# Calculus: Single Variable Part 2 – Differentiation Coursera Quiz Answers

## Get All Weeks Calculus: Single Variable Part 2 – Differentiation Coursera Quiz Answers

Calculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with an emphasis on conceptual understanding and applications.

The course is ideal for students beginning in the engineering, physical, and social sciences. Distinguishing features of the course include 1) the introduction and use of the Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus; 3) an emphasis on the conceptual over the computational; and 4) a clear, dynamic, unified approach.

In this second part–part two of five–we cover derivatives, differentiation rules, linearization, higher derivatives, optimization, differentials, and differentiation operators.

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### Week 01: Calculus: Single Variable Part 2 – Differentiation Coursera Quiz Answers

#### Main Quiz 01

Q1. Let f(x) = -x^2+6x-3f(x)=−x2+6x−3. Find f'(-2)f′(−2).

• -3−3
• -1−1
• 1010
• 66
• 44
• 00

Q2. Given the position function \displaystyle p(t) = 6 + \frac{1}{2}t +4t^2p(t)=6+21​t+4t2 of a particle as a function of time, what is the particle’s velocity at t=1t=1 ?

• \displaystyle \frac{1}{2}21​
• 66
• 11
• \displaystyle \frac{9}{2}29​
• 88
• \displaystyle \frac{17}{2}217​

Q3. A very rough model of population size PP for an ant species is P(t) = 2\ln(t+2)P(t)=2ln(t+2), where tt is time. What is the rate of change of the population at time t = 2t=2?

• 22
• \displaystyle \frac{1}{4}41​
• 11
• \displaystyle \frac{1}{3}31​
• 44
• \displaystyle \frac{1}{2}21​

Q4. Find \displaystyle \frac{dV}{dt}dtdV​ for \displaystyle V=\frac{1}{4}t^3V=41​t3

• \displaystyle \frac{dV}{dt} = \frac{3}{4}t^3dtdV​=43​t3
• \displaystyle \frac{dV}{dt} = 3t^2dtdV​=3t2
• \displaystyle \frac{dV}{dt} = \frac{1}{3}t^2dtdV​=31​t2
• \displaystyle \frac{dV}{dt} = \frac{3}{4}t^2dtdV​=43​t2
• \displaystyle \frac{dV}{dt} = 0dtdV​=0
• \displaystyle \frac{dV}{dt} = \frac{1}{4}t^2dtdV​=41​t2

Q5. If a car’s position is represented by s(t) = 4t^3s(t)=4t3, what is the car’s change in velocity from t=2t=2 to t=3t=3 ?

• 4848
• 1212
• 6060
• 2020
• 7676
• 108108

Q6. A particle’s position, pp, as a function of time, tt, is represented by \displaystyle p(t) = \frac{1}{3}t^3 – 3t^2 + 9tp(t)=31​t3−3t2+9t. When is the particle at rest?

• Never.
• At t=1t=1.
• At t=3t=3.
• At t = 6t=6.
• At t=0t=0.
• At \displaystyle t = \frac{1}{3}t=31​.

Q7. A rock is dropped from the top of a 320-foot building. The height of the rock at time tt is given s(t)=-8t^2+320s(t)=−8t2+320, where tt is measured in seconds. Find the speed (that is, the absolute value of the velocity) of the rock when it hits the ground in feet per second. Round your answer to one decimal place.

Hooke’s law states that the force FF exerted by an ideal spring displaced a distance xx from its equilibrium point is given by F(x) = -kxF(x)=−kx, where the constant kk is called the spring constant and varies from one spring to another. In real life, many springs are nearly ideal for small displacements; however, for large displacements, they might deviate from what Hooke’s law predicts.

Much of the confusion between nearly-ideal and non-ideal springs is clarified by thinking in terms of series: for xx near zero, F(x) = -kx + O(x^2)F(x)=−kx+O(x2).

Suppose you have a spring whose force follows the equation F(x) = – 2 \tan 3xF(x)=−2tan3x. What is its spring constant?

• 00
• 33
• 1212
• 11
• 22
• 66

#### Practice Quiz 01

Q1. The profit, PP, of a company that manufactures and sells NN units of a certain product is modeled by the function

P(N) = R(N) – C(N)P(N)=R(N)−C(N)

The revenue function, R(N)=S\cdot NR(N)=SN, is the selling price SS per unit times the number NN of units sold. The company’s cost, C(N)=C_0+C_\mathrm{op}(N)C(N)=C0​+Cop​(N), is a sum of two terms. The first is a constant C_0C0​ describing the initial investment needed to set up production. The other term, C_\mathrm{op}(N)Cop​(N), varies depending on how many units the company produces, and represents the operating costs.

Companies care not only about profit, but also marginal profit, the rate of change of profit with respect to NN.

Assume that S = \$50S=$50, C_0 = \$75,000C0​=$75,000, C_\mathrm{op}(N) = \$50 \sqrt{N}Cop​(N)=$50N​, and that the company currently sells N=100N=100 units. Compute the marginal profit at this rate of production. Round your answer to one decimal place.

Q2. In Economics, physical capital represents the buildings or machines used by a business to produce a product. The marginal product of physical capital represents the rate of change of output product with respect to physical capital (informally, if you increase the size of your factory a little, how much more product can you create?).

A particular model tells us that the output product YY is given, as a function of capital KK, by

Y = A K^{\alpha} L^{1-\alpha}Y=AKαL1−α

where AA is a constant, LL is units of labor (assumed to be constant), and \alphaα is a constant between 0 and 1. Determine the marginal product of physical capital predicted by this model.

• \displaystyle \frac{dY}{dK} = \alpha A \frac{L^{1-\alpha}}{K^{\alpha – 1}}dKdY​=αAKα−1L1−α
• \displaystyle \frac{dY}{dK} = (\alpha – 1)A \big( \frac{L}{K} \big)^{\alpha – 1}dKdY​=(α−1)A(KL​)α−1
• \displaystyle \frac{dY}{dK} = \alpha A K^{\alpha}L^{1-\alpha}dKdY​=αAKαL1−α
• \displaystyle \frac{dY}{dK} = \frac{A}{\alpha} \big( \frac{L}{K} \big)^{1-\alpha}dKdY​=αA​(KL​)1−α
• None of these.
• \displaystyle \frac{dY}{dK} = (1 – \alpha) A (KL)^{1-\alpha}dKdY​=(1−α)A(KL)1−α

#### Main Quiz 02

Q1. Find the derivative of f(x)= \sqrt{x}(2x^2-4x)f(x)=x​(2x2−4x).

• f'(x) = \sqrt{x}(5x^2-6x)f′(x)=x​(5x2−6x)
• \displaystyle f'(x) = \frac{2x-2}{\sqrt{x}}f′(x)=x​2x−2​
• f'(x) = 2x^{5/2}-4x^{3/2}f′(x)=2x5/2−4x3/2
• f'(x) = \sqrt{x}(5x-6)f′(x)=x​(5x−6)
• f'(x) = 2x^{3/2}-4x^{1/2}f′(x)=2x3/2−4x1/2
• f'(x) = 4\sqrt{x}(x-1)f′(x)=4x​(x−1)

Q2. Find the derivative of \displaystyle f(x) = 6x^4 -\frac{3}{x^2}-2\pif(x)=6x4−x23​−2π.

• \displaystyle f'(x) = 24x^3 – \frac{3}{x^2}f′(x)=24x3−x23​
• \displaystyle f'(x) = 24x^3-\frac{6}{x^2}f′(x)=24x3−x26​
• \displaystyle f'(x) = 24x^3-\frac{6}{x^3}f′(x)=24x3−x36​
• \displaystyle f'(x) = 24x^3 + \frac{3}{x^2}f′(x)=24x3+x23​
• \displaystyle f'(x) = 24x^3 – \frac{3}{2x}f′(x)=24x3−2x3​
• \displaystyle f'(x) = 24x^3+\frac{6}{x^3}f′(x)=24x3+x36​

Q3. Find the derivative of f(x) = 7(x^3+4x)^5 \cos xf(x)=7(x3+4x)5cosx.

• f'(x) = -45(x^3 + 4x)^4(3x^2 + 4)\sin xf′(x)=−45(x3+4x)4(3x2+4)sinx
• f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x – (x^3+4x)\sin x \big]f′(x)=7(x3+4x)4[5(3x2+4)cosx−(x3+4x)sinx]
• f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x – (x^3+4x)\sin x \big]f′(x)=7(x3+4x)4[5cosx−(x3+4x)sinx]
• f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x + (x^3+4x)\sin x \big]f′(x)=7(x3+4x)4[5(3x2+4)cosx+(x3+4x)sinx]
• f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x + (x^3+4x)\sin x \big]f′(x)=7(x3+4x)4[5cosx+(x3+4x)sinx]
• f'(x) = 45(x^3 + 4x)^4(3x^2 + 4)\cos xf′(x)=45(x3+4x)4(3x2+4)cosx

Q4. Find the derivative of f(x) = (e^x + \ln x)\sin xf(x)=(ex+lnx)sinx.

• f'(x) = (\sin x )(\ln x) + e^x\cos xf′(x)=(sinx)(lnx)+excosx
• \displaystyle f'(x) = \frac{\sin x}{x} + e^x\sin xf′(x)=xsinx​+exsinx
• \displaystyle f'(x) = \frac{\sin x}{x} + (\ln x)(\cos x) + e^x\sin x + e^x\cos xf′(x)=xsinx​+(lnx)(cosx)+exsinx+excosx
• \displaystyle f'(x) = \left( \frac{1}{x}+e^x \right) \cos xf′(x)=(x1​+ex)cosx
• \displaystyle f'(x) = \frac{e^x\sin x}{x}f′(x)=xexsinx
• f'(x) = e^x(\sin x + \cos x)f′(x)=ex(sinx+cosx)

Q5. Find the derivative of \displaystyle f(x) = \frac{\sqrt{x+3}}{x^2}f(x)=x2x+3​​.

• \displaystyle f'(x) = -\frac{3x+12}{2x \sqrt{x+3}}f′(x)=−2xx+3​3x+12​
• \displaystyle f'(x) = -\frac{3x+12}{2x^3 \sqrt{x+3}}f′(x)=−2x3x+3​3x+12​
• \displaystyle f'(x) = \frac{5x+12}{2x^3 \sqrt{x+3}}f′(x)=2x3x+3​5x+12​
• \displaystyle f'(x) = \frac{(x-4)\sqrt{x+3}}{2x^3}f′(x)=2x3(x−4)x+3​​
• \displaystyle f'(x) = \frac{5x+12}{2x \sqrt{x+3}}f′(x)=2xx+3​5x+12​
• \displaystyle f'(x) = \frac{1}{4x\sqrt{x+3}}f′(x)=4xx+3​1​

Q6. Find the derivative of \displaystyle f(x) = \frac{\ln x}{\cos x}f(x)=cosxlnx​.

• \displaystyle f'(x) = \frac{\cos x – \ln x\sin x}{x\sin^2 x}f′(x)=xsin2xcosx−lnxsinx
• \displaystyle f'(x) = \frac{\ln x\sin x}{x\cos^2 x}f′(x)=xcos2xlnxsinx
• \displaystyle f'(x) = \frac{\cos x + \ln x\sin x}{x\cos x}f′(x)=xcosxcosx+lnxsinx
• \displaystyle f'(x) = \frac{\cos x + x\ln x\sin x}{x\cos^2 x}f′(x)=xcos2xcosx+xlnxsinx
• \displaystyle f'(x) = \frac{\cos x – \ln x\sin x}{x\cos^2 x}f′(x)=xcos2xcosx−lnxsinx
• \displaystyle f'(x) = \frac{(1 + \ln x)\sin x}{x\cos^2 x}f′(x)=xcos2x(1+lnx)sinx

Q7. Find the derivative of \displaystyle f(x) = \frac{ \sqrt{x} – 4}{x^3}f(x)=x33x​−4​.

• \displaystyle f'(x) = \frac{10\sqrt{x} – 36}{3x^4}f′(x)=3x4103x​−36​
• \displaystyle f'(x) = \frac{10\sqrt{x} – 36}{x^4}f′(x)=x4103x​−36​
• \displaystyle f'(x) = \frac{12 – 2\sqrt{x}}{3x^4}f′(x)=3x412−23x​​
• \displaystyle f'(x) = \frac{36 – 8\sqrt{x}}{3x^4}f′(x)=3x436−83x​​
• \displaystyle f'(x) = \frac{36 – 8\sqrt{x}}{x^4}f′(x)=x436−83x​​
• \displaystyle f'(x) = \frac{12 – 2\sqrt{x}}{x^4}f′(x)=x412−23x​​

Q8. Find the derivative of f(x)=\sin^3 (x^3)f(x)=sin3(x3).

• f'(x) = 9x^2 \sin^3(x^3) \cos^2 (x^3)f′(x)=9x2sin3(x3)cos2(x3)
• f'(x) = 3 \sin^2(x^3) \cos(x^3)f′(x)=3sin2(x3)cos(x3)
• f'(x) = 9x^2 \sin^2 (x^3) \cos (x^3)f′(x)=9x2sin2(x3)cos(x3)
• f'(x) = 3\sin^2 (x^3)f′(x)=3sin2(x3)
• f'(x) = 9x^2 \sin^2 (x^2) \cos (3x^2)f′(x)=9x2sin2(x2)cos(3x2)
• f'(x) = 3\sin^2(3x^2)f′(x)=3sin2(3x2)

Q9. Find the derivative of f(x) = e^{-1/x^2}f(x)=e−1/x2.

• \displaystyle f'(x) = \frac{2}{x^3} e^{-1/x^2}f′(x)=x32​e−1/x2
• \displaystyle f'(x) = e^{2/x^3}f′(x)=e2/x3
• \displaystyle f'(x) = e^{-2/x^3}f′(x)=e−2/x3
• \displaystyle f'(x) = -\frac{1}{x^2} e^{-1/x^2}f′(x)=−x21​e−1/x2
• \displaystyle f'(x) = \frac{1}{x^2} e^{-1/x^2}f′(x)=x21​e−1/x2
• \displaystyle f'(x) = -\frac{2}{x^3} e^{-1/x^2}f′(x)=−x32​e−1/x2

### Week 02: Calculus: Single Variable Part 2 – Differentiation Coursera Quiz Answers

#### Main Quiz 01

Q1. Use a linear approximation to estimate \sqrt{67}367​. Round your answer to four decimal places.

Hint: remember that \sqrt{64} = 4364​=4. You can check the accuracy of this approximation by noting that \sqrt{67} \approx 4.0615367​≈4.0615.

Q2. Use a linear approximation to estimate the cosine of an angle of 66^\mathrm{o}66o. Round your answer to four decimal places.

Hint: remember that \displaystyle 60^\mathrm{o} = \frac{\pi}{3}60o=3π​, and hence \displaystyle 6^\mathrm{o} = \frac{\pi}{30}6o=30π​. You can check the accuracy of this approximation by noting that \cos 66^\mathrm{o} \approx 0.4067cos66o≈0.4067.

Q3. The golden ratio \displaystyle \varphi = \frac{1+\sqrt{5}}{2}φ=21+5​​ is a root of the polynomial x^2-x-1x2−x−1. If you use Newton’s method to estimate its value, what is the appropriate update rule for the sequence x_nxn​ ?

• \displaystyle x_{n+1} = x_n + \frac{2x_n – 1}{x_n^2 – x_n – 1}xn+1​=xn​+xn2​−xn​−12xn​−1​
• \displaystyle x_{n+1} = x_n – \frac{x_n^2 – x_n – 1}{2x_n – 1}xn+1​=xn​−2xn​−1xn2​−xn​−1​
• \displaystyle x_{n+1} = x_n – \frac{2x_n – 1}{x_n^2 – x_n – 1}xn+1​=xn​−xn2​−xn​−12xn​−1​
• \displaystyle x_{n+1} = \frac{x_n^2 – x_n – 1}{2x_n – 1}xn+1​=2xn​−1xn2​−xn​−1​
• \displaystyle x_{n+1} = x_n + \frac{x_n^2 – x_n – 1}{2x_n – 1}xn+1​=xn​+2xn​−1xn2​−xn​−1​
• \displaystyle x_{n+1} = \frac{2x_n – 1}{x_n^2 – x_n – 1}xn+1​=xn2​−xn​−12xn​−1​

Q4. To approximate \sqrt{10}10​ using Newton’s method, what is the appropriate update rule for the sequence x_nxn​ ?

• \displaystyle x_{n+1} = \frac{x_n}{2} + \frac{5}{x_n}xn+1​=2xn​​+xn​5​
• \displaystyle x_{n+1} = \frac{x_n}{2}xn+1​=2xn​​
• \displaystyle x_{n+1} = x_n + \frac{2x_n}{x_n^2 – 10}xn+1​=xn​+xn2​−102xn​​
• \displaystyle x_{n+1} = \frac{x_n}{2} – \frac{10}{x_n}xn+1​=2xn​​−xn​10​
• \displaystyle x_{n+1} = \frac{x_n}{2} – \frac{5}{x_n}xn+1​=2xn​​−xn​5​
• \displaystyle x_{n+1} = x_n – \frac{2x_n}{x_n^2 – 10}xn+1​=xn​−xn2​−102xn​​

Q5. You want to build a square pen for your new chickens, with an area of 1200\,\mathrm{ft}^21200ft2. Not having a calculator handy, you decide to use Newton’s method to approximate the length of one side of the fence. If your first guess is 30\,\mathrm{ft}30ft, what is the next approximation you will get?

• 3535
• 15.0515.05
• 4040
• -5−5
• 3030
• 30.0530.05

Q6. You are in charge of designing packaging materials for your company’s new product. The marketing department tells you that you must put them in a cube-shaped box. The engineering department says that you will need a box with a volume of 500\,\mathrm{cm}^3500cm3. What are the dimensions of the cubical box? Starting with a guess of 8\,\mathrm{cm}8cm for the length of the side of the cube, what approximation does one iteration of Newton’s method give you? Round your answer to two decimal places.

#### Practice Quiz 01

Q1. Without using a calculator, approximate 9.98^{98}9.9898. Here are some hints. First, 9.989.98 is close to 1010, and 10^{98}=1\,{\rm E}\,981098=1E98 in scientific notation. What does linear approximation give as an estimate when we decrease from 10^{98}1098 to 9.98^{98}9.9898?

• 1.000\,{\rm E}\,981.000E98
• 0.902\,{\rm E}\,980.902E98
• 1.960\,{\rm E}\,981.960E98
• 0.804\,{\rm E}\,980.804E98
• 0.9804\,{\rm E}\,980.9804E98
• 0.822\,{\rm E}\,980.822E98

Q2. A diving-board of length LL bends under the weight of a diver standing on its edge. The free end of the board moves down a distance

D = \frac{P}{3EI} L^3D=3EIPL3

where PP is the weight of the diver, EE is a constant of elasticity —that depends on the material from which the board is manufactured— and II is a moment of inertia. (These last two quantities will again make an appearance in Lectures 13 and 41, but do not worry about what exactly they mean now…)

Suppose our board has a length L = 2\,\mathrm{m}L=2m, and that it takes a deflection of D = 20\,\mathrm{cm}D=20cm under the weight of the diver. Use a linear approximation to estimate the deflection that it would take if its length was increased by 20\,\mathrm{cm}20cm

• 20.3\,\mathrm{cm}20.3cm
• 25.7\,\mathrm{cm}25.7cm
• 22\,\mathrm{cm}22cm
• 26\,\mathrm{cm}26cm
• 24.8\,\mathrm{cm}24.8cm
• 26.6\,\mathrm{cm}26.6cm

### Main Quiz 02

Q1. You are given the position, velocity and acceleration of a particle at time t = 0t=0. The position is p(0) = 2p(0)=2, the velocity v(0) = 4v(0)=4, and the acceleration a(0) = 3a(0)=3. Using this information, which Taylor series should they use to approximate p(t)p(t), and what is the estimated value of p(4)p(4) using this approximation?

• p(t) = 2 + 2t + 3 t^2 + O(t^3)p(t)=2+2t+3t2+O(t3), p(4) \simeq 58p(4)≃58.
• p(t) = 2 + 4t + 6 t^2 + O(t^3)p(t)=2+4t+6t2+O(t3), p(4) \simeq 114p(4)≃114.
• p(t) = 2 + 4t + 3 t^2 + O(t^3)p(t)=2+4t+3t2+O(t3), p(4) \simeq 66p(4)≃66.
• \displaystyle p(t) = 2 + 2t + \frac{3}{2} t^2 + O(t^3)p(t)=2+2t+23​t2+O(t3), p(4) \simeq 34p(4)≃34.
• \displaystyle p(t) = 2 + 4t + \frac{3}{2} t^2 + O(t^3)p(t)=2+4t+23​t2+O(t3), p(4) \simeq 42p(4)≃42.
• p(t) = 2 + 2t + 6 t^2 + O(t^3)p(t)=2+2t+6t2+O(t3), p(4) \simeq 106p(4)≃106.

Q2. If a particle moves according to the position function s(t) = t^3-6ts(t)=t3−6t, what are its position, velocity and acceleration at t=3t=3 ?

• s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 18a(3)=18
• s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 36a(3)=36
• s(3) = 21s(3)=21, v(3) = 18v(3)=18, a(3) = 6a(3)=6
• s(3) = 9s(3)=9, v(3) = 18v(3)=18, a(3) = 18a(3)=18
• s(3) = 9s(3)=9, v(3) = 21v(3)=21, a(3) = 9a(3)=9
• s(3) = 21s(3)=21, v(3) = 18v(3)=18, a(3) = 18a(3)=18

Q3. If the position of a car at time tt is given by the formula p(t) = t^4 – 24t^2p(t)=t4−24t2, for which times tt is its velocity decreasing?

• Never: the velocity always increases.
• -\sqrt{12} < t < \sqrt{12}−312​<t<312​
• t < -2t<−2
• -2 < t < 2−2<t<2
• -\sqrt{24} < t < \sqrt{24}−24​<t<24​
• t > 2t>2

Q4. What is a formula for the second derivative of f(t) = t^2\sin 2tf(t)=t2sin2t? Use this formula to compute f”(\pi/2)f′′(π/2).

• f”(t) = 4t\cos 2t + (2-4t^2)\sin 2tf′′(t)=4tcos2t+(2−4t2)sin2t, and f”(\pi/2) = -2\pif′′(π/2)=−2π
• f”(t) = -4t^2\sin 2tf′′(t)=−4t2sin2t, and f”(\pi/2) =0f′′(π/2)=0
• f”(t) = -8\sin 2tf′′(t)=−8sin2t, and f”(\pi/2) = 0f′′(π/2)=0
• f”(t) = 8t\cos 2t -4t^2\sin 2tf′′(t)=8tcos2t−4t2sin2t, and f”(\pi/2) = -4\pif′′(π/2)=−4π
• f”(t) = 4t\cos 2tf′′(t)=4tcos2t, and f”(\pi/2) = -2\pif′′(π/2)=−2π
• f”(t) = 8t\cos 2t + (2-4t^2)\sin 2tf′′(t)=8tcos2t+(2−4t2)sin2t, and f”(\pi/2) = -4\pif′′(π/2)=−4π

Q5. Use a Taylor series expansion to compute f^{(3)}(0)f(3)(0) for f(x) = \sin^3 \left(\ln(1+x) \right)f(x)=sin3(ln(1+x)).

• -3−3
• 66
• 1212
• 33
• 00
• -6−6

Q6. What is the curvature of the graph of the function f(x) = -2\sin(x^2)f(x)=−2sin(x2) at the point (0,0)(0,0)?

• 00
• 22
• \displaystyle \frac{1}{2}21​
• 11
• 44
• -4−4

#### Main Quiz 03

Q1. Find all the local maxima and minima of the function y=x e^{-x^2}y=xex2.

• The function has local minima at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​ and \displaystyle x = -\frac{\sqrt{2}}{2}x=−22​​, and a local maximum at x = 0x=0.
• The function has a local maximum at \displaystyle x = -\frac{\sqrt{2}}{2}x=−22​​, and a local minimum at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​.
• The function has local minima at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​ and \displaystyle x = -\frac{\sqrt{2}}{2}x=−22​​, but no local maxima.
• The function has a local maximum at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​, and a local minimum at \displaystyle x = – \frac{\sqrt{2}}{2}x=−22​​.
• The function has local maxima at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​ and \displaystyle x = -\frac{\sqrt{2}}{2}x=−22​​, and a local minimum at x = 0x=0.
• The function has local maxima at \displaystyle x = \frac{\sqrt{2}}{2}x=22​​ and \displaystyle x = -\frac{\sqrt{2}}{2}x=−22​​, but no local minima.

Q2. Which of the following statements is true about the function f(x) = e^{\sin(x^4)}\cos(x^2)f(x)=esin(x4)cos(x2) ?

• Its Taylor series expansion about x=0x=0 is \displaystyle 1 – \frac{x}{3} + O(x^2)1−3x​+O(x2). Hence x=0x=0 is not a critical point of f(x)f(x).
• Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^3}{2} + O(x^4)1+2x3​+O(x4). Hence x=0x=0 is a critical point of f(x)f(x) that is neither a local maximum nor a local minimum.
• Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2x4​+O(x5). Hence it has a local minimum at x=0x=0.
• Its Taylor series expansion about x=0x=0 is \displaystyle 1 – \frac{x^2}{2} + O(x^5)1−2x2​+O(x5). Hence it has a local minimum at x=0x=0.
• Its Taylor series expansion about x=0x=0 is \displaystyle 1 – \frac{x^2}{2} + O(x^5)1−2x2​+O(x5). Hence it has a local maximum at x=0x=0.
• Its Taylor series expansion about x=0x=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2x4​+O(x5). Hence it has a local maximum at x=0x=0.

Q3. Use a Taylor series about x=0x=0 to determine whether the function f(x) = \sin^3(x^3)f(x)=sin3(x3) has a local maximum or local minimum at the origin.

• x=0x=0 is a critical point of ff, but it is neither a local maximum nor a local minimum.
• x=0x=0 is not a critical point of ff.
• x=0x=0 is a local minimum of ff.
• x=0x=0 is a local maximum of ff.

Q4. Find the location of the global maximum and minimum of f(x) = x^3-6x^2+1f(x)=x3−6x2+1 on the interval [-1,7][−1,7].

• The global maximum is attained at x = 0x=0 and the global minimum at x = -1x=−1.
• The global maximum is attained at x = 0x=0 and the global minimum at x = 4x=4.
• The global maximum is attained at x = 7x=7, but there is no global minimum.
• The global maximum is attained at x = 7x=7 and the global minimum at x = 4x=4.
• The global maximum is attained at x = 7x=7 and the global minimum at x = -1x=−1.
• The global maximum is attained at x = 0x=0, but there is no global minimum.

Q5. Which of the following statements are true for the function \displaystyle f(x) = x^3 + \frac{48}{x^2}f(x)=x3+x248​ ? Select all that apply.

• x=-2x=−2 is the global maximum of ff in [-3, -1][−3,−1]
• x=-1x=−1 is the global maximum of ff in [-3, -1][−3,−1]
• x=2x=2 is the global maximum of ff in [-3, 3][−3,3]
• x=1x=1 is the global minimum of ff in [1, 3][1,3]
• x=2x=2 is the global minimum of ff in [1, 3][1,3]
• x=1x=1 is the global maximum of ff in [1, 3][1,3]

### Week 03: Calculus: Single Variable Part 2 – Differentiation Coursera Quiz Answers

#### Main Quiz 01

Q1. Use implicit differentiation to find \displaystyle \frac{dy}{dx}dxdy​ from the equation y^2 – y = \sin 2xy2−y=sin2x.

• \displaystyle \frac{dy}{dx} = \frac{y^2 – y}{2\cos 2x}dxdy​=2cos2xy2−y
• \displaystyle \frac{dy}{dx} = \frac{\sin 2x}{2y – 1}dxdy​=2y−1sin2x
• \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{2y – 1}dxdy​=2y−12cos2x
• \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{y^2 – y}dxdy​=y2−y2cos2x
• \displaystyle \frac{dy}{dx} = \frac{2y – 1}{\sin 2x}dxdy​=sin2x2y−1​
• \displaystyle \frac{dy}{dx} = \frac{2y – 1}{2\cos 2x}dxdy​=2cos2x2y−1​

Q2. Find the derivative \displaystyle \frac{dy}{dx}dxdy​ if xx and yy are related through xy = e^yxy=ey.

• \displaystyle \frac{dy}{dx} = \frac{e^y + x}{y}dxdy​=yey+x
• \displaystyle \frac{dy}{dx} = \frac{x – e^y}{y}dxdy​=yxey
• \displaystyle \frac{dy}{dx} = \frac{y}{e^y + x}dxdy​=ey+xy
• \displaystyle \frac{dy}{dx} = \frac{y}{x – e^y}dxdy​=xeyy
• \displaystyle \frac{dy}{dx} = \frac{y}{e^y – x}dxdy​=eyxy
• \displaystyle \frac{dy}{dx} = \frac{e^y – x}{y}dxdy​=yeyx

Q3. Use implicit differentiation to find \displaystyle \frac{dy}{dx}dxdy​ if \sin x = e^{-y\cos x}sinx=eycosx.

• \displaystyle \frac{dy}{dx} = y\cos x – e^{y\cos x}\sin xdxdy​=ycosxeycosxsinx
• \displaystyle \frac{dy}{dx} = \frac{y\sin x – e^{-y\cos x}}{\cos x}dxdy​=cosxysinxeycosx
• \displaystyle \frac{dy}{dx} = y\tan x – e^{y\cos x}dxdy​=ytanxeycosx
• \displaystyle \frac{dy}{dx} = -y\sin x + e^{-y\cos x}\cos xdxdy​=−ysinx+eycosxcosx
• \displaystyle \frac{dy}{dx} = e^{-y\cos x}(\cos x – y\sin x)dxdy​=eycosx(cosxysinx)
• \displaystyle \frac{dy}{dx} = \frac{y – e^{y\cos x}\tan x}{\sin x}dxdy​=sinxyeycosxtanx

Q4. Find the derivative \displaystyle \frac{dy}{dx}dxdy​ from the equation x\tan y – y^2\ln x = 4xtanyy2lnx=4.

• \displaystyle \frac{dy}{dx} = \frac{-y^2}{x^2\sec^2 y}dxdy​=x2sec2yy2​
• \displaystyle \frac{dy}{dx} = \tan y – \frac{y^2}{\sec^2 y}dxdy​=tany−sec2yy2​
• \displaystyle \frac{dy}{dx} = \frac{x\tan y – y^2}{2xy\ln x – x^2\sec^2 y}dxdy​=2xylnxx2sec2yxtanyy2​
• \displaystyle \frac{dy}{dx} = \frac{2xy\ln x – x^2\sec^2 y}{x\tan y – y^2}dxdy​=xtanyy22xylnxx2sec2y
• \displaystyle \frac{dy}{dx} = \frac{y^2 – \tan y}{x^2\sec^2 y – 2xy\ln x}dxdy​=x2sec2y−2xylnxy2−tany
• \displaystyle \frac{dy}{dx} = \frac{x\tan y}{2xy\ln x}dxdy​=2xylnxxtany

Q5. Model a hailstone as a round ball of radius RR. As the hailstone falls from the sky, its radius increases at a constant rate CC. At what rate does the volume VV of the hailstone change?

• \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C R^3dtdV​=34​πCR3
• \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C^3dtdV​=34​πC3
• \displaystyle \frac{dV}{dt} = 8\pi C RdtdV​=8πCR
• \displaystyle \frac{dV}{dt} = 4\pi C R^2dtdV​=4πCR2
• \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi R^3dtdV​=34​πR3
• \displaystyle \frac{dV}{dt} = 4\pi R^2dtdV​=4πR2

Q6. The volume of a cubic box of side-length LL is V = L^3V=L3. How are the relative rates of change of LL and VV related?

• \displaystyle \frac{dL}{L} = \frac{dV}{V}LdL​=VdV
• \displaystyle \frac{dV}{V} = 3 L^3 \frac{dL}{L}VdV​=3L3LdL
• \displaystyle \frac{dL}{L} = 3 \frac{dV}{V}LdL​=3VdV
• \displaystyle \frac{dV}{V} = -\frac{dL}{L}VdV​=−LdL
• \displaystyle \frac{dV}{V} = 0VdV​=0
• \displaystyle \frac{dV}{V} = 3 \frac{dL}{L}VdV​=3LdL

#### Practice Quiz 01

Q1. Consider a box of height hh with a square base of side length LL. Assume that LL is increasing at a rate of 10\%10% per day, but hh is decreasing at a rate of 10\%10% per day. Use a linear approximation to find at what (approximate) rate the volume of the box changing?

Hint: consider the relative rate of change of the volume of the box.

Hint^\mathbf{2}2: in this case you can very easily calculate the exact rate of change —8.9%—, so using linearization might seem like overkill. However, if you set up things right, you don’t even need a calculator to find out the approximate rate of change! Do you see why?

• Increasing at a rate of 5\%5% per day.
• Increasing at a rate of 10\%10% per day.
• Decreasing at a rate of 10\%10% per day.
• Increasing at a rate of 2.5\%2.5% per day.
• It does not change.
• Decreasing at a rate of 5\%5% per day.

Q2. A large tank of oil is slowly leaking oil into a containment tank surrounding it. The oil tank is a vertical cylinder with a diameter of 10 meters. The containment tank has a square base with side length of 15 meters and tall vertical walls. The bottom of the oil tank and the bottom of the containment tank are concentric (the round base inside the square base). Denote by h_oho​ the height of the oil inside of the oil tank, and by h_chc​ the height of the oil in the containment tank. How are the rates of change of these two quantities related?

Q2. \displaystyle dh_c = -\frac{225-25\pi}{25\pi} dh_odhc​=−25π225−25πdho

dh_c = (25\pi – 225) dh_odhc​=(25π−225)dho

\displaystyle dh_c = -\frac{25\pi}{225} dh_odhc​=−22525πdho

\displaystyle dh_c = (225 – 25\pi) dh_odhc​=(225−25π)dho

\displaystyle dh_c = -\frac{25\pi}{225-25\pi} dh_odhc​=−225−25π25πdho

\displaystyle dh_c = -\frac{225}{25\pi} dh_odhc​=−25π225​dho

Q3. The stopping distance D_\mathrm{stop}Dstop​ is the distance traveled by a vehicle from the moment the driver becomes aware of an obstacle in the road until the car stops completely. This occurs in two phases.

The first one, the reaction phase, spans from the moment the driver sees the obstacle until he or she has completely depressed the brake pedal. This entails taking the decision to stop the vehicle, lifting the foot from the gas pedal and onto the brake pedal, and pressing the latter down its full distance to obtain maximum braking power. The amount of time necessary to do all this is called the reaction time t_\mathrm{react}treact​, and is independent of the speed at which the vehicle was traveling. Although this quantity varies from driver to driver, it is typically between 1.5\,\mathrm{s}1.5s and 2.5\,\mathrm{s}2.5s. For the purposes of this problem, we will use an average value of 2\,\mathrm{s}2s. The distance traversed by the vehicle in this time is unsurprisingly called reaction distance D_\mathrm{react}Dreact​ and is given by the formula

D_\mathrm{react} = v t_\mathrm{react}Dreact​=vtreact​

where vv is the initial speed of the vehicle.

In the braking phase, the vehicle decelerates and comes to a complete stop. The braking distance D_\mathrm{brake}Dbrake​ that the vehicle covers in this phase is proportional to the square of the initial speed of the vehicle:

D_\mathrm{brake} = \alpha v^2Dbrake​=αv2

The constant of proportionality \alphaα depends on the vehicle type and condition, as well as on the road conditions. Consider a typical value of 10^{-2}\,\mathrm{s^2/m}10−2s2/m.

If the initial speed of the vehicle is 108\,\mathrm{km/h} = 30\,\mathrm{m/s}108km/h=30m/s, what is the ratio between the relative rate of change of the stopping distance and the relative rate of change of the initial speed?

• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{26}{23}dv/vdDstop​/Dstop​​=2326​
• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{24}{23}dv/vdDstop​/Dstop​​=2324​
• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{27}{23}dv/vdDstop​/Dstop​​=2327​
• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = 1dv/vdDstop​/Dstop​​=1
• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{28}{26}dv/vdDstop​/Dstop​​=2628​
• \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{25}{23}dv/vdDstop​/Dstop​​=2325​

Q4. Assume that you possess equal amounts of a product XX and YY, but you value them differently. Specifically, your utility function is of the form

U(X,Y) = C X^\alpha Y^\betaU(X,Y)=CXαYβ

for \alphaα, \betaβ, and CC positive constants. What is your marginal rate of substitution (MRS) of YY for XX?

Hint: recall that the MRS is equal to \displaystyle -\frac{dY}{dX}−dXdY​ along the indifference curve where UU is constant.

• \displaystyle \frac{\beta}{\alpha}αβ
• \displaystyle \frac{C}{\alpha\beta}αβC
• \displaystyle \frac{\alpha}{\beta}βα
• 11
• \displaystyle C\frac{\beta}{\alpha}Cαβ
• \displaystyle \frac{\alpha Y}{\beta X}βXαY

#### Main Quiz 02

Q1. Find the derivative of f(x) = (\cos x)^xf(x)=(cosx)x.

• f'(x) = \ln\cos x – x\tan xf′(x)=lncosxxtanx
• f'(x) = (\ln\cos x + x\cot x)(\cos x)^xf′(x)=(lncosx+xcotx)(cosx)x
• f'(x) = (\ln\cos x – x\tan x)(\cos x)^{x-1}f′(x)=(lncosxxtanx)(cosx)x−1
• f'(x) = – x (\cos x)^{x-1}\sin xf′(x)=−x(cosx)x−1sinx
• f'(x) = (\ln\cos x – x\tan x)(\cos x)^xf′(x)=(lncosxxtanx)(cosx)x
• f'(x) = -(\cos x)^{x-1}\sin xf′(x)=−(cosx)x−1sinx

Q2. Find the derivative of f(x) = (\ln x)^xf(x)=(lnx)x.

• \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{\ln x} + \ln(\ln x) \right)f′(x)=(lnx)x(lnx1​+ln(lnx))
• \displaystyle f'(x) = \frac{1}{\ln x} + \ln(\ln x)f′(x)=lnx1​+ln(lnx)
• \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{e^x} + e^x\ln x \right)f′(x)=(lnx)x(ex1​+exlnx)
• f'(x) = (\ln x)^x \ln(\ln x)f′(x)=(lnx)xln(lnx)
• \displaystyle f'(x) = (\ln x)^x \frac{\ln x}{x}f′(x)=(lnx)xxlnx
• \displaystyle f'(x) = \frac{1}{e^x} + e^x\ln xf′(x)=ex1​+exlnx

Q3. Find the derivative of f(x) = x^{\ln x}f(x)=xlnx.

• f'(x) = 2\ln xf′(x)=2lnx
• f'(x) = 2x^{\ln x} \ln xf′(x)=2xlnxlnx
• f'(x) = x^{\ln x} \ln xf′(x)=xlnxlnx
• f'(x) = x^{\ln(x) – 1} \ln xf′(x)=xln(x)−1lnx
• f'(x) = 2x^{\ln(x) – 1} \ln xf′(x)=2xln(x)−1lnx
• f'(x) = (\ln x + x) x^{\ln x}f′(x)=(lnx+x)xlnx

Q4. \displaystyle \lim_{x \to +\infty} \left( \frac{x+2}{x+3} \right)^{2x} =x→+∞lim​(x+3x+2​)2x=

Hint: write the fraction \displaystyle \frac{x+2}{x+3}x+3x+2​ as 1 + \text{something}1+something.

• e^{3/2}e3/2
• e^2e2
• e^{-2}e−2
• e^{4/3}e4/3
• e^{2/3}e2/3
• 11

Q5. \displaystyle \lim_{x \to 0^+} \left[ \ln(1+x) \right]^{x} =x→0+lim​[ln(1+x)]x=

• 11
• e^2e2
• The limit does not exist.
• 00
• \sqrt{e}e
• ee

Q6. \displaystyle \lim_{x \to 0} \left(1 + \arctan\frac{x}{2} \right)^{2/x} =x→0lim​(1+arctan2x​)2/x=

• e^2e2
• \sqrt{e}e
• 00
• 11
• ee
• +\infty+∞

#### Main Quiz 03

Q1. If f(x) = x^{2x}f(x)=x2x, compute \displaystyle \frac{df}{dx}dxdf​.

• 2 \ln \left( x^{2x} – 2x \right)2ln(x2x−2x)
• 2x^{2x}\left(1 + \ln x\right)2x2x(1+lnx)
• 2 \left[ x^x – \ln(2x-1) + 1 \right]2[xx−ln(2x−1)+1]
• x^{2x} \ln \left( x^{2x}+1 \right)x2xln(x2x+1)
• x^2 + (e^x)^2x2+(ex)2
• 2x^{2x-1}2x2x−1
• x^{2\ln x} – 2x^2x2lnx−2x2
• x^{2x} \ln 2xx2xln2x

Q2. Consider the function f(x) = \sqrt{3}\,x^2\,e^{1-x}f(x)=3​x2e1−x. Use the formula for curvature,

\kappa = \frac{|f”|}{ \left( 1+|f’|^2 \right)^{3/2}}κ=(1+∣f′∣2)3/2∣f′′∣​

to compute the curvature of the graph of ff at the point (1,\sqrt{3})(1,3​).

• \displaystyle -\frac{\sqrt{3}}{9}−93​​
• \displaystyle \frac{\sqrt{3}}{\left(\sqrt{1+\sqrt{3}}\right)^3}(1+3​​)33​​
• \displaystyle \frac{\sqrt{3}}{64}643​​
• \displaystyle \frac{2\sqrt{3}}{27}2723​​
• \displaystyle \frac{x^2-4x+2}{2x-x^2}2xx2x2−4x+2​
• \displaystyle \frac{2}{x} – 1x2​−1
• \sqrt{3}3​
• \displaystyle \frac{\sqrt{3}}{8}83​​

Q3. Assume that xx and yy are related by the equation y \ln x = e^{1-x} + y^3ylnx=e1−x+y3. Compute \displaystyle \frac{dy}{dx}dxdy​ evaluated at x = 1x=1.

• -3−3
• \displaystyle -\frac{1}{3}−31​
• \displaystyle \frac{e^2}{6}6e2​
• \displaystyle \frac{2 + e^2}{3}32+e2​
• \displaystyle \frac{-2 + e^{-2}}{6}6−2+e−2​
• 00
• \displaystyle \frac{2-e^2}{3}32−e2​
• \displaystyle \frac{1}{3}31​

Q4. Use the linear approximation of the function f(x) = \arctan\left(e^{3x}\right)f(x)=arctan(e3x) at x = 0x=0 to estimate the value of f(0.01)f(0.01).

Hint: remember that \displaystyle \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}dxd​arctan(x)=1+x21​.

• \displaystyle \frac{\pi}{4} + \frac{3}{2}4π​+23​
• \displaystyle \frac{\pi}{4} – \frac{3}{2}4π​−23​
• \displaystyle \frac{\pi}{4} + \frac{3}{200}4π​+2003​
• \displaystyle \frac{\pi}{4} – \frac{1}{20}4π​−201​
• \displaystyle \frac{\pi}{4} + \frac{1}{20}4π​+201​
• \displaystyle \frac{\pi}{4} – \frac{1}{200}4π​−2001​
• \displaystyle \frac{\pi}{4} – \frac{3}{200}4π​−2003​
• \displaystyle \frac{\pi}{4} + \frac{1}{200}4π​+2001​

Q5. A rectangular picture frame with total area 50000 \text{ cm}^250000 cm2 includes a border which is 1\text{ cm}1 cm thick at the top and the bottom and 5 \text{ cm}5 cm thick at the left and right side. What is the largest possible area of a picture that can be displayed in this frame?

• 85\text{ cm} \times 470\text{ cm}85 cm×470 cm
• 98\text{ cm} \times 490\text{ cm}98 cm×490 cm
• 80\text{ cm} \times 460\text{ cm}80 cm×460 cm
• 94\text{ cm} \times 475\text{ cm}94 cm×475 cm
• 96\text{ cm} \times 485\text{ cm}96 cm×485 cm
• 95\text{ cm} \times 499\text{ cm}95 cm×499 cm
• 99\text{ cm} \times 495\text{ cm}99 cm×495 cm
• 110\text{ cm} \times 450\text{ cm}110 cm×450 cm

Q6. Which of the following statements are true for the function \displaystyle f(x) = \frac{4}{x} + x^4f(x)=x4​+x4? In order to receive full credit for this problem, you must select all the true statements (there may be many) and none of the false statements.

1 point

• The global minimum of ff for \displaystyle \frac{1}{2}\leq x \leq 221​≤x≤2 is at x = 1x=1.
• ff is not differentiable at x=0x=0.
• The global maximum of ff for \displaystyle -1\leq x \leq-\frac{1}{2}−1≤x≤−21​ is at x = -1x=−1.
• The critical points of ff are at x = -1x=−1 and x = 1x=1.
• The global maximum of ff for -2\leq x \leq -1−2≤x≤−1 is at x = -2x=−2.
• The global maximum of ff for \displaystyle -\frac{3}{2}\leq x \leq 2−23​≤x≤2 is at x = -1x=−1.
• The global minimum of ff for -1\leq x \leq 2−1≤x≤2 is at x = 1x=1.
• The global minimum of ff for -2\leq x \leq 2−2≤x≤2 is at x = 1x=1.

Q7. To approximate \sqrt{15}315​ (the cube root of 1515) using Newton’s method, what is the appropriate update rule for the sequence x_nxn​?

• \displaystyle x_{n+1} = x_n + 3x_n^2xn+1​=xn​+3xn2​
• \displaystyle x_{n+1} = x_n + \frac{5}{x_n^2}xn+1​=xn​+xn2​5​
• \displaystyle x_{n+1} = \frac{2x_n}{3} – \frac{5}{x_n^2}xn+1​=32xn​​−xn2​5​
• \displaystyle x_{n+1} = \frac{2x_n}{3} + \frac{5}{x_n^2}xn+1​=32xn​​+xn2​5​
• \displaystyle x_{n+1} = x_n – \frac{3x_n^2}{x_n^3-15}xn+1​=xn​−xn3​−153xn2​​
• \displaystyle x_{n+1} = \frac{4x_n}{3} – \frac{5}{x_n^2}xn+1​=34xn​​−xn2​5​
• \displaystyle x_{n+1} = x_n + \frac{3x_n^2}{x_n^3-15}xn+1​=xn​+xn3​−153xn2​​
• \displaystyle x_{n+1} = \frac{2}{3}x_nxn+1​=32​xn

Q8. Fill in the blank:

\ln^2(x+h) = \ln^2 x + \underline{\qquad}\cdot h + O(h^2)ln2(x+h)=ln2x+​⋅h+O(h2)

Here, \ln^2 xln2x means \left(\ln x\right)^2(lnx)2.

• \displaystyle \frac{2}{x+h}\ln(x+h)x+h2​ln(x+h)
• 2\ln x2lnx
• \displaystyle \frac{2}{x}x2​
• \displaystyle \ln \frac{2}{x}lnx2​
• 22
• \displaystyle \ln \frac{1}{x}lnx1​
• \displaystyle 2\frac{\ln x}{x}2xlnx
• 2\ln(x+h)2ln(x+h)

Q9. Recall that the kinetic energy of a body is

K = \frac{1}{2}mv^2K=21​mv2

where mm is mass and vv is velocity. Compute the relative rate of change of kinetic energy, \displaystyle\frac{dK}{K}KdK​, given that the relative rate of change of mass is -7−7 and the relative rate of change of velocity is +5+5.

• \displaystyle\frac{dK}{K}=-2KdK​=−2
• \displaystyle\frac{dK}{K}=-\frac{7}{2}KdK​=−27​
• Not enough information is given to solve the problem.
• \displaystyle\frac{dK}{K}=\frac{3}{2}KdK​=23​
• \displaystyle\frac{dK}{K}=5KdK​=5
• \displaystyle\frac{dK}{K}=-7KdK​=−7
• \displaystyle\frac{dK}{K}=3KdK​=3
• \displaystyle\frac{dK}{K}=-9KdK​=−9

Q10. Compute the ninth derivative of (x-3)^{10}(x−3)10 with respect to xx.

• 10(x-3)^910(x−3)9
• \displaystyle\frac{1}{9!}(x-3)^99!1​(x−3)9
• 9!9!
• 11
• 9!(x-3)9!(x−3)
• 10!10!
• 10!(x-3)10!(x−3)
• 00
##### Conclusion:

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