Derivation of Some Differentiation Rules
These notes are intended to provide methods of deriving some of the formulas
used in differentiation which are different from those described in the textbook. We
will be making use of the limit definition ( “h-definition” ) of the derivative of a function,
f ' ( x ) = lim
h→0
f (x + h) − f (x)
h
.
Since we can (and will often need to) construct new functions by combining simpler
functions through arithmetic operations or by composition of one function on another,
we will need €
to know how to differentiate these newly-created functions.
Product Rule
We can create a function F ( x ) = f ( x ) · g ( x ) through multiplication of two
simpler functions. In calculating its derivative F ’( x ) , it will be convenient to define a
symbol for the change in a function by Δf = f ( x + h ) – f ( x ) , in order to save a bit of
writing in places. So we will have f ( x + h ) = f ( x ) + Δf and we will need to apply
binomial multiplication:
F ' ( x ) = [ f ( x ) g( x ) ]' = lim
h→0
= lim
[ f ( x ) + Δ f ] [ g( x ) + Δg ] − f ( x ) g( x )
h
= lim
[ f ( x ) g( x ) + f ( x ) Δg + g( x ) Δ f + Δ f Δg ] − f ( x ) g( x )
h
= lim
f ( x ) Δg + g( x ) Δ f + Δ f Δg
h
h→0
€ h→0
€
f ( x + h ) g( x + h ) − f ( x ) g( x )
h
h→0
€
.
At this point, we can now express this result as the sum of three separate limits
and write out explicitly the changes in the functions f and g :
€
[ f ( x ) g( x ) ]' = lim
h→0
€
= lim
f ( x )[ g( x + h ) − g( x ) ]
g( x )[ f ( x + h ) − f ( x ) ]
+ lim
h→0
h
h
+ lim
h→0
f ( x ) Δg
g( x ) Δ f
Δ f Δg
+ lim
+ lim
h→0
h→0
h
h
h
€
[ f ( x + h ) − f ( x ) ][ g( x + h ) − g( x ) ]
h→0
h
(continued)
€
= f ( x ) ⋅ lim
h→0
€ + lim
h→0
g( x + h ) − g( x )
f (x + h) − f (x)
+ g( x ) ⋅ lim
h→0
h
h
f (x + h) − f (x)
⋅[ g( x + h ) − g( x ) ] ,
h
where we have extracted the factor which does not depend on h in the first two of
these limit terms, and have simply separated one factor in the third limit. By applying
€
the limit definition
of a derivative function, we at last have
[ f ( x ) g( x ) ]' = f ( x ) ⋅ g ' ( x ) + g( x ) ⋅ f ' ( x )
⎡
f ( x + h ) − f ( x ) ⎤
+ ⎢ lim
g( x + h ) − g( x ) ]
⎥⎦ ⋅ [ hlim
→0
⎣ h → 0
h
€
“the limit of a product is the product of the limits”
= f ( x ) ⋅ g ' ( x ) + g( x ) ⋅ f ' ( x ) + f ' ( x ) ⋅[ g( x + 0) − g( x ) ] *
€
= f ( x ) ⋅ g ' ( x ) + g( x ) ⋅ f ' ( x ) + f ' ( x ) ⋅ 0 = f ( x ) ⋅ g ' ( x ) + g( x ) ⋅ f ' ( x ) .
€ * We would obtain a similar result if we separated out the factor involving f instead.
€ More simply, the Product Rule is often expressed as ( f g ) ’ = f ‘ · g + f · g ’ . Notice
that this Rule can be repeatedly applied to work out the derivative for a product of more
than two functions; for three functions, for instance,
( f ⋅ g ⋅ h)' = ( f ⋅ g)' ⋅ h + ( f ⋅ g) ⋅ h'
= ( f ' ⋅ g + f ⋅ g ' ) ⋅ h + ( f ⋅ g ) ⋅ h ' = f ' ⋅ g h + f ⋅ g ' ⋅ h + f g ⋅ h ' . In other words, the derivative of any product of functions can be expressed by a set of
terms in which each function is differentiated in turn and multiplied by all the other
€ in the set. The Product Rule applies at those values of x for which every one
functions
of the functions in the product is continuous.
€
Quotient Rule
We take a similar approach here with a new function defined by the ratio of two
functions, G ( x ) =
f (x)
g( x ) . Naturally, we expect the algebra here to be a little more
complicated.
⎡ g( x ) ⋅ f ( x + h ) − g( x + h ) ⋅ f ( x ) ⎤
f (x + h)
f (x)
−
⎢⎣
⎥⎦
g( x + h ) ⋅ g( x )
⎡ f ( x ) ⎤'
g( x + h )
g( x )
G€
' ( x ) = ⎢
=
lim
=
lim
⎣ g( x ) ⎥⎦
h→0
h→0
h
h
subtracting fractions in the numerator
= lim
€
h→0
€
g( x ) ⋅[ f ( x ) + Δ f ] − f ( x ) ⋅[ g( x ) + Δg ]
h ⋅ g( x + h ) ⋅ g( x )
(continued) =
€
1
f ( x ) g( x ) + g( x ) Δ f − f ( x ) g( x ) − f ( x ) Δg
⋅ lim
g( x ) h → 0
h ⋅ g( x + h )
using our expression for f ( x + h ) and g ( x + h )
and extracting a factor which does not involve h
=
1
g( x ) Δ f − f ( x ) Δg
⋅ lim
g( x ) h → 0
h ⋅ g( x + h )
.
We will now write out the changes in the functions f and g again, so that we can apply
the limit definition of derivative:
€
⎡
⎧ g( x ) ⋅ [ f ( x + h ) − f ( x ) ] − f ( x )[ g( x + h ) − g( x ) ] ⎫ ⎤
⎡ f ( x ) ⎤'
1
1
⋅ ⎨
⎬ ⎥ ⎢
⎢⎣ g( x ) ⎥⎦ = g( x ) ⋅ hlim
→ 0 ⎣ g( x + h )
h
⎩
⎭ ⎦
=
€
⎤ ⎡
1 ⎡
1
f (x + h) − f (x)
g( x + h ) − g( x ) ⎤
⋅ ⎢ lim
− lim f ( x ) ⋅
⎥ ⋅ ⎢ lim g( x ) ⋅
⎥ h→0
g( x ) ⎣ h → 0 g( x + h ) ⎦ ⎣ h → 0
h
h
⎦
=
€
=
€
⎤ ⎡
1 ⎡
1
f (x + h) − f (x)
g( x + h ) − g( x ) ⎤
⋅ ⎢ lim
− f ( x ) ⋅ lim
⎥ ⋅ ⎢ g( x ) ⋅ lim
⎥ h→0
h→0
g( x ) ⎣ h → 0 g( x + h ) ⎦ ⎣
h
h
⎦
1
g( x )
extracting factors which do not involve h
⎡
⎤
1
g( x ) ⋅ f ' ( x ) − f ( x ) ⋅ g ' ( x )
⋅ ⎢
⎥ ⋅ [ g( x ) ⋅ f ' ( x ) − f ( x ) ⋅ g ' ( x )] =
g(
x
+
0
)
[ g( x ) ]2
⎣
⎦
.
The Quotient Rule applies at those values of x for which both f ( x ) and g ( x ) are
continuous and where g ( x ) ≠ 0 (that is, where f ( x )/g( x ) is defined and thus
€
continuous) .
Chain Rule
It is a bit more of a challenge to differentiate a composite function, which is
formed by taking the result of one function and subjecting it to the operation of a
second function. So we need to be somewhat careful about what the changes in the two
functions mean. Applying the limit definition of derivative to the composite function
H ( x ) = f ( g ( x ) ) , we have
H ' ( x ) = [ f ( g( x ) ) ] ' = lim
h→0
= lim
€
h→0
f ( g( x + h ) ) − f ( g( x ) )
h
f ( g( x ) + Δg ) − f ( g( x ) )
h
.
We write the last expression in this way as a reminder that the change in the composite
function f ( g ( x ) ) is connected to the change in the function g ( x ) . When we then
use our
€ way of showing the shift in the value of the first term of the numerator to write
(continued)
[ f ( g( x ) ) ] ' = lim
h→0
[ f ( g( x ) ) + Δ f ] − f ( g( x ) )
,
h
it is then perhaps easier to keep in mind that this change in the function f , Δf , is
dependent upon the change in the function g , Δg (whereas in our derivations of the
€ Product and Quotient Rules above, these changes were not connected). We can now say
[ f ( g( x ) ) ] ' = lim
h→0
€
f ( g( x ) ) + Δ f − f ( g( x ) )
Δf
= lim
h→0 h
h
= lim
Δf
Δg
Δf
Δg
⋅
= lim
⋅ lim
h → 0 Δg
h→0 h
Δg
h
= lim
Δf
g( x + h ) − g( x )
Δf
⋅ lim
= lim
⋅ g'(x) .
h → 0 Δg
Δg h → 0
h
h→0
h→0
€
What remains to be understood is this first limit term. Since it is certainly the
case that Δg approaches zero as h approaches zero, we can think of this limit as
€
lim
h→0
Δf
Δf
f ( g( x ) + Δg ) − f ( g( x ) )
= lim
= lim
Δg → 0 Δg
Δg → 0
Δg
Δg
, reverting the numerator to a form it had earlier. But this resembles the limit definition
€ for f ‘ ( x ) , lim
h→0
f (x + h) − f (x)
, with Δg standing in for h and g ( x ) in place
h
of x . This limit in question then gives the derivative function f ‘ ( u ) evaluated at the
value u = g ( x ) . This permits us to write the Chain Rule for differentiation of a
composite function,
€
[ f ( g( x ) ) ] ' = lim
h→0
Δf
⋅ g' ( x ) = f ' ( u )
Δg
u = g(x)
⋅ g' ( x ) ,
or, as it is often more simply written, [ f ( g( x ) ) ] ' = f ' ( g( x ) ) ⋅ g' ( x ) .
The Chain Rule applies at those values of x for which both g ( x ) and f ( g ( x ) ) are
€
continuous.
€
Derivatives of f ( x ) = sin x
and g ( x ) = cos x
These are the first of the elementary functions we encounter where something
more than simple algebra is required in order to work out their derivative functions. We
will need to construct a couple of new “limit laws” for the purpose.
The first of these is to find the value for
lim
x→0
sin x
. One method of
x
calculating this is provided in the textbook (Stewart, 6th ed., pp. 190-191). A couple of
others are shown here to offer alternative approaches.
€
For any of these methods, we must consider a wedge of a circle of radius 1 ,
with center at point O and the angle ∠AOB having measure (size) θ . The area of this
wedge is Aw = ½ · r2 ·
θ = ½ · 12 · θ = ½ θ .
We can extend a line downward from point A which is perpendicular to the line
OB and meets it at point C to form the right triangle ΔOCA . From trigonometry, we
know that, since the hypotenuse OA is a radius of the circle and so has a length of 1 ,
then OC has length cos θ and AC has length sin θ . The segments OC and AC are
the base and altitude of the right triangle
ΔOCA , so its area is AOCA = ½ · cos θ · sin θ .
We can then also extend a line upward from point B which is perpendicular to
the line OB , and we will also extend the segment OA . These lines meet at a point D ,
allowing us to make another right triangle ΔOBD . Since OB is a radius of the circle, it
has a length of 1 . Again, from trigonometry, the altitude of this triangle BD has a
height h , thus h / 1 = tan θ ⇒ h = tan θ . As the segments OB and BD are the
base and altitude of this right triangle, its area is AOBD = ½ · 1 · tan
θ = ½ tan θ .
The wedge of the circle is enclosed between these two right triangles, so we can
write the inequality for the areas of these geometrical figures as
AOCA < Aw < AOBD
⇒
1
1
1
cos θ sin θ < θ < tan θ
2
2
2
If we now divide the inequality through by ½ sin
the angle approaches zero, we have
€
.
θ and take the limit of the terms as
(continued)
1
2
cos θ sin θ
1
2
<
sin θ
⇒
€
1
2
1
θ
2
sin θ
1
2
1
2
<
lim + cos θ <
θ →0
tan θ
⇒ cos θ <
sin θ
lim +
θ →0
θ
<
sin θ
lim +
θ →0
θ
1
<
sin θ
cos θ
1
cos θ
.
Upon evaluating the limits at each end of the inequality, we find 1 <
€
and therefore, by the “Squeeze Theorem”,
lim +
θ →0
lim +
θ →0
θ
1
<
,
sin θ
1
θ
= 1 . By another of the limit
sin θ
laws, we can now write
€
lim +
θ →0
sin θ
=
θ €
1
lim
θ → 0+
θ
sin θ
=
1
= 1 ,
1
giving us our new trigonometric limit law.
€
Another method involves lengths of lines and arcs, rather than the areas of
wedges and triangles. We start once again with the wedge of the unit circle, OAB . Since
the angle ∠AOB has measure θ , the length of the arc AB is sw = θ · r = θ · 1 = θ .
We again “drop” a perpendicular line from point A to the line OB to form the
right triangle ΔOCA . This time, we are interested in the length of this line, which is
the altitude of the triangle we earlier found to be LAC = sin
θ.
We will now make a new circular wedge using the segment OC as the radius.
The angle ∠DOC must also have measure θ . We know that OC has length cos θ , so
the length of the arc CD is sCD =
θ · rCD = θ · cos θ .
The way in which the altitude of the right triangle falls between the arcs of the
two wedges gives us the inequality
sCD < LAC < sw
€
⇒ θ cos θ < sin θ < θ .
(continued)
We divide this inequality through by the angle
angle approaches zero:
θ cos θ
sin θ
θ
<
<
θ
θ
θ
which gives us
€
lim +
θ →0
lim + cos θ <
⇒
θ →0
lim +
θ →0
θ and take the limits of the terms as this
sin θ
<
θ
lim + 1 ⇒ 1 <
θ →0
lim +
θ →0
sin θ
< 1 ,
θ
sin θ
= 1 by the “Squeeze Theorem”.
θ
We can proceed from this result to the other limit law we will need. We can
make a€product of certain limits and then use the already known limit laws to write
θ →0
⎛ 1 ⎞
sin θ
1
⋅ lim + sin θ ⋅ lim +
= 1 ⋅ 0 ⋅ ⎜
⎟ = 0
θ
θ →0
θ → 0 1 + cos θ
⎝ 1 + 1 ⎠
⇒
lim +
sin θ
1
⋅ sin θ ⋅
=
θ
1 + cos θ
lim +
(1 − cos 2 θ )
= 0 ⇒
θ (1 + cos θ )
lim +
€
⇒
€ € θ →0
θ →0
lim +
θ →0
θ →0
lim +
θ →0
applying the Pythagorean Identity
⇒
lim +
sin 2 θ
= 0
θ (1 + cos θ )
(1 − cos θ ) ⋅ (1 + cos θ )
= 0 θ (1 + cos θ )
factoring difference of two squares
(1 − cos θ )
= 0 .
θ
safe to divide through, since
lim (1 + cos θ ) ≠ 0
θ →0
€ We now have the trigonometric limit laws we need to calculate the derivative
€
functions for sin x and cos x . Using the “angle-addition formulas” for sine and cosine
(discussed in another Note), we have the limits
[ sin x ]' = lim
h→0
= lim
€
h→0
€
( cos h − 1)
sin h
+ lim cos x ⋅
h→0
h
h
= lim sin x ⋅
⎡
( cos h − 1) ⎤ ⎡
sin h ⎤
= ⎢ lim sin x ⋅ lim
+ ⎢ lim cos x ⋅ lim
⎥
⎥ h→0
h → 0 h ⎦
⎣ h → 0
⎦ ⎣ h → 0
h
= (sin x ⋅ 0) + ( cos x ⋅ 1) = cos x
€
€
(sin x cos h − sin x )
( cos x sin h )
+ lim
h→0
h
h
€
sin ( x + h ) − sin x
(sin x cos h + cos x sin h ) − sin x
= lim
h
→
0
h
h
h→0
and [ cos x ]' = lim
h→0
= lim
€
h→0
( cos x cos h − cos x )
(− sin x sin h )
+ lim
h→0
h
h
( cos h − 1)
sin h
− lim sin x ⋅
h→0
h
h
= lim cos x ⋅
⎡
( cos h − 1) ⎤ ⎡
sin h ⎤
= ⎢ lim cos x ⋅ lim
− ⎢ lim sin x ⋅ lim
⎥
⎥ h→0
h → 0 h ⎦
⎣ h → 0
⎦ ⎣ h → 0
h
= ( cos x ⋅ 0) − (sin x ⋅ 1) = − sin x
h→0
€
cos ( x + h ) − cos x
( cos x cos h − sin x sin h ) − cos x
= lim
h→0
h
h
€
. €
Derivative of the general exponential function
€
This is another function that requires some specific handling and also touches
upon topics beyond the scope of Calculus I. We can apply the limit definition of
derivative to the general exponential function f ( x ) = ax , with a > 0 , to obtain
[ a x ]' = lim
h→0
ax + h − ax
(ax ⋅ ah ) − ax
ah − 1
= lim
= lim a x ⋅
h→0
h→0
h
h
h
applying properties of exponents
€
= a x ⋅ lim
h→0
ah − 1
. h
extracting factor which
does not involve h
We are
€ not in a position to evaluate this last limit (we will know how to do that in
Calculus II), but we can recognize that this is the point derivative for our function,
f ‘ ( 0 ) , the slope of the tangent line to the exponential function y = ax at x = 0 (as
discussed in Stewart, 6th ed., pp. 178-79).
By experimenting with different values of a > 0 , we find that this limit has a
value which depends upon the value of a . Mathematicians basically assign a name to
the value at which this limit is exactly 1 ; that number is called ‘ e ‘ . (This is to say
that we don’t prove that e is the number for which this limit is 1 ; instead, we prove
that there must be such a number and the value at which this occurs is approximately
2.718281828… , which is designated as the constant ‘ e ‘ .) So we can say that
lim
h→0
eh − 1
= 1
h
and thus [ e x ]' = e x ⋅ lim
h→0
eh − 1
= ex ⋅ 1 = ex
h
. The function ex is thus a function which is its own derivative function; in fact, it is the
only (non-constant) function for which that is the case. Because it emerges directly from
€ the structure of mathematics, ex is called the “natural exponential function”.
We can take this a bit further by looking at the function g ( x ) =
the limit definition of derivative yields
[ e kx ]' = lim
h→0
€
ekx , for which
ek ( x + h ) − ek x
ek h − 1
= e k x ⋅ lim
,
h→0
h
h
following the argument we used above for ax . Now if k is a positive integer, we can
write the numerator of the ratio in the limit expression as ( eh )k – 1 , and apply the
so-called “geometric expression”,
x k − 1 = ( x − 1) ⋅ ( x k −1 + x k − 2 + K + x 2 + x + 1 ) ,
144444424444443
k terms
to re-write the derivative function as
kx
[ e kx ]'
€ = e ⋅ lim
h→0
= e k x ⋅ lim
h→0
€
€
= e
kx
(e h ) k − 1
h
(e h − 1) ⋅ ([ e h ]k −1 + [ e h ]k − 2 + K + [ e h ]2 + e h + 1)
h
(e h − 1)
⋅ lim
⋅ ([ e h ]k −1 + [ e h ]k − 2 + K + [ e h ]2 + e h + 1 ) 144444444
42444444444
3
h→0
h
k terms
= e k x ⋅ 1 ⋅ ([ 1]k −1 + [ 1]k − 2 + K + [ 1]2 + 1 + 1 ) = e k x ⋅ 1 ⋅ k .
14444444244444443
k terms
€
kx
kx
Hence, we have shown that [ e ]' = k ⋅ e
, at least when k is a positive
integer. This is akin to the proof we’ve given earlier in the course that
€ [ x n ]' = n ⋅ x n −1
, where n is a positive integer (see, for example, Stewart, p. 174).
We can now show immediately that for a =
[ a x ]' = k ⋅ a x
€
€
€
ek , with k being a positive integer , that
. But from what we’ve learned prior to this course,
x
x
a = e
⇒ k = ln a . So we can argue plausibly that [ a ]' = ( ln a ) ⋅ a , even
though we have really only so far shown it to be true when ln a is a positive integer.
We will be able to demonstrate (elsewhere) the derivative rule for ax more generally
using the Chain Rule. From the discussion earlier, we have also shown that the slope of
k
the tangent line to f ( x ) = ax at x = 0 is
€
€
f ' ( 0 ) = lim
h→0
ah − 1
= ln a .
h
-- G. Ruffa
May – June 2010