Questions and Answers #63 Derivatives

Questions and Answers Sheet 63 Derivatives Question #1 How do you find the instantaneous rate of change of 𝑓(𝑡) = (2𝑡 3 − 3𝑡 + 4) when 𝑡 = 2? Answer: The instantaneous rate of change can be found by 𝑓 ′ (2) = 21. By taking the derivative, 𝑓(𝑡) = 6𝑡 2 − 3 By pluging in 𝑡 = 2, 𝑓 ′ (2) = 6(2)2 − 3 = 21 Question #2 How do you find the instantaneous rate of change of 𝑓(𝑥) = 𝑥 2 −2 𝑥+4 at 𝑥 = −1 Answer: You evaluate the derivative of your function then substitute the value of x in it: Derivative (quotient Rule) 𝑓 ′ (𝑥) = = 2𝑥(𝑥 + 4) − (𝑥 2 − 2) (𝑥 + 4)2 2𝑥 2 + 8𝑥 − 𝑥 2 + 2 𝑥 2 + 8𝑥 + 2 = (𝑥 + 4)2 (𝑥 + 4)2 Substitute 𝑥 = −1 𝑓 ′ (−1) = 1−8+2 5 =− 2 (−1 + 4) 9 Question #3 How do you find the instantaneous rate of change of 𝑓(𝑥) = 4𝑥 2 + 4𝑥 − 3 when 𝑥 = 4? Answer: The instantaneous rate of change is equal to the slope of the tangent at the required point or, equivalently, the derivative of the function evaluated at that point. If 𝑓(𝑥) = 4𝑥 2 + 4𝑥 − 3 then the derivative is 𝑓 ′ (𝑥) = 8𝑥 + 4 Evaluating this derivative at 𝑥 = 4 gives the instantaneous rate of change 8(4) + 4 = 36 Question #4 2 How do you find the instantaneous rate of change of 𝑓(𝑥) = 𝑥 2 − + 4 at 𝑥 = −1? 𝑥 Answer: When you calculate a function's derivative, you obtain an other function representing the variations of the first function's curve's slope. A curve's slope is the instantaneous variation rate of the curve's function at a given point. Therefore, if you are looking for the instantaneous variation rate of a function at a given point, you should calculate this function's derivative at said point. In your case: 2 𝑓(𝑥) = 𝑥 2 − + 4 variation rate at 𝑥 = −1? 𝑥 Calculating the derivative: 𝑓 ′ (𝑥) 2 𝑑(𝑥 2 ) 𝑑 (𝑥 ) 𝑑4 = − + 𝑑𝑥 𝑑𝑥 𝑑𝑥 = 2𝑥 − (− 2 2 ) + 0 = 2𝑥 + 2 𝑥2 𝑥 Now, you just need to replace x in f'(x) with its given value, 𝑥 = −1 2 𝑓 ′ (−1) = 2(−1) + (−1)2 = −2 + 2 = 0 The derivative is null, therefore the instantaneous change rate is null and the function doesn't increase or decrease at this specific point. Question #5 How do you find the instantaneous rate of change of 𝑓(𝑥) = 𝑥 2 + 2𝑥 4 at 𝑥 = 1? Answer: Explanation: We know that, the Instantaneous Rate of Change of f at 𝑥 = 1 is given by f'(1) 𝑓(𝑥) = 𝑥 2 + 2𝑥 4 → 𝑓 ′ (𝑥) = 2𝑥 + 2(4𝑥 3 ) ∴ The redq. Rate = 𝑓 ′ (1) = 2 + 8 = 10 𝑢𝑛𝑖𝑡/𝑠𝑒𝑐. Question #6 How do you find the instantaneous rate of change of 𝑔(𝑡) = 3𝑡 2 + 6 at 𝑡 = 4? Answer: Explanation: Compute the first derivative 𝑔′ (𝑡) = 6𝑡 Evaluate it at 𝑡 = 4 𝑔′ (4) = 24 Question #7 How do you find the instantaneous rate of change of 𝑦 = 4𝑥 3 + 2𝑥 − 3 at 𝑥 = 2? Answer: The instantanous rate of change of the function f(x) in 𝑥 = 𝑎 is the slope of the tangent line to the graph of the function in that point: 𝑚 = 𝑦 ′ (𝑎). So: 𝑦 ′ = 12𝑥 2 + 2 and 𝑦 ′ (2) = 12 ∗ 4 + 2 = 50. The answer is: 50. Question #8 How do you find the instantaneous rate of change of 𝑦 = 𝑥 2 − 4𝑥 at 𝑥 = 1 by using the limit method? Answer: Explanation: The instantaneous rate of change of 𝑦 = 𝑓(𝑥) = 𝑥 2 − 4𝑥 at 𝑥 = 1 is the derivative f'(1). Using the definition of the derivative in terms of limits, we have: 𝑓 ′ (1) = lim ℎ→0 𝑓(1 + ℎ) − 𝑓(1) ℎ (1 + ℎ)2 − 4(1 + ℎ) − (1 − 4) ℎ→0 ℎ = lim 1 + 2ℎ + ℎ2 − 4 − 4ℎ + 3 ℎ→0 ℎ = lim ℎ(ℎ − 2) = lim (ℎ − 2) = 0 − 2 = −2 ℎ→0 ℎ→0 ℎ = lim Question #9 How do you find the instantaneous rate of change of revenue when 1000 units are produced if the revenue (in thousands of dollars) from producing x units of an item is 𝑅(𝑥) = 12𝑥 − .005𝑥 2 ? Answer: Just take the derivative of R and plug in 𝑥 = 1000. Doing so gives 𝑅′ (𝑥) = 12 − 0.01𝑥 so that 𝑅′ (1000) = 12 − 10 = 2 thousand dollars per unit produced. Question #10 How do you find the instantaneous rate of change of the function 𝑓(𝑥) = 𝑥 2 + 3𝑥 + 4 when 𝑥 = 2? Answer: Have you learned differentiation formulas yet, or do you need to do this using the definition? Using differentiation formulas, find f(x) and the evaluate at 𝑥 = 2 𝑓 ′ (𝑥) = 2𝑥 + 3, so the instantaneous rate of change when 𝑥 = 2 is 𝑓 ′ (2) = 2(2) = 3 = 7 Question #11 1 𝑧 𝑧 2 How do you find the instantaneous rate of change of w with respect to z for 𝑤 = + ? Answer: 𝑑𝑤 𝑑 1 𝑧 = ( + ) 𝑑𝑧 𝑑𝑧 𝑧 2 Initial set-up. 𝑑𝑤 𝑑 1 𝑑 𝑧 = ( )+ ( ) 𝑑𝑧 𝑑𝑧 𝑧 𝑑𝑧 2 The derivative of a sum is equal to the sum of the derivatives. 𝑑𝑤 𝑑 −1 1𝑑 (𝑧 ) + (𝑧) = 𝑑𝑧 𝑑𝑧 2 𝑑𝑧 A function 𝑓(𝑧) = 𝑑 𝑑𝑧 𝑐𝑓(𝑧) = 𝑐 𝑑 𝑑𝑧 𝑐 𝑧𝑛 with c constant can also be written as 𝑓(𝑧) = 𝑐𝑧 −𝑛 Second part: 𝑓(𝑧) if c is constant. 𝑑𝑤 1 = −1 ∗ 𝑧 −2 + ∗ 1 𝑑𝑧 2 Use of the power rule: 𝑑 𝑑𝑧 𝑧 𝑛 = 𝑛𝑧 𝑛−1 . Then 𝑑 𝑑𝑧 𝑧= 𝑑 𝑑𝑧 𝑧1 = 𝑧 0 = 1 𝑑𝑤 1 = −𝑧 −2 + 𝑑𝑧 2 Multiplicative identity postulate. 𝑑𝑤 1 1 =− 2+ 𝑑𝑧 𝑧 2 A function written as 𝑓(𝑧) = 𝑐𝑧 −𝑛 can also be written 𝑓(𝑧) = 𝑐 𝑧𝑛 Question #12 How do you find the instantaneous velocity at 𝑡 = 0 for the position function 𝑠(𝑡) = 6𝑡 2 + 8𝑡 ? Answer: The position is given by the function 𝑠(𝑡) = 6𝑡 2 + 8𝑡 The velocity is the rate of change of displacement over time, so it'll be the derivative of the position function, i.e. 𝑣(𝑡) = 𝑠 ′ (𝑡) = 12𝑡 + 8 So, at 𝑡 = 0, the instantaneous velocity will be: 𝑠 ′ (0) = 12 ∗ 0 + 8 =0+8 =8 Question #13 How do you find the instantaneous velocity at 𝑡 = 2 for the position function 𝑠(𝑡) = 𝑡 3 + 8𝑡 2 − 𝑡 ? Answer: Explanation: The instantaneous velocity is given by Since s(𝑡) = 𝑡 3 + 8𝑡 2 − 𝑡, 𝑑𝑠 At 𝑡 = 2, [ ] 𝑑𝑡 𝑡=2 𝑑𝑠 𝑑𝑡 𝑑𝑠 𝑑𝑡 . = 3𝑡 2 + 16𝑡 − 1. = 3 ∗ 22 + 16 ∗ 2 − 1 = 43. Question #14 How do you find the numbers b such that the average value of 𝑓(𝑥) = 7 + 10𝑥 − 9𝑥 2 on the interval [0, b] is equal to 8? Answer: The average value of 𝑓(𝑥) = 7 + 10𝑥 − 9𝑥 2 on the interval [0,b] is 1 7𝑏+5𝑏 2 −3𝑏 3 𝑏 𝑏 Which can be evaluated to, [7𝑥 + 5𝑥 2 − 3𝑥 3 ]𝑏0 = 𝑏 1 ∫ (7 𝑏−0 0 + 10𝑥 − 9𝑥 2 )𝑑𝑥 = 7 + 5𝑏 − 3𝑏 2 We want the average value to be 8, so we need to solve 7 + 5𝑏 − 3𝑏 2 = 8 . Use the formula or complete the square to get two solutions: 𝑏 = 5−√13 6 and 𝑏 = 5+√13 6 Since both are positive, the interval [0,b] exists for either of these. Question #15 5 How do you find the point on the graph of 𝑦 = 𝑥 2 where the curve has a slope ? 3 Answer: The curve will have an instantaneous rate of change of Let's start by finding the derivative. 𝑦 ′ = 2𝑥 5 = 2𝑥 3 5 𝑥=3 2 𝑥= 5 6 5 2 25 6 36 The corresponding y-coordinate is 𝑦 = ( ) = . 5 3 when the point 𝑥 = 𝑎 is put into the derivative. Answer: 5 25 The point ( , ) 6 36 Question #16 How do you find the slope of a demand curve? Answer: The slope of a demand curve can be found just like the slope of any other line. Remember, in order to find a slope, you must divide rise by run. In the case of a demand curve, this means dividing change in price by change in quantity demanded. Mathematically, this looks like 𝑃2 −𝑃1 𝑄2 −𝑄1 Note that in order to calculate this slope, you need two points that you know are on the demand curve. The ordered pairs of course being (𝑄1 , 𝑃1 ) and (𝑄2 , 𝑃2 ) . Question #17 How do you find the slope of a tangent line to the curve 𝑦 = 𝑥 − 𝑥 5 at the point (1, 0)? Answer: The slope of the tangent line to a curve in a point is the first derivative in that point, so: 𝑦 ′ = 1 − 5𝑥 4 and: 𝑚 = 𝑦 ′ (1) = 1 − 5 ∗ 1 = −4. Question #18 How do you find the slope of a tangent line to the graph of the function at a given point 𝑓(𝑥) = 3 − 2𝑥 at (−1,5)? Answer: The given function 𝑓(𝑥) = 3 − 2𝑥 is the function of a straight line so its "tangent" is the line itself (at all points on the line) The function definition could be re-written in "slope-intercept" form 𝑦 = −2𝑥 + 3 where the slope is (-2) (and the Y-intercept is 3) Question #19 How do you find the slope of a tangent line to the graph of the function 𝑓(𝑥) = 3 − 5𝑥 at (-1, 8)? Answer: Given: 𝑓(𝑥) = 3 − 5𝑥 at (−1,8) Usually we find the slope by finding the first derivative and evaluating it at the x value of the point: f'(-1) In the given case, we have a linear equation: 𝑦 = 𝑚𝑥 + 𝑏 The slope of the line is given as 𝑚 = −5. This slope is the same everywhere along the line. If you find 𝑓 ′ (𝑥) = −5. 𝑓 ′ (−1) = −5 since it is a constant slope. Question #20 How do you find the slope of a tangent line to the graph of the function √𝑥 at (4,2)? Answer: Given a function f(x), the slope of the tangent line to the graph of f at a point c is given by f'(c). 𝑓(𝑥) = √𝑥 Slope at 𝑥 = 4 is equal to 𝑓 ′ (4) = 1 2√4 = 1 4 Answer: The slope of the tangent line to the graph of √𝑥 at (4,2) is 1 4 .