The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
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V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Notes
Section 2.5
The Chain Rule
V63.0121.041, Calculus I
New York University
October 5, 2010
Announcements
Quiz 2 in recitation next week (October 11-15)
Midterm in class on §§1.1–2.5
Announcements
Notes
Quiz 2 in recitation next
week (October 11-15)
Midterm in class on
§§1.1–2.5
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 2 / 36
Objectives
Notes
Given a compound
expression, write it as a
composition of functions.
Understand and apply the
Chain Rule for the derivative
of a composition of
functions.
Understand and use
Newtonian and Leibnizian
notations for the Chain Rule.
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 3 / 36
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2.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Compositions
See Section 1.2 for review Notes
Deﬁnition
If f and g are functions, the composition (f ◦ g )(x) = f (g (x)) means “do
g ﬁrst, then f .”
x g (x) f (g (x))
g f ◦ g f
Our goal for the day is to understand how the derivative of the composition
of two functions depends on the derivatives of the individual functions.
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 4 / 36
Outline
Notes
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 5 / 36
Analogy
Notes
Think about riding a bike. To go
faster you can either:
pedal faster
change gears
radius of front sprocket
The angular position (ϕ) of the back wheel depends on the position of the
front sprocket (θ):
Rθ
ϕ(θ) =
r
And so the angular speed of the back wheel depends on the derivative of
this function and the speed of the front sprocket.
radius of back sprocket
Image credit: SpringSun
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 6 / 36
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3.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
The Linear Case
Notes
Question
Let f (x) = mx + b and g (x) = m x + b . What can you say about the
composition?
Answer
f (g (x)) = m(m x + b ) + b = (mm )x + (mb + b)
The composition is also linear
The slope of the composition is the product of the slopes of the two
functions.
The derivative is supposed to be a local linearization of a function. So
there should be an analog of this property in derivatives.
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 7 / 36
The Nonlinear Case
Notes
Let u = g (x) and y = f (u). Suppose x is changed by a small amount ∆x.
Then
∆y ≈ f (y )∆u
and
∆u ≈ g (u)∆x.
So
∆y
∆y ≈ f (y )g (u)∆x =⇒ ≈ f (y )g (u)
∆x
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 8 / 36
Outline
Notes
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 9 / 36
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4.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Theorem of the day: The chain rule
Notes
Theorem
Let f and g be functions, with g diﬀerentiable at x and f diﬀerentiable at
g (x). Then f ◦ g is diﬀerentiable at x and
(f ◦ g ) (x) = f (g (x))g (x)
In Leibnizian notation, let y = f (u) and u = g dy Then
du
(x).
dx
du
dy dy du
=
dx du dx
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 10 / 36
Observations
Notes
Succinctly, the derivative of a
composition is the product of
the derivatives
The only complication is where
these derivatives are evaluated:
at the same point the functions
are
In Leibniz notation, the Chain
Rule looks like cancellation of
(fake) fractions
Image credit: ooOJasonOoo
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 11 / 36
Outline
Notes
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 17 / 36
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5.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Example
Notes
Example
let h(x) = 3x 2 + 1. Find h (x).
Solution
√
First, write h as f ◦ g . Let f (u) = u and g (x) = 3x 2 + 1. Then
−1/2
f (u) = 1 u
2 , and g (x) = 6x. So
3x
h (x) = 1 u −1/2 (6x) = 2 (3x 2 + 1)−1/2 (6x) = √
2
1
3x 2 + 1
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 18 / 36
Corollary
Notes
Corollary (The Power Rule Combined with the Chain Rule)
If n is any real number and u = g (x) is diﬀerentiable, then
d n du
(u ) = nu n−1 .
dx dx
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 19 / 36
Order matters!
Notes
Example
d d
Find (sin 4x) and compare it to (4 sin x).
dx dx
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 20 / 36
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6.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Example
3
2 Notes
Let f (x) = x5 − 2 + 8 . Find f (x).
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 21 / 36
A metaphor
Notes
Think about peeling an onion:
2
3
f (x) = x 5 −2 +8
5
√
3
+8
2
− 2)−2/3 (5x 4 )
3 1 5
f (x) = 2 x5 − 2 + 8 3 (x
Image credit: photobunny
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 22 / 36
Combining techniques
Notes
Example
d
Find (x 3 + 1)10 sin(4x 2 − 7)
dx
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 23 / 36
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7.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Your Turn
Notes
Find derivatives of these functions:
1. y = (1 − x 2 )10
√
2. y = sin x
√
3. y = sin x
4. y = (2x − 5)4 (8x 2 − 5)−3
z −1
5. F (z) =
z +1
6. y = tan(cos x)
7. y = csc2 (sin θ)
8. y = sin(sin(sin(sin(sin(sin(x))))))
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 24 / 36
Solution to #1
Notes
Example
Find the derivative of y = (1 − x 2 )10 .
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 25 / 36
Solution to #2
Notes
Example
√
Find the derivative of y = sin x.
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 26 / 36
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8.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Solution to #3
Notes
Example
√
Find the derivative of y = sin x.
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 27 / 36
Solution to #4
Notes
Example
Find the derivative of y = (2x − 5)4 (8x 2 − 5)−3
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 28 / 36
Solution to #5
Notes
Example
z −1
Find the derivative of F (z) = .
z +1
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 29 / 36
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9.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Solution to #6
Notes
Example
Find the derivative of y = tan(cos x).
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 30 / 36
Solution to #7
Notes
Example
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 31 / 36
Solution to #8
Notes
Example
Find the derivative of y = sin(sin(sin(sin(sin(sin(x)))))).
Solution
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 32 / 36
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10.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Outline
Notes
Heuristics
Analogy
The Linear Case
The chain rule
Examples
Related rates of change
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 33 / 36
Related rates of change at the Deli
Notes
Question
Suppose a deli clerk can slice a stick of pepperoni (assume the tapered
ends have been removed) by hand at the rate of 2 inches per minute, while
a machine can slice pepperoni at the rate of 10 inches per minute. Then
dV dV
for the machine is 5 times greater than for the deli clerk. This is
dt dt
explained by the
A. chain rule
B. product rule
C. quotient Rule
D. addition rule
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 34 / 36
Related rates of change in the ocean
Notes
Question
The area of a circle, A = πr 2 ,
changes as its radius changes. If
the radius changes with respect
to time, the change in area with
respect to time is
dA
A. = 2πr
dr
dA dr
B. = 2πr +
dt dt
dA dr
C. = 2πr
dt dt
D. not enough information
Image credit: Jim Frazier
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 35 / 36
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11.
V63.0121.041, Calculus I Section 2.5 : The Chain Rule October 5, 2010
Summary
Notes
The derivative of a
composition is the product
of derivatives
In symbols:
(f ◦ g ) (x) = f (g (x))g (x)
V63.0121.041, Calculus I (NYU) Section 2.5 The Chain Rule October 5, 2010 36 / 36
Notes
Notes
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