Calculus Of One Real Variable – By Pheng Kim Ving Chapter 3: Rules Of Differentiation – Section 3.3: Differentiation Of Compositions Of Functions – The Chain Rule 3.3 Differentiation Of Compositions Of Functions – The Chain Rule

 1. Compositions Of Functions

Consider an example. Let f(x) = x2 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 3
x + 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 composite function, of f and g is the function from D to range( f ) that is denoted by f o g, pronounced “ f circle g ” or “ f  round g ”, and that is defined by:   ( f o g)(x) = f( g(x))   for all x in D. Note that D = dom( f o g).

 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) is differentiable at x and:   ( 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.

# A Simplified Notation

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).

### Example 2.1

Differentiate g(x) = (3x + 4)2.

## Solutiong'(x) = 2(3x + 4)(3) = 6(3x + 4). EOS

We think of 3x + 4 as u (u = 3x + 4) and f(u) as u2 ( f(u) = u2). 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 = 9x2 + 24x + 16;
thus g'(x) = 18x + 24 = 6(3x + 4), the same as found by the chain rule.

 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 xn is nxn–1: (d/dx)xn = nxn–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

### Example 3.1

Differentiate g(x) = (3x + 4)2.

## Solutiong'(x) = 2(3x + 4)(3) = 6(3x + 4). EOS

This is the same function g as in Example 2.1, where we used the chain rule directly to differentiate g.

 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) xa = axa–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.

### Example 4.1 ## Solution EOS

 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.

### Example 5.1 ## Solution EOS

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.

 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 Solution  2.  Let y = f(u) = (u2 + 3u4)3/2 and u = g(x) = x33. 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 + x3|.

Solution b.  f '(x) = (sgn (2 + x3))(3x2) = 3x2 sgn (2 + x3). 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 sin3 x.

b.  Find v' if v = cos2 (5 – 4y3).

c.  Calculate (d/dt) ln (a ln(bt + c)).

Solution

a.  f '(x) = (cos cos sin3 x)(– sin sin3 x)(3 (sin2 x)(cos x)) = –3 sin2 x cos x cos cos sin3 x sin sin3 x.

b.  v' = 2 (cos (5 – 4y3))(sin (5 – 4y3))(–12y2) = 24y2 cos (5 – 4y3) sin (5 – 4y3).  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 = (–1/22)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(ab) < 0 < n(ba),
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.