Differentiating
“Combined” Functions
Deriving the Product Rule for
Differentiation.
Algebraic Combinations
 We have seen that it is fairly easy to compute the
derivative of a “simple” function using the
definition of the derivative.
 More complicated functions can be difficult or
impossible to differentiate using this method.
 If we know the derivatives of two fairly simple
functions, we candeduce the derivative of an
algebraic combination (e.g. the sum, product,
quotient) of these functions without going back
to the difference quotient.
A Gambier Example
LARGE RATS!!!!
A Gambier Example
The rate at which the deer gobble my hostas is
proportional to the product of the number of deer and
the number of hostas. So we have a gobble function:
g (t )  k RL (t ) H (t )
What can we say about the rate of change of this function?
What if over a short period of time t, from t to t + t the deer population
increases by a small amount by RL and the hosta population increases by
H. How much does the gobble rate change between time t and time t + t ?
A Gambier Example
g (t )  k RL (t ) H (t )
So we need to compute g (t  t )  g (t ) :
g (t  t )  k  RL (t )  RL  H (t )  H 
 k RL (t ) H (t )  k RL (t )H  k RL H (t )  k RL H
g (t  t )  g (t )  k RL (t )H  k RL H (t )  k RL H
A Gambier Example
g (t )  k RL (t ) H (t )
The change in g has three relevant pieces:
k RL (t )H
k RL H (t )
k RL H
Old Rats eating poor baby hostas
“Cute” baby rats eating vulnerable old hostas
“Cute” baby rats eating poor baby hostas
A Gambier Example
Now we consider g(t)= lim
t 0
g (t  t )  g (t )
:
t
k RL (t )H  k RL H (t )  k RL H
g (t  t )  g (t )
lim
 lim
t 0
t 0
t
t
 lim
t 0
k RL (t )H
k RL H (t )
k RL H
 lim
 lim
t 0
t 0
t
t
t
 k RL (t ) lim
t 0
RL
H
H
 k H (t ) lim
 lim k RL lim
t 0 t
t 0
t 0 t
t
0
SO, the rate of change of
gobble is given by . . .
g (t  t )  g (t )
lim
 k RL (t ) H (t )  k H (t ) RL (t )
t 0
t
The rate at which the number of
hostas is changing times the
number of large rats.
The rate at which the number of
large rats is changing times the
number of hostas.
The Product Rule for
Derivatives
So the rate at which a product changes is not merely
the product of the changing rates.
The nature of the interaction between the functions,
causes the overall rate of change to depend on the
size of the quantities themselves.
In general, we have…
f ( x  h) g ( x  h )    f ( x ) g ( x ) 

d
 f ( x) g ( x)   lim
h 0
dx
h
f ( x  h ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x )
 lim
h 0
h
g ( x  h)  f ( x  h)  f ( x)   f ( x)  g ( x  h)  g ( x) 

 lim
h 0
h
f ( x  h)  f ( x )
g ( x  h)  g ( x )
 lim g ( x  h) lim
 f ( x) lim
h 0
h 0
h 0
h
h
 f ( x) g ( x)  f ( x) g ( x)
Continuity Required!
In the course of the calculation above, we said that
lim g ( x  h)  g ( x)
h 0
Is this actually true? Is it ALWAYS true?
lim g ( x  h)
h 0
g ( x)
x
x+h
The Limit of the Product of
Two Functions
f ( x  h) g ( x  h )    f ( x ) g ( x ) 

d
 f ( x) g ( x)   lim
h 0
dx
h
f ( x  h ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x )
 lim
h 0
h
g ( x  h)  f ( x  h)  f ( x)   f ( x)  g ( x  h)  g ( x) 

 lim
h 0
h
f ( x  h)  f ( x )
g ( x  h)  g ( x )
 lim g ( x  h) lim
 f ( x) lim
h 0
h 0
h 0
h
h
 f ( x) g ( x)  f ( x) g ( x)
Would be zero if g
were continuous at
x=a. Is it?
The Limit of the Product of
Two Functions
f ( x  h) g ( x  h )    f ( x ) g ( x ) 

d
 f ( x) g ( x)   lim
h 0
dx
h
f ( x  h ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x  h)  f ( x ) g ( x )
 lim
h 0
h
g ( x  h)  f ( x  h)  f ( x)   f ( x)  g ( x  h)  g ( x) 

 lim
h 0
h
f ( x  h)  f ( x )
g ( x  h)  g ( x )
 lim g ( x  h) lim
 f ( x) lim
h 0
h 0
h 0
h
h
 f ( x) g ( x)  f ( x) g ( x)
g is continuous at
x=a because g is
differentiable at x=a.