Math 1210-001
Tuesday Feb 9
WEB LilO
2.5: Chain rule
The chain rule tells us how to differentiate compositions of ffinctions. In other words, it tells us how to
compute the rate of change of a composition two fhnctions, in terms of the individual rates of change for
each function. The chain rule malces intuitive sense when you think of real-world examples of rates of
change, and keep track of units, so we’ll start with two of those. Then we’ll write the chain rule
mathematically, see the mathematical reason it’s true in general, and begin practicing it. You’ll need to do a
lot of practice outside of class as well, in order to develop proficiency in computations.
miles
Exercise 1) Suppose you are drivmg south at 75 hour , and the outside temperature is mcreasmg at a
.
rate of 0.04 derees in the southerly direction. How fast is the outside temperature you’re experiencing
changing?
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Remark Notice in the example above, that there is an underlying composition of two functions: Your
location south is a function, say g(t) miles south (of e.g. Salt Lake City) at time t hours, with g’ (t) = 75
miles
hour And there is a temperature function T(y) degrees at location y miles south,
.
.
.
with T’(y)
.04 derees
Youwereaskedto fmd D~(T(g(t))).
Exercise 2 We consider a small town’s reservoir and spring run-off. The volume Vof the town’s reservoir
(in acre-feet) depends on the water depth u feet, measured from the deepest point in the reservoir, i.e.
V f( is). The depth of the reservoir at day I of the calendar year is a function oft, is g(I). Suppose that
at! 100 (March something) we have
100
day
u 20fi
V 5000 acre—f!.
Consult the two graphs below to answer questions ~&. Use secant approximations for ~and tangent
approximations for ç..
a)
much does the water depth increase between! = 100, 101?
b)
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much does the reservoir volume increase between I = 100, 101?
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(44-
c) About how fast is the reservoir volume changing at!
track of units.)
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The moral of Exercises 1,2, is that when functions are composed, the rate of change of the composition is
the product of the rates of change for each function being composed. This is what the chain rule says.
Let’s work our way to the precise statement and its proof using the A (“change in”) notation.
For u = g(t) we write “At” for “change in t”, instead of ‘h”.
and we write “Au” for the corresponding change in u as t increments to t + At,
Au=g(t+At) —g(t).
So, converting notations,
h) —g(t)
Au
D~g(t)= lim g(t+
=
lim
At
0 &
h
—
—‘
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Use these notation conventions, and consider the general composition
ug(t)
g’(t)
Vf(u)
f’(u)
Au
At—’ 0At
AV
lim
Au
0Au
lim
—4
Then
D1f(g(t))= lim
At—’ 0 &
AVAu
(assuming Au # 0)
AI4OAu
&
f’(u)~g’(t)
Chain Rule!
D~(f(g(t)) =f’(g(tflg’(t)
‘The derivative of the composition function is the derivative of the outside function evaluated at the inside
function, times the derivative of the inside function.” OR “The rate of change of the composition of two
functions is the product of their rates of change,” evaluated at the appropriate input values.
Exercise 3) Write each composition asfo g, and use the chain rule to fmd the derivative of the
composition function.
F(x)
=
(x3 +
(check answer with “foil)
‘3
1) f
U
‘1
i-pc’)
(;Z~1
z3x~ j
1- K
+(L4J~Lct
3~)
F(t) = sin(2 t) (understand your answer geometrically by looking at the graph below, from
Monday’s notes).
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ía
it
3m
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1
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0.5
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it
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/2
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