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


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