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Concepts and Vocabulary
The derivative of a composite function (f∘g)(x) can be found using the ______ Rule.
Chain
True or False If y=f(u) and u=g(x) are differentiable functions, then y=f(g(x)) is differentiable.
True
True or False If y=f(g(x)) is a differentiable function, then y′= f′(g′(x)).
False
To find the derivative of y=tan(1+cosx), using the Chain Rule, begin with y = ______ and u = ______.
tanu; 1+cosx
If y=(x3+4x+1)100, then y′= ______.
100(x3+4x+1)99(3x2+4)
If f(x)=e3x2+5, then f′(x)= ______.
6xe3x2+5
True or False The Chain Rule can be applied to multiple composite functions.
True
ddxsinx2= ______.
2xcos(x2)
Skill Building
In Problems 9–14, write y as a function of x. Find dydx using the Chain Rule.
y=u5, u=x3+1
y = (x3 + 1)5; dydx=15x2(x3+1)4
y=u3, u=2x+5
y=uu+1, u=x2+1
y=u−1u, u=x2−1
y=(u+1)2, u=1x
y=(u2−1)3, u=1x+2
In Problems 15–32, find the derivative of each function using the Power Rule for Functions.
f(x)=(3x+5)2
f′(x)=6(3x+5)
f(x)=(2x−5)3
f(x)=(6x−5)−3
f′(x)=−18(6x−5)4
f(t)=(4t+1)−2
g(x)=(x2+5)4
g′(x)=8x(x2+5)3
F(x)=(x3−2)5
f(u)=(u−1u)3
f′(u)=3(u−1u)2(1+1u2)
f(x)=(x+1x)3
g(x)=(4x+ex)3
g′(x)=3(4x+ex)2(4+ex)
F(x)=(ex−x2)2
f(x)=tan2x
f′(x)=2tanxsec2x
f(x)=sec3x
f(z)=(tanz+cosz)2
f′(z)=2(tanz+cosz)(sec2z−sinz)
f(z)=(ez+2sinz)3
y=(x2+4)2(2x3−1)3
y′=4x(x2+4)(2x3−1)3+18x2(x2+4)2(2x3−1)2=2x(x2+4)(2x3−1)2(13x3+36x−2)
y=(x2−2)3(3x4+1)2
y=(sinxx)2
y′=2sinx(xcosx−sinx)x3
y=(x+cosxx)5
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In Problems 33–54, find y′.
y=sin(4x)
y′=4cos(4x)
y=cos(5x)
y=2sin(x2+2x−1)
y′=2cos(x2+2x−1)⋅(2x+2)
y=12cos(x3−2x+5)
y=sin1x
y′=cos1x⋅(−1x2)
y=sin3x
y=sec(4x)
y′=4sec(4x)tan(4x)
y=cot(5x)
y=e1/x
y′=e1/x(−1x2)
y=e1/x2
y=1x4−2x+1
y′=−4x3−2(x4−2x+1)2
y=3x5+2x2−3
y=1001+99e−x
y′=9900e−x(1+99e−x)2
y=11+2e−x
y=2sinx
y′=(ln2)2sinxcosx
y=(√3)cosx
y=6secx
y′=(ln6)6secxsecxtanx
y=3tanx
y=5xe3x
y′=5e3x+15xe3x
y=x3e2x
y=x2sin(4x)
y′=2xsin(4x)+4x2cos(4x)
y=x2cos(4x)
In Problems 55–58, find y′ (a and b are constants).
y=e−axsin(bx)
y′=−ae−axsin(bx)+be−axcos(bx)
y=eaxcos(−bx)
y=eax−1eax+1
y′=2aeax(eax+1)2
y=e−ax+1ebx−1
In Problems 59–62, write y as a function of x. Find dydx using the Chain Rule.
y=u3, u=3v2+1, v=4x2
y=3u, u=3v2−4, v=1x
y=u2+1, u=4v, v=x2
y=u3−1, u=−2v, v=x3
In Problems 63–70, find y′.
y=e−2xcos(3x)
y′=−2e−2xcos(3x)−3e−2xsin(3x)
y=eπxtan(πx)
y=cos(ex2)
y′=−2xex2sin(ex2)
y=tan(ex2)
y=ecos(4x)
y′=ecos(4x)(−4sin(4x))
y=ecsc2x
y=4sin2(3x)
y′=24sin(3x)cos(3x)
y=2cos2(x2)
In Problems 71 and 72, find the derivative of each function by:
y=(x3+1)2
y=(x2−2)3
In Problems 73–78:
f(x)=(x2−2x+1)5 at (1,0)
f(x)=(x3−x2+x−1)10 at (0,1)
f(x)=x(x2−1)3 at (2,227)
f(x)=x2(x2−1)2 at (2,49)
f(x)=sin(2x)+cosx2 at (0,1)
f(x)=sin2x+cos3x at (π2,1)
In Problems 79 and 80, find the indicated derivative.
d2dx2cos(x5)
d2ydx2=−25x8cos(x5)−20x3sin(x5)
d3dx3sin3x
Suppose h=f∘g. Find h′(1) if f′(2)=6, f(1)=4, g(1)=2, and g′(1)=−2.
h′(1)=−12
Suppose h=f∘g. Find h′(1) if f′(3)=4, f(1)=1, g(1)=3, and g′(1)=3.
Suppose h=g∘f. Find h′(0) if f(0)=3, f′(0)=−1, g(3)=8, and g′(3)=0.
h′(0)=0
Suppose h=g∘f. Find h′(2) if f(1)=2, f′(1)=4, f(2)=−3, f′(2)=4, g(−3)=1, and g′(−3)=3.
If y=u5+u and u=4x3+x−4, find dydx at x=1.
78
If y=eu+3u and u=cosx, find dydx at x=0.
Applications and Extensions
In Problems 87–94, find the indicated derivative.
ddxf(x2+1) (Hint: Let u=x2+1.)
dfdx=f′(x2+1)(2x)
ddxf(1−x2)
ddxf(x+1x−1)
dfdx=f′(x+1x−1)(−2(x−1)2)
ddxf(1−x1+x)
ddxf(sinx)
dfdx=f′(sinx)cosx
ddxf(tanx)
d2dx2f(cosx)
d2fdx2=f″
\dfrac{d^{2}}{dx^{2}} f(e^{x})
Rectilinear Motion The distance s, in meters, of an object from the origin at time t ≥ 0 seconds is given by s=s( t) = A\cos (\omega t+\phi ), where A, \omega , and \phi are constant.
(a)
(b)
(c)
(d)
Rectilinear Motion A bullet is fired horizontally into a bale of paper. The distance s (in meters) the bullet travels into the bale of paper in t seconds is given by s=s( t) =8-(2-t)^{3},\quad 0 ≤ t≤ 2.
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Rectilinear Motion Find the acceleration a of a car if the distance s, in feet, it has traveled along a highway at time t≥ 0 seconds is given by s( t) =\frac{80}{3}\left[ {{t+\frac{3}{\pi }\sin }}\left( {{\frac{\pi }{6}t}}\right) \right]
Rectilinear Motion An object moves in rectilinear motion so that at time t≥ 0 seconds, its distance from the origin is s(t) =\sin e^{t}, in feet.
Resistance The resistance R (measured in ohms) of an 80-meter-long electric wire of radius x (in centimeters) is given by the formula R=R(x) =\dfrac{0.0048}{x^{2}}. The radius x is given by x=0.1991 + 0.000003T where T is the temperature in Kelvin. How fast is R changing with respect to T when T=320{\,{\rm{K}}}?
{dR\over dT} = -3.597\times 10^{-6} ohms/K
Pendulum Motion in a Car The motion of a pendulum swinging in the direction of motion of a car moving at a low, constant speed, can be modeled by s=s(t)=0.05\sin (2t)+3t \ \ \ \ \ \ 0≤ t≤ \pi where s is the distance in meters and t is the time in seconds.
Source: Mathematics students at Trine University.
Economics The function A(t) =102-90\,e^{-0.21t} represents the relationship between A, the percentage of the market penetrated by second-generation smart phones, and t, the time in years, where t=0 corresponds to the year 2010.
Meteorology The atmospheric pressure at a height of x meters above sea level is P(x)=10^{4}e^{-0.00012x} kilograms per square meter. What is the rate of change of the pressure with respect to the height at x=500 m? At x=750 m?
Hailstones Hailstones originate at an altitude of about 3000{\,{\rm{m}}}, although this varies. As they fall, air resistance slows down the hailstones considerably. In one model of air resistance, the speed of a hailstone of mass m as a function of time t is given by v(t)=\dfrac{mg}{k}( 1-e^{-kt/m}) {\,{\rm{m}}}/\!{\,{\rm{s}}}, where g=9.8{\,{\rm{m}}}/\!{\,{\rm{s}}}^{2} is the acceleration due to gravity and k is a constant that depends on the size of the hailstone and the conditions of the air.
Mean Earnings The mean earnings E, in dollars, of workers 18 years and over are given in the table below:
Year | 1975 | 1980 | 1985 | 1990 | 1995 | 2000 | 2005 | 2010 |
Mean Earnings | 8,552 | 12,665 | 17,181 | 21,793 | 26,792 | 32,604 | 41,231 | 49,733 |
Rectilinear Motion An object moves in rectilinear motion so that at time t>0 its distance s from the origin is s=s( t) . The velocity v of the object is v=\dfrac{ds}{dt}, and its acceleration is a=\dfrac{dv}{dt}=\dfrac{d^{2}s}{dt^{2}}. If the velocity v=v ( s) is expressed as a function of s, show that the acceleration a can be expressed as a=v\dfrac{dv}{ds}.
See Student Solutions Manual.
Student Approval Professor Miller’s student approval rating is modeled by the function Q( t) =21+\dfrac{10\sin \left(\dfrac{2\pi t}{7}\right) }{\sqrt{t}-\sqrt{20}}, where 0≤ t≤ 16 is the number of weeks since the semester began.
Source: Mathematics students at Millikin University, Decatur, Illinois.
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Angular Velocity If the disk in the figure is rotated about the vertical through an angle \theta , torsion in the wire attempts to turn the disk in the opposite direction. The motion \theta at time t (assuming no friction or air resistance) obeys the equation \theta ( t) =\dfrac{\pi }{3}\cos \left( \dfrac{1}{2}\sqrt{\dfrac{2k}{5}}t\right) where k is the coefficient of torsion of the wire.
(a)
(b)
Harmonic Motion A weight hangs on a spring making it 2{\,{\rm{m}}} long when it is stretched out (see the figure). If the weight is pulled down and then released, the weight oscillates up and down, and the length l of the spring after t seconds is given by l( t) =2+\cos \left(2\pi t\right) .
Find F^\prime (1) if f(x)=\sin x and F(t)=f(t^{2}-1).
F'(1)=2
Normal Line Find the point on the graph of y=e^{-x} where the normal line to the graph passes through the origin.
Use the Chain Rule and the fact that {\cos x=\sin \left(\dfrac{\pi }{2}-x\right) } to show that \dfrac{d}{dx}\cos x=-\sin x.
See Student Solutions Manual.
If y=e^{2x}, show that y^{\prime\prime} -4y=0.
If y=e^{-2x}, show that y^{\prime\prime} -4y=0.
See Student Solutions Manual.
If y=Ae^{2x}+Be^{-2x}, where A and B are constants, show that y^{\prime\prime} -4y=0.
If y=Ae^{ax}+Be^{-ax}, where A, B, and a are constants, show that y^{\prime\prime} -a^{2}y=0.
See Student Solutions Manual.
If y=Ae^{2x}+Be^{3x}, where A and B are constants, show that y^{\prime\prime} -5y^\prime +6y=0.
If y=Ae^{-2x}+Be^{-x}, where A and B are constants, show that y^{\prime\prime} +3y^\prime +2y=0.
See Student Solutions Manual.
If y=A~\sin ( \omega t) +B~\cos (\omega t) , where A, B, and \omega are constants, show that y^{\prime \prime }+\omega ^{2}y=0.
Show that \dfrac{d}{dx}f(h(x))=2xg(x^{2}) , if \dfrac{d}{dx}f(x)=g(x) and h(x)=x^{2}.
See Student Solutions Manual.
Find the nth derivative of f(x)=(2x+3)^{n}.
Find a general formula for the nth derivative of y.
If y=e^{-at}~[ A~\sin ( \omega t) +B~\cos ( \omega t) ] , where A, B, a, and \omega are constants, find y^\prime and y^{\prime\prime} .
Show that if a function f has the properties:
then f^\prime =f.
See Student Solutions Manual.
Challenge Problems
Find the nth derivative of f(x)=\dfrac{1}{3x-4}.
Let f_{1}(x),\ldots ,f_{n}(x) be n differentiable functions. Find the derivative of y=f_{1}(f_{2}(f_{3} (\ldots (f_{n}(x)\ldots )))).
See Student Solutions Manual.
Let f(x)=\left\{ \begin{array}{c@{\qquad}ll} x^{2}\sin {\dfrac{1}{x}} & \hbox{if} & x\;≠ 0 \\ 0 & \hbox{if} & x\;=0 \end{array} \right.
Show that f^\prime (0) exists, but that f^\prime (x) is not continuous at 0.
Define f by f(x)=\left\{ \begin{array}{l@{\quad}ll} e^{-1/x^{2}} & \hbox{if}& x\;≠ 0 \\ 0 & \hbox{if}& x\;=0 \end{array} \right.
Show that f is differentiable on (-\infty ,\infty ) and find f^\prime (x) for each value of x. [Hint: To find f^\prime (0), use the definition of the derivative. Then show that 1< x^{2}e^{1/x^{2}} for x≠ 0.]
See Student Solutions Manual.
Suppose f( x) =x^{2} and g( x) =\left\vert x-1\right\vert . The functions f and g are continuous on their domains, the set of all real numbers.
Suppose f( x) =x^{4} and g( x) =x^{1/3}. The functions f and g are continuous on their domains, the set of all real numbers.
See Student Solutions Manual.
The function f(x)=e^{x} has the property f^\prime (x)=f(x). Give an example of another function g(x) such that g(x) is defined for all real x, g^{\prime }(x)=g(x), and g(x)≠ f(x).
Harmonic Motion The motion of the piston of an automobile engine is approximately simple harmonic. If the stroke of a piston (twice the amplitude) is 10{\,{\rm{cm}}} and the angular velocity \omega is 60 revolutions per second, then the motion of the piston is given by s( t) =5\sin ( 120\pi t ) {\,{\rm{cm}}}.