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Math 181: Examples 2
8. Chain Rule
Chain Rule Review
A few scenarios for the chain rule:
20( )y stuff19' 20( ) 'y stuff stuff
' 'stuffy e stuff stuffy e
sin( )y stuff ' cos( ) 'y stuff stuff
ln( )y stuff'
'stuff
ystuff
Example
• Find the derivative of
34 5y x
1
3 32
1
3 22
' [(4 5) ]' (4 5) '
1(4 5) (12 )
2
y x x
x x
Example
• Let’s use the chain rule to find the derivative of
• You can think of this as the equation below:
• So the chain rule would give us
• Now we just put the x2 + 1 back in where the stuff is.
2 20( 1)y x
20(stuff)y
19' 20( ) 'y stuff stuff
Example cont.
2 19 2
2 19
2 19
'( ) 20( 1) ( 1)
20( 1) (2 )
40 ( 1)
dy x x x
dx
x x
x x
Example
Find the derivative of the following function
With the multiplication present we know we’ll need the product rule and when we have to find the derivative of we’ll need to use the chain rule.
3 5(2 6)y x x
5(2 6)x
Example cont.
Applying the product rule:
Now we’ll use the chain rule on the last part:
2 5 3 5' 3 (2 6) [(2 6) ]'y x x x x
2 5 3 4
2 5 3 4
' 3 (2 6) 5(2 6) 2
' 3 (2 6) 10 (2 6)
y x x x x
y x x x x
9. Implicit Differentiation & More Rates of Change
Example
Find y’ using implicit differentiation. (Remember: y’ and dy/dx are the same thing.)
Use the chain rule on the y2, the product rule on the xy, and the power rule when needed.
Now simplify and solve for y’.
2 3 9y xy x
2 ' 1 ' 3 0yy y x y
2 ' ' 3
(2 ) ' 3
3'
2
yy xy y
y x y y
yy
y x
Example
• Homework #7: Use implicit differentiation to find the differential equation for 2 24 1x y
2 8 ' 0
8 ' 2
'4
x yy
yy x
xy
y
Example
• p. 136 – Find the equation of the tangent line to the circle
at the point (4,3).
• Start by finding y’, which we know is
• Then plug in 4 for x and 3 for y to find the slope.
• Then plug the slope and the point into the point-
slope equation to find the line.
2 2 25x y dy x
dx y
4
3m
43 ( 4)
3
25 4
3 3
y x
y x
10. Rates of Change & Exponential Functions
Example
• p. 141 The total cost in making x tennis rackets (in thousands of rackets) is given by
in thousands of dollars. What is the marginal cost of making 9,000 tennis rackets?
2( ) 20 1.2 0.01C x x x
Example cont.
• Take the derivative of the given cost function.
• Now plug in 9 for x since we want to know the marginal cost of making 9,000 rackets.
• Now interpret the results using the units.
( ) 1.2 0.02C x x
(9) 1.2 0.02(9) 1.02C
(9) 1.02 1.02thousandsofdollars dollars
Cthousandrackets racket
Example
• Homework #13: Given , ,
and , find .
3y x x 1dx
dt
1x dy
dtdy dy dx
dt dx dt
2(1 3 )( 1)
@x 1
x
2dy
dt
Mechanics Example
• Simplify to an expression involving a single exponential.
• Since it’s the same base (in this case e), you just add the exponents when you distribute.
2 2(e )x x xe e
2 2( )x x xe e e
0xe e
1xe
11. Natural Logarithm, Sine and Cosine
Example
• p. 160 Find k given that and
• First we substitute the 2 in for the t and the 7 in for the left side of the equation.
• Now take the natural log of both sides to get the 2k out of the exponent (Since ln is the inverse of e).
3 kty t e
(2) 7y
27 3 ke2 7
3
ke
2 7ln( ) ln( )
3
72 ln( )
3
ke
k
7ln( )
3 0.42364892
k
Example
• p. 102 – Find the derivative of
• Take the natural log of both sides.
• Simplify using the properties of logs.
• Take the derivative of both sides.
• Isolate y’.
2 1
xy
x
Example
• p. 168 – Find y’() when y() = sin (4 ).
• Treat just like x. So we’ll need to use the chain rule.
sin( ) cos( ) ( )d d
innerstuff innerstuff innerstuffd d
sin(4 ) cos(4 ) (4 ) 4cos(4 )d d
d d
12. Trig & Additional Functions
Example
• p. 178 – Find f’(x) for csc(x2 + 1).
• Remember
• Since the inner stuff is not just x, we’ll be using the chain rule again.
csc csc cotd
x x xdx
csc( ) csc( )cot( ) ( )d d
innerstuff innerstuff innerstuff innerstuffdx dx
2 2 2 2
2 2
cot( 1) csc( 1)cot( 1) ( 1)
2 csc( 1)cot( 1)
d dx x x x
dx dx
x x x
Example
• HW16, #13: Find y’(t) for y(t) = sin3(t)
• First, realize that y(t) is the same as this:
• This will be a case for the chain rule (stuff)3
with stuff equaling sin(t) in this case.
3( ) (sin( ))y t t
2'( ) 3(sin( )) cos( )y t t t
Example
• HW16, #27: Find the period, frequency, and amplitude of
and find the rate of change of
• First, use the period (p171)
Since = , T = 2
• The frequency is given by
So the frequency is ½.
• The amplitude is given by
For this problem, a = 5 and b = 0.
( ) 5cos( )y t t1
3t 2
T
1f
T
2 2
mA a b
5mA
Example cont.
• Now just take the derivative and plug in 1/3 for t.
( ) 5cos( )y t t
( ) 5sin( )
1 3( ) 53 2
y t t
y
sin( )3
You can find using the unit circle.
Image from Jim.belk, Wikimedia Commons
Example
• HW17, #21: Find the first derivative of
• Using the formulas from the previous slide, we will rewrite
• Now find the derivative:
ln(x)( ) 10f x
ln(x) ln(10)( )f x e
ln( ) ln(10)'(x) 10 xf
x
13. Mean Value Theorem & L’Hopital’s Rule
Example
• p. 194: Find the differential dy for the curve
• Take the derivative of the function
• So the differential dy is
3 2y x x 2' 3 2y x
2(3 2)dy x dx
Example
• HW18, #13: Find dy and y at x = 2 and x = dx = 0.1 For f(x) = x3 + x2.
2 2' 3 2 (3 2 )y x x dy x x dx
0.1, 2 1.6dx xdy
3 2 3 2
0.1, 2 ((2.1) (2.1) ) ((2) (2) ) 1.671x xy
Example
• HW18, #21: A certain rectangle has a length which is twice its width. The length is 1 cm measured to within a tolerance of 0.1 cm. Estimate the maximum error in calculating the area of the rectangle using this length.
12
2L W W L
21, (L)
2
dAA L W L L dA dL
dL
2
1, 0.1 0.1L dLdA cm
Example
• p. 202 (3.11): Evaluate the limit
• So just plugging in infinity and evaluating the limit, we get the form /.
• So we’ll apply L’Hopital’s Rule to take the derivative of the top and the bottom. In this particular case, it still results in /.
• So we’ll apply L’Hopital’s Rule one final time to discover that the limit is 1.
4
4
1lim
2 3x
x
x x
14. L’Hopital’s Rule & Absolute & Relative Extrema
Example
Evaluate the limit
• Just plug in infinity and you’ll get ∙0 which isn’t a case we can use L’Hopital’s rule on
• Instead, simply rewrite the e-x as 1/ex
• Use L’Hopital’s rule twice to get the answer
2lim( )x
xx e
2
lim( )xx
x
e
2 2 2lim( ) lim lim 0x
x xx x x
xx e
e e
Example
Evaluate the limit
Start by taking the ln of the limit.
Using Theorem 8.2 from the text:
1/
0lim( 1) x
xx
1
0 0 0
1 ln( 1)limln( 1) lim ln( 1) limx
x x x
xx x
x x
1
0 0 0
1ln( 1)
1limln( 1) lim lim 11
x
x x x
dx
dx xxd
xdx
1
1
0lim( 1) x
xx e e
0
0
Example
Find the extrema of
Find f’.
Set it equal to 0 and solve for x.
So there are peaks/troughs at 0 and 3.
You can graph it to see the nature of the 2 critical points.
4 3( ) 4 1f x x x 3 2'( ) 4 12f x x x
3 2
2
4 12 0
4 ( 3) 0
0,3
x x
x x
x
Extreme Value Theorem
The Extreme Value Theorem (p. 212) says that if a function f(x) is continuous on [a,b], then it attains its absolute maximum and its absolute minimum on [a,b], although an absolute extremum may occur at an endpoint of [a,b]. – So start by taking the derivative, just like finding local
extrema – Set it equal to 0 – Solve for x to find the critical points – Now just find the function value at the critical points
and the endpoints. The highest is the absolute maximum. Likewise the smallest value is the absolute minimum.
Example
HW20 #17: Find the critical points of
and determine the absolute extrema of f(x) over [0,4]
Now evaluate f(x) at 2, 0, and 4.
So x = 2 is an absolute minimum and x = 0 & x = 4 are maximums
2( ) 4f x x x
'( ) 2 4
2 4 0
2( 2) 0
2
f x x
x
x
x
(0) 0, (2) 4, (4) 0f f f
Revenue
• Finding extrema is a useful task in many applications, including revenue problems.
• Revenue = price ∙ number sold
• Example 6, p. 214: Suppose Acme Airlines charges a $300 base price for a seat on an airplane, and suppose they add $2 for each unsold seat on that airplane. Determine how many seats should be sold on a 200 seat airplane in order to maximize revenue.
Revenue Example cont.
300 2(200 ) 700 2p x x
2(700 2 )x 700 2R x x x
' 700 4
700 4 0 175
R x
x x
Now evaluate R at 0,175, and 200 (0) 0, (175) 61250, (200) 60000R R R
The absolute maximum is at x = 175 seats
15. Monotonicity & Concavity
1st Derivative Test
• Formal definition: If f’(x) > 0 for all x in (p,q), then f(x) is increasing over (p,q), Likewise, if f’(x) < 0 for all x in (p,q), then f(x) is decreasing over (p,q).
• This is how you use it:
– Find the critical points by setting the derivative = 0.
– Write the critical points on a number line.
– Pick a value between the critical points and plug it in to the derivative of the function.
– Mark on the number line whether it’s positive or negative in the regions between each critical point.
Example
• p. 219: Determine the intervals of monotonicity and identify the extrema of
• Take the derivative.
• Set it equal to 0 & solve for the critical points.
3( ) 3f x x x 2'( ) 3 3f x x
2
2
3 3 0
3( 1) 0
3( 1)( 1) 0
1, 1
x
x
x x
x
Example cont.
• Test points between the critical points by plugging them into the derivative.
From - to -1 it’s increasing.
From -1 to 1, it’s decreasing.
From 1 to , it’s increasing.
• Finally, sketch the curve.
2
2
2
3( 2) 3 9
3(0) 3 3
3(2) 3 9
Example
• HW21 #25: Determine the intervals of monotonicity and the location of the extrema of the function f(x) given the graph of its derivative f’(x).
x = 1
+
Decreasing (-, 1) Increasing (1, ) So x = 1 in a minimum
2nd Derivative Test
• Similar to the information we get from the 1st derivative test, we can use the 2nd derivative test to see whether a function is concave up or concave down at a critical point.
• Where the 2nd derivative equals 0, there is an inflection point.
• If x = a is a critical point where f’(a) = 0, then 1. If f’’(a) > 0, we have a local minimum at x = a.
2. If f’’(a) < 0, we have a local maximum at x = a.
2. Local Max 1. Local Min
Example
• HW22 #3: For find the intervals of concavity and the inflection points.
• Find the critical points by taking the derivative and setting it equal to 0.
• Now find the 2nd derivative and set that equal to 0 to find any inflection points.
3 2( ) 3f x x x
2
2
'( ) 3 6
3 6 0
3 ( 2) 0
0,2
f x x x
x x
x x
x
Example cont.
• Evaluate the location of the critical points within the 2nd derivative.
''( ) 6 6
6 6 0
1
f x x
x
x
''(0) 6
''(2) 6
f
f
x = 1
+