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**Calculus: Single Variable Part 2 – Differentiation Quiz Answers**

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

#### Main Quiz 01

Q1. Let f(x) = -x^2+6x-3*f*(*x*)=−*x*2+6*x*−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^2*p*(*t*)=6+21*t*+4*t*2 of a particle as a function of time, what is the particle’s velocity at t=1*t*=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 P*P* for an ant species is P(t) = 2\ln(t+2)*P*(*t*)=2ln(*t*+2), where t*t* is time. What is the rate of change of the population at time t = 2*t*=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^3*V*=41*t*3

- \displaystyle \frac{dV}{dt} = \frac{3}{4}t^3
*dtdV*=43*t*3 **\displaystyle \frac{dV}{dt} = 3t^2***dtdV*=3*t*2- \displaystyle \frac{dV}{dt} = \frac{1}{3}t^2
*dtdV*=31*t*2 - \displaystyle \frac{dV}{dt} = \frac{3}{4}t^2
*dtdV*=43*t*2 - \displaystyle \frac{dV}{dt} = 0
*dtdV*=0 - \displaystyle \frac{dV}{dt} = \frac{1}{4}t^2
*dtdV*=41*t*2

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

- 4848
- 1212
- 6060
**2020**- 7676
- 108108

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

- Never.
- At t=1
*t*=1. - At t=3
*t*=3. **At t = 6***t*=6.- At t=0
*t*=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 t*t* is given s(t)=-8t^2+320*s*(*t*)=−8*t*2+320, where t*t* 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 F*F* exerted by an **ideal** spring displaced a distance x*x* from its equilibrium point is given by F(x) = -kx*F*(*x*)=−*k**x*, where the constant k*k* 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 x*x* near zero, F(x) = -kx + O(x^2)*F*(*x*)=−*k**x*+*O*(*x*2).

Suppose you have a spring whose force follows the equation F(x) = – 2 \tan 3x*F*(*x*)=−2tan3*x*. What is its spring constant?

- 00
- 33
**1212**- 11
- 22
- 66

#### Practice Quiz 01

Q1. The profit, P*P*, of a company that manufactures and sells N*N* 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 N*R*(*N*)=*S*⋅*N*, is the selling price S*S* per unit times the number N*N* of units sold. The company’s cost, C(N)=C_0+C_\mathrm{op}(N)*C*(*N*)=*C*0+*C*op(*N*), is a sum of two terms. The first is a constant C_0*C*0 describing the initial investment needed to set up production. The other term, C_\mathrm{op}(N)*C*op(*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 N*N*.

Assume that S = \$50*S*=$50, C_0 = \$75,000*C*0=$75,000, C_\mathrm{op}(N) = \$50 \sqrt{N}*C*op(*N*)=$50*N*, and that the company currently sells N=100*N*=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 Y*Y* is given, as a function of capital K*K*, by

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

where A*A* is a constant, L*L* is units of labor (assumed to be constant), and \alpha*α* 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*−1*L*1−*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*=*αAKαL*1−*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*(2*x*2−4*x*).

- f'(x) = \sqrt{x}(5x^2-6x)
*f*′(*x*)=*x*(5*x*2−6*x*) - \displaystyle f'(x) = \frac{2x-2}{\sqrt{x}}
*f*′(*x*)=*x*2*x*−2 - f'(x) = 2x^{5/2}-4x^{3/2}
*f*′(*x*)=2*x*5/2−4*x*3/2 - f'(x) = \sqrt{x}(5x-6)
*f*′(*x*)=*x*(5*x*−6) **f'(x) = 2x^{3/2}-4x^{1/2}***f*′(*x*)=2*x*3/2−4*x*1/2- f'(x) = 4\sqrt{x}(x-1)
*f*′(*x*)=4*x*(*x*−1)

Q2. Find the derivative of \displaystyle f(x) = 6x^4 -\frac{3}{x^2}-2\pi*f*(*x*)=6*x*4−*x*23−2*π*.

- \displaystyle f'(x) = 24x^3 – \frac{3}{x^2}
*f*′(*x*)=24*x*3−*x*23 - \displaystyle f'(x) = 24x^3-\frac{6}{x^2}
*f*′(*x*)=24*x*3−*x*26 - \displaystyle f'(x) = 24x^3-\frac{6}{x^3}
*f*′(*x*)=24*x*3−*x*36 - \displaystyle f'(x) = 24x^3 + \frac{3}{x^2}
*f*′(*x*)=24*x*3+*x*23 **\displaystyle f'(x) = 24x^3 – \frac{3}{2x}***f*′(*x*)=24*x*3−2*x*3- \displaystyle f'(x) = 24x^3+\frac{6}{x^3}
*f*′(*x*)=24*x*3+*x*36

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

- f'(x) = -45(x^3 + 4x)^4(3x^2 + 4)\sin x
*f*′(*x*)=−45(*x*3+4*x*)4(3*x*2+4)sin*x* - f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x – (x^3+4x)\sin x \big]
*f*′(*x*)=7(*x*3+4*x*)4[5(3*x*2+4)cos*x*−(*x*3+4*x*)sin*x*] - f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x – (x^3+4x)\sin x \big]
*f*′(*x*)=7(*x*3+4*x*)4[5cos*x*−(*x*3+4*x*)sin*x*] - f'(x) = 7(x^3+4x)^4 \big[ 5(3x^2+4) \cos x + (x^3+4x)\sin x \big]
*f*′(*x*)=7(*x*3+4*x*)4[5(3*x*2+4)cos*x*+(*x*3+4*x*)sin*x*] **f'(x) = 7(x^3+4x)^4 \big[ 5 \cos x + (x^3+4x)\sin x \big]***f*′(*x*)=7(*x*3+4*x*)4[5cos*x*+(*x*3+4*x*)sin*x*]- f'(x) = 45(x^3 + 4x)^4(3x^2 + 4)\cos x
*f*′(*x*)=45(*x*3+4*x*)4(3*x*2+4)cos*x*

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

- f'(x) = (\sin x )(\ln x) + e^x\cos x
*f*′(*x*)=(sin*x*)(ln*x*)+*ex*cos*x* - \displaystyle f'(x) = \frac{\sin x}{x} + e^x\sin x
*f*′(*x*)=*x*sin*x*+*ex*sin*x* - \displaystyle f'(x) = \frac{\sin x}{x} + (\ln x)(\cos x) + e^x\sin x + e^x\cos x
*f*′(*x*)=*x*sin*x*+(ln*x*)(cos*x*)+*ex*sin*x*+*ex*cos*x* - \displaystyle f'(x) = \left( \frac{1}{x}+e^x \right) \cos x
*f*′(*x*)=(*x*1+*ex*)cos*x* **\displaystyle f'(x) = \frac{e^x\sin x}{x}***f*′(*x*)=*xex*sin*x*- f'(x) = e^x(\sin x + \cos x)
*f*′(*x*)=*ex*(sin*x*+cos*x*)

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

- \displaystyle f'(x) = -\frac{3x+12}{2x \sqrt{x+3}}
*f*′(*x*)=−2*xx*+33*x*+12 - \displaystyle f'(x) = -\frac{3x+12}{2x^3 \sqrt{x+3}}
*f*′(*x*)=−2*x*3*x*+33*x*+12 - \displaystyle f'(x) = \frac{5x+12}{2x^3 \sqrt{x+3}}
*f*′(*x*)=2*x*3*x*+35*x*+12 **\displaystyle f'(x) = \frac{(x-4)\sqrt{x+3}}{2x^3}***f*′(*x*)=2*x*3(*x*−4)*x*+3- \displaystyle f'(x) = \frac{5x+12}{2x \sqrt{x+3}}
*f*′(*x*)=2*xx*+35*x*+12 - \displaystyle f'(x) = \frac{1}{4x\sqrt{x+3}}
*f*′(*x*)=4*xx*+31

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

- \displaystyle f'(x) = \frac{\cos x – \ln x\sin x}{x\sin^2 x}
*f*′(*x*)=*x*sin2*x*cos*x*−ln*x*sin*x* - \displaystyle f'(x) = \frac{\ln x\sin x}{x\cos^2 x}
*f*′(*x*)=*x*cos2*x*ln*x*sin*x* - \displaystyle f'(x) = \frac{\cos x + \ln x\sin x}{x\cos x}
*f*′(*x*)=*x*cos*x*cos*x*+ln*x*sin*x* - \displaystyle f'(x) = \frac{\cos x + x\ln x\sin x}{x\cos^2 x}
*f*′(*x*)=*x*cos2*x*cos*x*+*x*ln*x*sin*x* - \displaystyle f'(x) = \frac{\cos x – \ln x\sin x}{x\cos^2 x}
*f*′(*x*)=*x*cos2*x*cos*x*−ln*x*sin*x* **\displaystyle f'(x) = \frac{(1 + \ln x)\sin x}{x\cos^2 x}***f*′(*x*)=*x*cos2*x*(1+ln*x*)sin*x*

Q7. Find the derivative of \displaystyle f(x) = \frac{ \sqrt[3]{x} – 4}{x^3}*f*(*x*)=*x*33*x*−4.

- \displaystyle f'(x) = \frac{10\sqrt[3]{x} – 36}{3x^4}
*f*′(*x*)=3*x*4103*x*−36 **\displaystyle f'(x) = \frac{10\sqrt[3]{x} – 36}{x^4}***f*′(*x*)=*x*4103*x*−36- \displaystyle f'(x) = \frac{12 – 2\sqrt[3]{x}}{3x^4}
*f*′(*x*)=3*x*412−23*x* - \displaystyle f'(x) = \frac{36 – 8\sqrt[3]{x}}{3x^4}
*f*′(*x*)=3*x*436−83*x* - \displaystyle f'(x) = \frac{36 – 8\sqrt[3]{x}}{x^4}
*f*′(*x*)=*x*436−83*x* - \displaystyle f'(x) = \frac{12 – 2\sqrt[3]{x}}{x^4}
*f*′(*x*)=*x*412−23*x*

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

- f'(x) = 9x^2 \sin^3(x^3) \cos^2 (x^3)
*f*′(*x*)=9*x*2sin3(*x*3)cos2(*x*3) - f'(x) = 3 \sin^2(x^3) \cos(x^3)
*f*′(*x*)=3sin2(*x*3)cos(*x*3) - f'(x) = 9x^2 \sin^2 (x^3) \cos (x^3)
*f*′(*x*)=9*x*2sin2(*x*3)cos(*x*3) - f'(x) = 3\sin^2 (x^3)
*f*′(*x*)=3sin2(*x*3) **f'(x) = 9x^2 \sin^2 (x^2) \cos (3x^2)***f*′(*x*)=9*x*2sin2(*x*2)cos(3*x*2)- f'(x) = 3\sin^2(3x^2)
*f*′(*x*)=3sin2(3*x*2)

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

**\displaystyle f'(x) = \frac{2}{x^3} e^{-1/x^2}***f*′(*x*)=*x*32*e*−1/*x*2- \displaystyle f'(x) = e^{2/x^3}
*f*′(*x*)=*e*2/*x*3 - \displaystyle f'(x) = e^{-2/x^3}
*f*′(*x*)=*e*−2/*x*3 - \displaystyle f'(x) = -\frac{1}{x^2} e^{-1/x^2}
*f*′(*x*)=−*x*21*e*−1/*x*2 - \displaystyle f'(x) = \frac{1}{x^2} e^{-1/x^2}
*f*′(*x*)=*x*21*e*−1/*x*2 - \displaystyle f'(x) = -\frac{2}{x^3} e^{-1/x^2}
*f*′(*x*)=−*x*32*e*−1/*x*2

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

#### Main Quiz 01

Q1. Use a linear approximation to estimate \sqrt[3]{67}367. Round your answer to four decimal places.

**Hint:** remember that \sqrt[3]{64} = 4364=4. You can check the accuracy of this approximation by noting that \sqrt[3]{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=3*π*, and hence \displaystyle 6^\mathrm{o} = \frac{\pi}{30}6o=30*π*. 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-1*x*2−*x*−1. If you use Newton’s method to estimate its value, what is the appropriate update rule for the sequence x_n*xn* ?

**\displaystyle x_{n+1} = x_n + \frac{2x_n – 1}{x_n^2 – x_n – 1}***xn*+1=*xn*+*xn*2−*xn*−12*xn*−1- \displaystyle x_{n+1} = x_n – \frac{x_n^2 – x_n – 1}{2x_n – 1}
*xn*+1=*xn*−2*xn*−1*xn*2−*xn*−1 - \displaystyle x_{n+1} = x_n – \frac{2x_n – 1}{x_n^2 – x_n – 1}
*xn*+1=*xn*−*xn*2−*xn*−12*xn*−1 - \displaystyle x_{n+1} = \frac{x_n^2 – x_n – 1}{2x_n – 1}
*xn*+1=2*xn*−1*xn*2−*xn*−1 - \displaystyle x_{n+1} = x_n + \frac{x_n^2 – x_n – 1}{2x_n – 1}
*xn*+1=*xn*+2*xn*−1*xn*2−*xn*−1 - \displaystyle x_{n+1} = \frac{2x_n – 1}{x_n^2 – x_n – 1}
*xn*+1=*xn*2−*xn*−12*xn*−1

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

**\displaystyle x_{n+1} = \frac{x_n}{2} + \frac{5}{x_n}***xn*+1=2*xn*+*xn*5- \displaystyle x_{n+1} = \frac{x_n}{2}
*xn*+1=2*xn* - \displaystyle x_{n+1} = x_n + \frac{2x_n}{x_n^2 – 10}
*xn*+1=*xn*+*xn*2−102*xn* - \displaystyle x_{n+1} = \frac{x_n}{2} – \frac{10}{x_n}
*xn*+1=2*xn*−*xn*10 - \displaystyle x_{n+1} = \frac{x_n}{2} – \frac{5}{x_n}
*xn*+1=2*xn*−*xn*5 - \displaystyle x_{n+1} = x_n – \frac{2x_n}{x_n^2 – 10}
*xn*+1=*xn*−*xn*2−102*xn*

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 L*L* 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^3*D*=3*E**I**P**L*3

where P*P* is the weight of the diver, E*E* is a constant of elasticity —that depends on the material from which the board is manufactured— and I*I* 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 = 0*t*=0. The position is p(0) = 2*p*(0)=2, the velocity v(0) = 4*v*(0)=4, and the acceleration a(0) = 3*a*(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+2*t*+3*t*2+*O*(*t*3), p(4) \simeq 58*p*(4)≃58.- p(t) = 2 + 4t + 6 t^2 + O(t^3)
*p*(*t*)=2+4*t*+6*t*2+*O*(*t*3), p(4) \simeq 114*p*(4)≃114. - p(t) = 2 + 4t + 3 t^2 + O(t^3)
*p*(*t*)=2+4*t*+3*t*2+*O*(*t*3), p(4) \simeq 66*p*(4)≃66. - \displaystyle p(t) = 2 + 2t + \frac{3}{2} t^2 + O(t^3)
*p*(*t*)=2+2*t*+23*t*2+*O*(*t*3), p(4) \simeq 34*p*(4)≃34. - \displaystyle p(t) = 2 + 4t + \frac{3}{2} t^2 + O(t^3)
*p*(*t*)=2+4*t*+23*t*2+*O*(*t*3), p(4) \simeq 42*p*(4)≃42. - p(t) = 2 + 2t + 6 t^2 + O(t^3)
*p*(*t*)=2+2*t*+6*t*2+*O*(*t*3), p(4) \simeq 106*p*(4)≃106.

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

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

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

- Never: the velocity always increases.
- -\sqrt[3]{12} < t < \sqrt[3]{12}−312<
*t*<312 - t < -2
*t*<−2 **-2 < t < 2−2<***t*<2- -\sqrt{24} < t < \sqrt{24}−24<
*t*<24 - t > 2
*t*>2

Q4. What is a formula for the second derivative of f(t) = t^2\sin 2t*f*(*t*)=*t*2sin2*t*? Use this formula to compute f”(\pi/2)*f*′′(*π*/2).

- f”(t) = 4t\cos 2t + (2-4t^2)\sin 2t
*f*′′(*t*)=4*t*cos2*t*+(2−4*t*2)sin2*t*, and f”(\pi/2) = -2\pi*f*′′(*π*/2)=−2*π* - f”(t) = -4t^2\sin 2t
*f*′′(*t*)=−4*t*2sin2*t*, and f”(\pi/2) =0*f*′′(*π*/2)=0 **f”(t) = -8\sin 2t***f*′′(*t*)=−8sin2*t*, and f”(\pi/2) = 0*f*′′(*π*/2)=0- f”(t) = 8t\cos 2t -4t^2\sin 2t
*f*′′(*t*)=8*t*cos2*t*−4*t*2sin2*t*, and f”(\pi/2) = -4\pi*f*′′(*π*/2)=−4*π* - f”(t) = 4t\cos 2t
*f*′′(*t*)=4*t*cos2*t*, and f”(\pi/2) = -2\pi*f*′′(*π*/2)=−2*π* - f”(t) = 8t\cos 2t + (2-4t^2)\sin 2t
*f*′′(*t*)=8*t*cos2*t*+(2−4*t*2)sin2*t*, and f”(\pi/2) = -4\pi*f*′′(*π*/2)=−4*π*

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(*x*2) 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*=*xe*−*x*2.

- 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 = 0*x*=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 = 0*x*=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*)=*e*sin(*x*4)cos(*x*2) ?

**Its Taylor series expansion about x=0***x*=0 is \displaystyle 1 – \frac{x}{3} + O(x^2)1−3*x*+*O*(*x*2). Hence x=0*x*=0 is not a critical point of f(x)*f*(*x*).- Its Taylor series expansion about x=0
*x*=0 is \displaystyle 1 + \frac{x^3}{2} + O(x^4)1+2*x*3+*O*(*x*4). Hence x=0*x*=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=0
*x*=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2*x*4+*O*(*x*5). Hence it has a local minimum at x=0*x*=0. - Its Taylor series expansion about x=0
*x*=0 is \displaystyle 1 – \frac{x^2}{2} + O(x^5)1−2*x*2+*O*(*x*5). Hence it has a local minimum at x=0*x*=0. - Its Taylor series expansion about x=0
*x*=0 is \displaystyle 1 – \frac{x^2}{2} + O(x^5)1−2*x*2+*O*(*x*5). Hence it has a local maximum at x=0*x*=0. - Its Taylor series expansion about x=0
*x*=0 is \displaystyle 1 + \frac{x^4}{2} + O(x^5)1+2*x*4+*O*(*x*5). Hence it has a local maximum at x=0*x*=0.

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

**x=0***x*=0 is a critical point of f*f*, but it is neither a local maximum nor a local minimum.- x=0
*x*=0 is not a critical point of f*f*. - x=0
*x*=0 is a local minimum of f*f*. - x=0
*x*=0 is a local maximum of f*f*.

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

- The global maximum is attained at x = 0
*x*=0 and the global minimum at x = -1*x*=−1. - The global maximum is attained at x = 0
*x*=0 and the global minimum at x = 4*x*=4. **The global maximum is attained at x = 7***x*=7, but there is no global minimum.- The global maximum is attained at x = 7
*x*=7 and the global minimum at x = 4*x*=4. **The global maximum is attained at x = 7***x*=7 and the global minimum at x = -1*x*=−1.- The global maximum is attained at x = 0
*x*=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*)=*x*3+*x*248 ? Select all that apply.

- x=-2
*x*=−2 is the global maximum of f*f*in [-3, -1][−3,−1] **x=-1***x*=−1 is the global maximum of f*f*in [-3, -1][−3,−1]**x=2***x*=2 is the global maximum of f*f*in [-3, 3][−3,3]- x=1
*x*=1 is the global minimum of f*f*in [1, 3][1,3] - x=2
*x*=2 is the global minimum of f*f*in [1, 3][1,3] - x=1
*x*=1 is the global maximum of f*f*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 2x*y*2−*y*=sin2*x*.

- \displaystyle \frac{dy}{dx} = \frac{y^2 – y}{2\cos 2x}
*dxdy*=2cos2*xy*2−*y* - \displaystyle \frac{dy}{dx} = \frac{\sin 2x}{2y – 1}
*dxdy*=2*y*−1sin2*x* - \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{2y – 1}
*dxdy*=2*y*−12cos2*x* - \displaystyle \frac{dy}{dx} = \frac{2\cos 2x}{y^2 – y}
*dxdy*=*y*2−*y*2cos2*x* **\displaystyle \frac{dy}{dx} = \frac{2y – 1}{\sin 2x}***dxdy*=sin2*x*2*y*−1- \displaystyle \frac{dy}{dx} = \frac{2y – 1}{2\cos 2x}
*dxdy*=2cos2*x*2*y*−1

Q2. Find the derivative \displaystyle \frac{dy}{dx}*dxdy* if x*x* and y*y* are related through xy = e^y*xy*=*ey*.

- \displaystyle \frac{dy}{dx} = \frac{e^y + x}{y}
*dxdy*=*yey*+*x* - \displaystyle \frac{dy}{dx} = \frac{x – e^y}{y}
*dxdy*=*yx*−*ey* - \displaystyle \frac{dy}{dx} = \frac{y}{e^y + x}
*dxdy*=*ey*+*xy* - \displaystyle \frac{dy}{dx} = \frac{y}{x – e^y}
*dxdy*=*x*−*eyy* **\displaystyle \frac{dy}{dx} = \frac{y}{e^y – x}***dxdy*=*ey*−*xy*- \displaystyle \frac{dy}{dx} = \frac{e^y – x}{y}
*dxdy*=*yey*−*x*

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

- \displaystyle \frac{dy}{dx} = y\cos x – e^{y\cos x}\sin x
*dxdy*=*y*cos*x*−*ey*cos*x*sin*x* **\displaystyle \frac{dy}{dx} = \frac{y\sin x – e^{-y\cos x}}{\cos x}***dxdy*=cos*xy*sin*x*−*e*−*y*cos*x*- \displaystyle \frac{dy}{dx} = y\tan x – e^{y\cos x}
*dxdy*=*y*tan*x*−*ey*cos*x* - \displaystyle \frac{dy}{dx} = -y\sin x + e^{-y\cos x}\cos x
*dxdy*=−*y*sin*x*+*e*−*y*cos*x*cos*x* - \displaystyle \frac{dy}{dx} = e^{-y\cos x}(\cos x – y\sin x)
*dxdy*=*e*−*y*cos*x*(cos*x*−*y*sin*x*) - \displaystyle \frac{dy}{dx} = \frac{y – e^{y\cos x}\tan x}{\sin x}
*dxdy*=sin*xy*−*ey*cos*x*tan*x*

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

- \displaystyle \frac{dy}{dx} = \frac{-y^2}{x^2\sec^2 y}
*dxdy*=*x*2sec2*y*−*y*2 - \displaystyle \frac{dy}{dx} = \tan y – \frac{y^2}{\sec^2 y}
*dxdy*=tan*y*−sec2*yy*2 **\displaystyle \frac{dy}{dx} = \frac{x\tan y – y^2}{2xy\ln x – x^2\sec^2 y}***dxdy*=2*xy*ln*x*−*x*2sec2*yx*tan*y*−*y*2- \displaystyle \frac{dy}{dx} = \frac{2xy\ln x – x^2\sec^2 y}{x\tan y – y^2}
*dxdy*=*x*tan*y*−*y*22*xy*ln*x*−*x*2sec2*y* - \displaystyle \frac{dy}{dx} = \frac{y^2 – \tan y}{x^2\sec^2 y – 2xy\ln x}
*dxdy*=*x*2sec2*y*−2*xy*ln*xy*2−tan*y* - \displaystyle \frac{dy}{dx} = \frac{x\tan y}{2xy\ln x}
*dxdy*=2*xy*ln*xx*tan*y*

Q5. Model a hailstone as a round ball of radius R*R*. As the hailstone falls from the sky, its radius increases at a constant rate C*C*. At what rate does the volume V*V* of the hailstone change?

- \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C R^3
*dtdV*=34*πCR*3 - \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi C^3
*dtdV*=34*πC*3 **\displaystyle \frac{dV}{dt} = 8\pi C R***dtdV*=8*πCR*- \displaystyle \frac{dV}{dt} = 4\pi C R^2
*dtdV*=4*πCR*2 - \displaystyle \frac{dV}{dt} = \frac{4}{3}\pi R^3
*dtdV*=34*πR*3 - \displaystyle \frac{dV}{dt} = 4\pi R^2
*dtdV*=4*πR*2

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

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

#### Practice Quiz 01

Q1. Consider a box of height h*h* with a square base of side length L*L*. Assume that L*L* is increasing at a rate of 10\%10% per day, but h*h* 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_o*ho* the height of the oil inside of the oil tank, and by h_c*hc* 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_o*d**h**c*=−25*π*225−25*π**d**h**o*

dh_c = (25\pi – 225) dh_o*d**h**c*=(25*π*−225)*d**h**o*

**\displaystyle dh_c = -\frac{25\pi}{225} dh_o dhc=−22525πdho**

\displaystyle dh_c = (225 – 25\pi) dh_o*d**h**c*=(225−25*π*)*d**h**o*

\displaystyle dh_c = -\frac{25\pi}{225-25\pi} dh_o*d**h**c*=−225−25*π*25*π**d**h**o*

\displaystyle dh_c = -\frac{225}{25\pi} dh_o*dhc*=−25*π*225*dho*

Q3. The *stopping distance* D_\mathrm{stop}*D*stop 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}*t*react, 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}*D*react and is given by the formula

D_\mathrm{react} = v t_\mathrm{react}*D*react=*v**t*react

where v*v* 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}*D*brake that the vehicle covers in this phase is proportional to the square of the initial speed of the vehicle:

D_\mathrm{brake} = \alpha v^2*D*brake=*α**v*2

The constant of proportionality \alpha*α* 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*/*vdD*stop/*D*stop=2326- \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{24}{23}
*dv*/*vdD*stop/*D*stop=2324 - \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{27}{23}
*dv*/*vdD*stop/*D*stop=2327 - \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = 1
*dv*/*vdD*stop/*D*stop=1 - \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{28}{26}
*dv*/*vdD*stop/*D*stop=2628 - \displaystyle \frac{dD_\mathrm{stop} / D_\mathrm{stop}}{dv / v} = \frac{25}{23}
*dv*/*vdD*stop/*D*stop=2325

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

U(X,Y) = C X^\alpha Y^\beta*U*(*X*,*Y*)=*C**X**α**Y**β*

for \alpha*α*, \beta*β*, and C*C* positive constants. What is your marginal rate of substitution (MRS) of Y*Y* for X*X*?

**Hint:** recall that the MRS is equal to \displaystyle -\frac{dY}{dX}−*dXdY* along the *indifference curve* where U*U* 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)^x*f*(*x*)=(cos*x*)*x*.

- f'(x) = \ln\cos x – x\tan x
*f*′(*x*)=lncos*x*−*x*tan*x* - f'(x) = (\ln\cos x + x\cot x)(\cos x)^x
*f*′(*x*)=(lncos*x*+*x*cot*x*)(cos*x*)*x* **f'(x) = (\ln\cos x – x\tan x)(\cos x)^{x-1}***f*′(*x*)=(lncos*x*−*x*tan*x*)(cos*x*)*x*−1- f'(x) = – x (\cos x)^{x-1}\sin x
*f*′(*x*)=−*x*(cos*x*)*x*−1sin*x* - f'(x) = (\ln\cos x – x\tan x)(\cos x)^x
*f*′(*x*)=(lncos*x*−*x*tan*x*)(cos*x*)*x* - f'(x) = -(\cos x)^{x-1}\sin x
*f*′(*x*)=−(cos*x*)*x*−1sin*x*

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

- \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{\ln x} + \ln(\ln x) \right)
*f*′(*x*)=(ln*x*)*x*(ln*x*1+ln(ln*x*)) **\displaystyle f'(x) = \frac{1}{\ln x} + \ln(\ln x)***f*′(*x*)=ln*x*1+ln(ln*x*)- \displaystyle f'(x) = (\ln x)^x \left(\frac{1}{e^x} + e^x\ln x \right)
*f*′(*x*)=(ln*x*)*x*(*ex*1+*ex*ln*x*) - f'(x) = (\ln x)^x \ln(\ln x)
*f*′(*x*)=(ln*x*)*x*ln(ln*x*) - \displaystyle f'(x) = (\ln x)^x \frac{\ln x}{x}
*f*′(*x*)=(ln*x*)*xx*ln*x* - \displaystyle f'(x) = \frac{1}{e^x} + e^x\ln x
*f*′(*x*)=*ex*1+*ex*ln*x*

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

**f'(x) = 2\ln x***f*′(*x*)=2ln*x*- f'(x) = 2x^{\ln x} \ln x
*f*′(*x*)=2*x*ln*x*ln*x* - f'(x) = x^{\ln x} \ln x
*f*′(*x*)=*x*ln*x*ln*x* - f'(x) = x^{\ln(x) – 1} \ln x
*f*′(*x*)=*x*ln(*x*)−1ln*x* - f'(x) = 2x^{\ln(x) – 1} \ln x
*f*′(*x*)=2*x*ln(*x*)−1ln*x* - f'(x) = (\ln x + x) x^{\ln x}
*f*′(*x*)=(ln*x*+*x*)*x*ln*x*

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

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

- e^{3/2}
*e*3/2 **e^2***e*2- e^{-2}
*e*−2 - e^{4/3}
*e*4/3 - e^{2/3}
*e*2/3 - 11

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

**11**- e^2
*e*2 - The limit does not exist.
- 00
- \sqrt{e}
*e* - e
*e*

Q6. \displaystyle \lim_{x \to 0} \left(1 + \arctan\frac{x}{2} \right)^{2/x} =*x*→0lim(1+arctan2*x*)2/*x*=

- e^2
*e*2 **\sqrt{e}***e*- 00
- 11
- e
*e* - +\infty+∞

#### Main Quiz 03

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

- 2 \ln \left( x^{2x} – 2x \right)2ln(
*x*2*x*−2*x*) - 2x^{2x}\left(1 + \ln x\right)2
*x*2*x*(1+ln*x*) - 2 \left[ x^x – \ln(2x-1) + 1 \right]2[
*xx*−ln(2*x*−1)+1] - x^{2x} \ln \left( x^{2x}+1 \right)
*x*2*x*ln(*x*2*x*+1) - x^2 + (e^x)^2
*x*2+(*ex*)2 - 2x^{2x-1}2
*x*2*x*−1 - x^{2\ln x} – 2x^2
*x*2ln*x*−2*x*2 - x^{2x} \ln 2x
*x*2*x*ln2*x*

Q2. Consider the function f(x) = \sqrt{3}\,x^2\,e^{1-x}*f*(*x*)=3*x*2*e*1−*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 f*f* 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}2
*x*−*x*2*x*2−4*x*+2 - \displaystyle \frac{2}{x} – 1
*x*2−1 - \sqrt{3}3
- \displaystyle \frac{\sqrt{3}}{8}83

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

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

Q4. Use the linear approximation of the function f(x) = \arctan\left(e^{3x}\right)*f*(*x*)=arctan(*e*3*x*) at x = 0*x*=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+*x*21.

- \displaystyle \frac{\pi}{4} + \frac{3}{2}4
*π*+23 - \displaystyle \frac{\pi}{4} – \frac{3}{2}4
*π*−23 - \displaystyle \frac{\pi}{4} + \frac{3}{200}4
*π*+2003 - \displaystyle \frac{\pi}{4} – \frac{1}{20}4
*π*−201 - \displaystyle \frac{\pi}{4} + \frac{1}{20}4
*π*+201 - \displaystyle \frac{\pi}{4} – \frac{1}{200}4
*π*−2001 - \displaystyle \frac{\pi}{4} – \frac{3}{200}4
*π*−2003 - \displaystyle \frac{\pi}{4} + \frac{1}{200}4
*π*+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^4*f*(*x*)=*x*4+*x*4? 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 f
*f*for \displaystyle \frac{1}{2}\leq x \leq 221≤*x*≤2 is at x = 1*x*=1. - f
*f*is not differentiable at x=0*x*=0. - The global maximum of f
*f*for \displaystyle -1\leq x \leq-\frac{1}{2}−1≤*x*≤−21 is at x = -1*x*=−1. - The critical points of f
*f*are at x = -1*x*=−1 and x = 1*x*=1. - The global maximum of f
*f*for -2\leq x \leq -1−2≤*x*≤−1 is at x = -2*x*=−2. - The global maximum of f
*f*for \displaystyle -\frac{3}{2}\leq x \leq 2−23≤*x*≤2 is at x = -1*x*=−1. - The global minimum of f
*f*for -1\leq x \leq 2−1≤*x*≤2 is at x = 1*x*=1. - The global minimum of f
*f*for -2\leq x \leq 2−2≤*x*≤2 is at x = 1*x*=1.

Q7. To approximate \sqrt[3]{15}315 (the cube root of 1515) using Newton’s method, what is the appropriate update rule for the sequence x_n*xn*?

- \displaystyle x_{n+1} = x_n + 3x_n^2
*xn*+1=*xn*+3*xn*2 - \displaystyle x_{n+1} = x_n + \frac{5}{x_n^2}
*xn*+1=*xn*+*xn*25 - \displaystyle x_{n+1} = \frac{2x_n}{3} – \frac{5}{x_n^2}
*xn*+1=32*xn*−*xn*25 - \displaystyle x_{n+1} = \frac{2x_n}{3} + \frac{5}{x_n^2}
*xn*+1=32*xn*+*xn*25 - \displaystyle x_{n+1} = x_n – \frac{3x_n^2}{x_n^3-15}
*xn*+1=*xn*−*xn*3−153*xn*2 - \displaystyle x_{n+1} = \frac{4x_n}{3} – \frac{5}{x_n^2}
*xn*+1=34*xn*−*xn*25 - \displaystyle x_{n+1} = x_n + \frac{3x_n^2}{x_n^3-15}
*xn*+1=*xn*+*xn*3−153*xn*2 - \displaystyle x_{n+1} = \frac{2}{3}x_n
*xn*+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*)=ln2*x*+⋅*h*+*O*(*h*2)

Here, \ln^2 xln2*x* means \left(\ln x\right)^2(ln*x*)2.

- \displaystyle \frac{2}{x+h}\ln(x+h)
*x*+*h*2ln(*x*+*h*) - 2\ln x2ln
*x* - \displaystyle \frac{2}{x}
*x*2 - \displaystyle \ln \frac{2}{x}ln
*x*2 - 22
- \displaystyle \ln \frac{1}{x}ln
*x*1 - \displaystyle 2\frac{\ln x}{x}2
*x*ln*x* - 2\ln(x+h)2ln(
*x*+*h*)

Q9. Recall that the kinetic energy of a body is

K = \frac{1}{2}mv^2*K*=21*m**v*2

where m*m* is mass and v*v* 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}=-2
*KdK*=−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}=5
*KdK*=5 - \displaystyle\frac{dK}{K}=-7
*KdK*=−7 - \displaystyle\frac{dK}{K}=3
*KdK*=3 - \displaystyle\frac{dK}{K}=-9
*KdK*=−9

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

- 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

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