3.6E: Exercises for Section 3.6
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- Jan 2, 2022
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In exercises 1 - 6, given y=f(u) and u=g(x), find dydx by using Leibniz’s notation for the chain rule: dydx=dydududx.
1) y=3u−6,u=2x2
2) y=6u3,u=7x−4
- Answer
- dydx=18u2⋅7=18(7x−4)2⋅7=126(7x−4)2
3) y=sinu,u=5x−1
4) y=cosu,u=−x8
- Answer
- dydx=−sinu⋅(−18)=18sin(−x8)
5) y=tanu,u=9x+2
6) y=√4u+3,u=x2−6x
- Answer
- dydx=8x−242√4u+3=4x−12√4x2−24x+3
For each of the following exercises,
a. decompose each function in the form y=f(u) and u=g(x), and
b. find dydx as a function of x.
7) y=(3x−2)6
8) y=(3x2+1)3
- Answer
- a. f(u)=u3,u=3x2+1;
b. dydx=18x(3x2+1)2
9) y=sin5(x)
10) y=(x7+7x)7
- Answer
- a. f(u)=u7,u=x7+7x;
b. dydx=7(x7+7x)6⋅(17−7x2)
11) y=tan(secx)
12) y=csc(πx+1)
- Answer
- a. f(u)=cscu,u=πx+1;
b. dydx=−πcsc(πx+1)⋅cot(πx+1)
13) y=cot2x
14) y=−6sin−3x
- Answer
- a. f(u)=−6u−3,u=sinx;
b. dydx=18sin−4x⋅cosx
In exercises 15 - 24, find dydx for each function.
15) y=(3x2+3x−1)4
16) y=(5−2x)−2
- Answer
- dydx=4(5−2x)3
17) y=cos3(πx)
18) y=(2x3−x2+6x+1)3
- Answer
- dydx=6(2x3−x2+6x+1)2⋅(3x2−x+3)
19) y=1sin2(x)
20) y=(tanx+sinx)−3
- Answer
- dydx=−3(tanx+sinx)−4⋅(sec2x+cosx)
21) y=x2cos4x
22) y=sin(cos7x)
- Answer
- dydx=−7cos(cos7x)⋅sin7x
23) y=√6+secπx2
24) y=cot3(4x+1)
- Answer
- dydx=−12cot2(4x+1)⋅csc2(4x+1)
25) Let y=[f(x)]3 and suppose that f′(1)=4 and dydx=10 for x=1. Find f(1).
26) Let y=(f(x)+5x2)4 and suppose that f(−1)=−4 and dydx=3 when x=−1. Find f′(−1)
- Answer
- f′(−1)=1034
27) Let y=(f(u)+3x)2 and u=x3−2x. If f(4)=6 and dydx=18 when x=2, find f′(4).
28) [T] Find the equation of the tangent line to y=−sin(x2) at the origin. Use a calculator to graph the function and the tangent line together.
- Answer
- y=−12x
29) [T] Find the equation of the tangent line to y=(3x+1x)2 at the point (1,16). Use a calculator to graph the function and the tangent line together.
30) Find the x -coordinates at which the tangent line to y=(x−6x)8 is horizontal.
- Answer
- x=±√6
31) [T] Find an equation of the line that is normal to g(θ)=sin2(πθ) at the point (14,12). Use a calculator to graph the function and the normal line together.
For exercises 32 - 39, use the information in the following table to find h′(a) at the given value for a.
x | f(x) | f′(x) | g(x) | g′(x) |
0 | 2 | 5 | 0 | 2 |
1 | 1 | −2 | 3 | 0 |
2 | 4 | 4 | 1 | −1 |
3 | 3 | −3 | 2 | 3 |
32) h(x)=f(g(x));a=0
- Answer
- h′(0)=10
33) h(x)=g(f(x));a=0
34) h(x)=(x4+g(x))−2;a=1
- Answer
- h′(1)=−18
35) h(x)=(f(x)g(x))2;a=3
36) h(x)=f(x+f(x));a=1
- Answer
- h′(1)=−4
37) h(x)=(1+g(x))3;a=2
38) h(x)=g(2+f(x2));a=1
- Answer
- h′(1)=−12
39) h(x)=f(g(sinx));a=0
40) [T] The position function of a freight train is given by s(t)=100(t+1)−2, with s in meters and t in seconds. At time t=6 s, find the train’s
a. velocity and
b. acceleration.
c. Considering your results in parts a. and b., is the train speeding up or slowing down?
- Answer
- a. v(6)=−200343 m/s,
b. a(6)=6002401m/s2,
c. The train is slowing down since velocity and acceleration have opposite signs.
41) [T] A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where t is measured in seconds and s is in inches:
s(t)=−3cos(πt+π4).
a. Determine the position of the spring at t=1.5 s.
b. Find the velocity of the spring at t=1.5 s.
42) [T] The total cost to produce x boxes of Thin Mint Girl Scout cookies is C dollars, where C=0.0001x3−0.02x2+3x+300. In t weeks production is estimated to be x=1600+100t boxes.
a. Find the marginal cost C′(x).
b. Use Leibniz’s notation for the chain rule, dCdt=dCdx⋅dxdt, to find the rate with respect to time t that the cost is changing.
c. Use your result in part b. to determine how fast costs are increasing when t=2 weeks. Include units with the answer.
- Answer
- a. C′(x)=0.0003x2−0.04x+3
b. dCdt=100⋅(0.0003x2−0.04x+3)=100⋅(0.0003(1600+100t)2−0.04(1600+100t)+3)=300t2+9200t+70700
c. Approximately $90,300 per week
43) [T] The formula for the area of a circle is A=πr2, where r is the radius of the circle. Suppose a circle is expanding, meaning that both the area A and the radius r (in inches) are expanding.
a. Suppose r=2−100(t+7)2 where t is time in seconds. Use the chain rule dAdt=dAdr⋅drdt to find the rate at which the area is expanding.
b. Use your result in part a. to find the rate at which the area is expanding at t=4 s.
44) [T] The formula for the volume of a sphere is S=43πr3, where r (in feet) is the radius of the sphere. Suppose a spherical snowball is melting in the sun.
a. Suppose r=1(t+1)2−112 where t is time in minutes. Use the chain rule dSdt=dSdr⋅drdt to find the rate at which the snowball is melting.
b. Use your result in part a. to find the rate at which the volume is changing at t=1 min.
- Answer
- a. dSdt=−8πr2(t+1)3=−8π(1(t+1)2−112)2(t+1)3
b. The volume is decreasing at a rate of −π36ft3/min
45) [T] The daily temperature in degrees Fahrenheit of Phoenix in the summer can be modeled by the function T(x)=94−10cos[π12(x−2)], where x is hours after midnight. Find the rate at which the temperature is changing at 4 p.m.
46) [T] The depth (in feet) of water at a dock changes with the rise and fall of tides. The depth is modeled by the function D(t)=5sin(π6t−7π6)+8, where t is the number of hours after midnight. Find the rate at which the depth is changing at 6 a.m.
- Answer
- 2.3 ft/hr