 = 2 ± √12 𝑟 = 2 ± 2√3 Result: 50th derivative of 𝑦 = −250 cos(2𝑥) Question #6 The graphs of f and g are shown. Let ℎ(𝑥) = 𝑓(𝑔(𝑥)) and 𝑠(𝑥) = 𝑔(𝐽(𝑥)). Find each derivative , if it exists. If the derivative ve does not exist, explain why. Find h'( 1). Answer: To find the derivative of the function we will use Chain Rule: 𝑑 [𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥)) = 𝑔′ (𝑥)] 𝑑𝑥 ℎ′ (𝑥) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥) Plug in 𝑥 = 1 to find h'(1). ℎ′ (1) = 𝑓 ′ (𝑔(1))𝑔′ (1) = 𝑓 ′ (4)𝑔′ (1) 1 = (−1) ∗ (− ) 2 = 1 2 To estimate f'(4), there is a part of the picture where we can see that the tangent line goes down 1 unit as the variable x moves one unit to the right. This means that it has slope -1, thus 𝑓 ′ (4) = −1. To estimate g'(1), there is a part of the picture where we can see that the tangent line goes down 1 unit as the 1 1 variable x moves two units to the right. This means that it has slope − , thus 𝑔′ (1) = − . 2 2 Result: 𝑎= 1 2 Question #7 The graphs of f and g are shown. Let ℎ(𝑥) = 𝑓(𝑔(𝑥)) and 𝑠(𝑥) = 𝑔(𝐽(𝑥)). Find each derivative , if it exists. If the derivative ve does not exist, explain why. Find 𝑠 ′ (5). Answer: To find the derivative of the function we will use Chain Rule: 𝑑 [𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥)) = 𝑔′ (𝑥)] 𝑑𝑥 𝑠 ′ (𝑥) = 𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥) Plug in 𝑥 = 5 to find s'(5). 𝑠 ′ (5) = 𝑔′ (𝑓(5))𝑓 ′ (5) = 𝑔′ (6)𝑓 ′ (5) As wee see on graph of g(x) there is a sharp corner at 𝑥 = 6, so g(x) is not differentiable at 𝑥 = 6. s'(5) does not exist because g is not differetiable at 6. Result: s'(5) doesn't exist Question #8 What's the derivative of sec 2 (𝑥)? Answer: 𝑑 (sec 2 (𝑥)) = 2 sec(𝑥) × sec(𝑥) tan(𝑥) = 2 sec 2 (𝑥) tan(𝑥) 𝑑𝑥 Question #9 What is the derivative of x? Answer: Use the Power Rule first: 𝑓(𝑥) = 𝑥 = 𝑥 1 𝑓 ′ (𝑥) = 1𝑥 1−1 = 1𝑥 0 = 1 × 1 = 1 And now we use difference quotient: lim = ℎ⇒0 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑥 + ℎ − 𝑥 ℎ = = =1 ℎ ℎ ℎ Question #10 How to find the derivative of 1 𝑥2 ? Answer: Use the power rule: 1 = 𝑥 −2 𝑥2 Then, we have −2𝑥 −2−1 = −2𝑥 −3 = − 2 𝑥3 Question #11 What is the derivative of 2? Answer: The derivative is the measure of the rate of change of a function and 2 is a constant whose value never cahnges. It means the derivative of 2 is 0. Question #12 𝑥 What is the derivative of ? 2 Answer: The function can be expressed as The derivative of 1 2 𝑥 is Question #13 1 2 1 2 𝑥 What is the derivative of ln(2𝑥) ? Answer: Using the chain rule: ln(2𝑥) = 1 1 ×2= 2𝑥 𝑥 Question #14 How to find the derivative of a fraction? Answer: Use the quotient rule to differentiate algebraic fractions or any other function written as quotient or fraction of two functions or expressions. Question #15 Tell, what is the antiderivative of 𝑒 2𝑥 ? Answer: 𝑙 ∫ 𝑒 2𝑥 = 1 ∫ 2𝑒 2𝑥 𝑑𝑥 2 If 2𝑑𝑥 = 𝑑(2𝑥), 2 is the derivative of 2x And we have: 1 ∫ 𝑒 2𝑥 𝑑(2𝑥) 2 Let 𝑢 = 2𝑥 Thus, 1 2 1 1 ∫ 𝑒 𝑢 𝑑𝑢 = 2 𝑒 𝑢 = 2 𝑒 2𝑥 Question #16 Find the derivative of 𝑓(𝑥) = tan−1 (𝑥) . Answer: 𝑦 = arctan(𝑥) tan(𝑦) = 𝑥 sec 2 𝑦 𝑑𝑦 =1 𝑑𝑥 𝑑𝑦 1 = 𝑑𝑥 sec 2 𝑦 Since tan(𝑦) = 𝑥 1 and √12 + 𝑥 2 = √1 + 𝑥 2 , 2 sec 2 𝑦 = ( ⇒ √1 + 𝑥 2 ) = 1 + 𝑥2 1 𝑑𝑦 1 = 𝑑𝑥 1 + 𝑥 2 Question #17 Find the first derivative of 𝑒 𝑥 2 Answer: 𝑒𝑥 2 The derivative of 𝑦 = 𝑒 𝑓(𝑥) is 𝑑𝑦 𝑑𝑥 = 𝑓 ′ (𝑥)𝑒 𝑓(𝑥) 𝑓(𝑥) = 𝑥 2 in this case, and the derivative of 𝑥 2 is 2x ∴ 𝑓 ′ (𝑥) = 2𝑥 , and 𝑑𝑦 𝑑𝑥 = 2𝑥𝑒 𝑥 2 Question #18 Find the derivative of 𝑒 2𝑥 . Answer: 𝑑 2𝑥 (𝑒 ) 𝑑𝑥 Apply the Chain Rule: 𝑒 2𝑥 If 𝑑 (2𝑥) 𝑑𝑥 𝑑 𝑑𝑥 (2𝑥) = 2 , then we have = 𝑒 2𝑥 × 2 𝑑 2𝑥 (𝑒 ) = 𝑒 2𝑥 × 2 𝑑𝑥 Question #19 How to find the derivative of a function at a given point? Answer: Firstly find the derivative of your function 𝑓(𝑥) = 𝑥 3 𝑓 ′ (𝑥) = 3𝑥 2 And if you want to find the derivative of f(x) when 𝑥 = 5 , then you just have to substitute that value into f'(x) 𝑓 ′ (5) = 3(5)2 = 3 × 25 = 75 Question #20 Find the slope of the tan line to the graph of the function ℎ(𝑡) = 𝑡 2 + 3 at (−2; 7) Answer: ℎ(𝑡) = 𝑡 2 + 3 𝑦 = 𝑡2 + 3 𝑑𝑦 = 2𝑡 𝑑𝑥 Slope at the curve exactly at 𝑥 = −2 is 𝑑𝑦 = 2(−2) = −4 𝑑𝑥 THe tangent is passing through the point (-2;7) 𝑚 = −4 𝑥 = −2 𝑦=7 𝑚𝑥 + 𝑐 = 𝑦 (−4)(−2) + 𝑐 = 7 8+𝑐 =7 𝑐 = 7 − 8 = −1 𝑦 = −4𝑥 − 1 - the equation of the tangent