Calculus Of One Real Variable – By Pheng Kim Ving 
3.3 
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1.
Compositions Of Functions 
Consider an example. Let f(x) = x^{2} and g(x) = 3x + 1. Then f( g(x)) = ( g(x))^{2} = (3x + 1)^{2}. We've obtained a new
function, whose value at x is (3x + 1)^{2}. This new
function is obtained by combining or composing functions f and g, and
thus is called the composition, or the composite function, of f and g. It's denoted by f o g, pronounced “ f circle g ” or
“ f round g ”. Hence, ( f o g)(x) = f( g(x)) = (3x + 1)^{2}.
In the notation f( g(x)), f is the outside
function and g the inside one. To compute ( f o g)(x) = (3x + 1)^{2}, first we
compute 3x + 1 = g(x), then we compute (3x + 1)^{2} = f( g(x)) = ( f o g)(x). Therefore, the
inside function g precedes

Fig. 1.1 Composition Of Functions f And g. 
the outside function f in the order of
computation of ( f o g)(x) or f( g(x)). See Fig. 1.1. Keep in mind that here we have
three different functions: f, g, and f o g. The domain of f o g is the set of all
x such that f( g(x)) exists or makes sense.
So it's the set of all x in the domain of g such that g(x) is in the domain
of f.
Definition 1.1
Let f and g be
functions and D the set of all x in dom( g) such that g(x) is in dom( f ). Then
the composition, or (
f o g)(x) = f( g(x)) for all x in D. Note that D = dom( f o g). 
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2. The Chain
Rule 
We're
going to establish in the following theorem the formula for the derivative of
the composition f o g of differentiable
functions f and g in terms of the
derivatives of f and g. Let u = g(x) and y = f(u) = f( g(x)) = ( f o g)(x). See Fig. 2.1.
Recall the interpretation of the derivative as the rate of change as discussed
in Section
2.3. Suppose du/dx = 3 and

Fig. 2.1 y = f(u) = f( g(x)) = ( f o g)(x). 
dy/du = 2. As x changes, u changes 3 times as fast as x does and y changes 2 times
as fast as u does, thus y changes
2 x 3 = 6 times as fast as x does. We've just observed
intuitively that the rate of change of y with respect to x is equal to
the rate of change of y with respect to u multiplied by the rate of change of u with respect to x: dy/dx = (dy/du)(du/dx).
That's
the same as to say that the derivative of f o g with respect to x is equal to the derivative of f
with respect to g(x)
multiplied by the derivative of g with respect to x: ( f o g)'(x) = f '( g(x))g'(x).
Theorem 2.1 – The Chain Rule
If u
= g(x)
is differentiable at the point x and y = f(u) is differentiable at the point g(x), then y = f( g(x)) = ( f o g)'(x) = f '( g(x))g'(x). In Leibniz notation:

Proof
Hence:
EOP
Note On
The Proof
Remark 2.1
In the Leibniz notation:
the du's appear to
cancel from the numerator and denominator of the two fractions. The notations dy/dx, dy/du, and
du/dx appear as
normal fractions. This is useful in remembering the formula.
The formula ( f o g)'(x) = f '(g(x))g'(x) can be written as ( f(g(x)))' = f '(g(x))g'(x). Now let u = u(x) = g(x). Then
this last formula can be written in the following somehow simplified
form, which is perhaps easier to remember:
( f(u))' = f '(u)u'(x), 
where ( f(u))' = (d/dx) f(u) (derivative of f(u) with repsect to x),
f '(u) = (d/du) f(u) (derivative of f(u) with respect to
u), and u'(x) = g'(x).
Differentiate g(x) = (3x + 4)^{2}.
We think of 3x
+ 4 as u (u
= 3x + 4) and f(u) as u^{2} ( f(u) = u^{2}). Then g(x) = f(u) and so g'(x) = (d/dx) f(u) = ( f(u))' =
f '(u)u'(x) = 2u(3) = 6(3x + 4). For this particular simple function g we can check: g(x) = (3x + 4)^{2} = 9x^{2} + 24x + 16;
thus g'(x) = 18x + 24 = 6(3x + 4), the
same as found by the chain rule.
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3. The Power
Rule – Differentiation Of Integer Powers Of Functions 
In Section
3.2 Corollary 4.1 we have that for any integer n,
the derivative of x^{n} is nx^{n}^{–1}: (d/dx)x^{n} = nx^{n}^{–1}. That's the
derivative of integer powers of variables. We now extend that to integer powers
of functions.
Corollary 3.1 – The Power Rule For Integer Exponents

Proof
Let y = (u(x))^{n}. Using the chain rule
we obtain:
EOP
Differentiate g(x) = (3x + 4)^{2}.
This is the same function g as
in Example 2.1, where we used the chain
rule directly to differentiate g.
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4. The Power
Rule – Differentiation Of Rational Powers Of Functions 
We now extend the power rule to
rational powers of functions. Recall that a rational number is a number
that can be
written as a ratio or fraction m/n, where m is an
integer and n a positive integer.
Corollary 4.1 – The Power Rule For Rational Exponents

Proof
There exist an integer m and a
positive integer n such that r = m/n. Utilizing the power rule for integer exponents
we
have:
EOP
We'll see in Section
6.3 Eq. [4.1] that (d/dx) x^{a}
= ax^{a}^{–1} for any real
number a,
from which we obtain, by the chain rule,
(d/dx) (u(x))^{a}
= a(u(x))^{a}^{–1}(du/dx) for any real
number a,
rational or irrational.
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5.
Differentiation Of Square Roots Of Functions 
In summary, for all x > 0 or for all
x where u(x) > 0:

Remark that the first formula was also
obtained in Section
3.2 Corollary 2.1.
This function h(t) was also
differentiated in Example 4.1 using the
power rule. As a matter of fact for the square root
function the square root rule as seen here is simpler than the power rule.
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6. Applying The Chain Rule More Than
Once In One Step 
Example 6.1
Solution
1
EOS
In the above solution, we apply
the chain rule twice in two different steps: first to differentiate the 10th
power, and then
to differentiate the 15th power. We can and it's better to apply all the instances of
the chain rule in just one step, as
shown in Solution 2 below.
Solution
2
EOS
Problems & Solutions 
1. Differentiate the following functions. You needn't simplify your
answers.
Solution
2. Let y = f(u) = (u^{2} + 3u – 4)^{3/2} and u = g(x) = x^{3} – 3. Find
( f o g)'(2) by:
a. Expressing y directly as a function of x and
differentiating.
b. Using the chain rule.
Solution
3. a. Show that (d/dx) x = sgn x, where sgn is the signum function defined by:
b. Find f '(x) if f(x) = 2 + x^{3}.
Solution
b. f '(x) = (sgn (2 + x^{3}))(3x^{2}) = 3x^{2} sgn (2 + x^{3}).
4. Use the formulas (d/dx)sin x = cos x, (d/dx) cos x = – sin x, and (d/dx) ln x = 1/x.
a. Find f '(x) if f(x) = sin cos sin^{3} x.
b. Find v' if v = cos^{2}^{
}(5 – 4y^{3}).
c. Calculate (d/dt) ln (a ln(bt + c)).
Solution
a. f '(x) = (cos cos sin^{3} x)(– sin sin^{3} x)(3 (sin^{2} x)(cos x)) = –3 sin^{2} x cos x cos cos sin^{3} x sin sin^{3} x.
b. v' = 2 (cos (5 – 4y^{3}))(–
sin (5 – 4y^{3}))(–12y^{2}) = 24y^{2}^{ }cos (5 – 4y^{3}) sin (5 – 4y^{3}).
5. Let y = (u + 1)/u. Suppose u = g(x) and g(3) = 2, where g is a differentiable
function, and suppose (dy/dx)_{x}_{=}_{3}
= – 5.
Find g'(3).
Solution
We have:
or – 5 = (–1/2^{2})g'(3) = (–1/4)g'(3). Thus, g'(3) = 20.
6. Let f(x) = (x – a)^{m}(x – b)^{n}, where a < b
and m and n are
positive integers. Prove that there exists
c where
a < c < b
such that the derivative of f vanishes
at c.
Solution
We have:
Using the facts that a
< b, m >
0, and n > 0 we get:
a – b < 0 < b – a,
m(a – b) < 0 < n(b – a),
ma – mb < 0
< nb – na,
ma – mb + na – na < 0
< nb – na + mb – mb,
a(m + n) – (mb + na) <
0 < b(m + n) – (mb + na),
a(m + n) < mb + na < b(m + n),
completes the proof.
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