B2.2 & B2.3 - Product Rule
and Quotient Rules
MCB4U & IB Math HL/SL Santowski
(A) Review
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The derivative of a sum/difference is
simply the sum/difference of the
derivatives i.e. (f + g)` = f ` + g`
The power rule tells us how to find the
derivative of any power function y = xn
which works for any real value of n
(B) Product Rule – An Investigation
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Now the question concerns products of functions
 is the derivative of a product of functions the
same as the product of the derivatives? i.e. is
(fg)` = f` x g` ??
Let's investigate with a product function h(x) =
f(x)g(x) where f(x) = x² and g(x) = x² - 2x.
Thus h(x) = x²(x² - 2x)
(B) Product Rule – An Investigation
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Trial 1  if we go with our idea that (fg)` = f` x g` 
then (fg)` = (2x)(2x - 2) = 4x² - 4x
We can graph h(x) on the GC, graph the derivative and
then program in 4x² - 4x and compare it to the
calculated derivative from the GC. We will find that the
two do not match up!!!
If we simply tried expanding h(x) = x4 - 2x3 and then
taking the derivative, we would get 4x3 – 6x2. If we
program 4x3 – 6x2 into the GC, we find that we match
the GC generated derivative exactly.
(B) Product Rule – An Investigation
(C) Power Rule - Derivation
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So why does (fg)` = f ` x g` not work?
Let’s go back to limits and basic principles to find what
the differentiation technique should be if we wish to find
a derivative of a product
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Let K(x) = [f(x)] x [g(x)]
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Then K `(x) = lim h0 1/h[K(x + h) – K(x)]
And K `(x) = lim h0 1/h [ f(x+h)g(x+h) – f(x)g(x)]
which gets simplified as on the next slide:
(C) Power Rule - Derivation
K ( x)  lim
h 0
K ( x)  lim
K ( x  h)  K ( x )
h
f ( x  h)  g ( x  h)  f ( x )  g ( x )
h 0
K ( x)  lim
h
f ( x  h)  g ( x  h)  f ( x  h) g ( x )  f ( x  h) g ( x )  f ( x )  g ( x )
h 0
h
g ( x  h)  g ( x )
f ( x  h)  f ( x ) 

K ( x)  lim  f ( x  h) 
 g ( x) 

h 0
h
h


K ( x)  lim f ( x  h)  lim
h 0
h 0
g ( x  h)  g ( x )
h
K ( x)  f ( x)  g ( x)  g ( x)  f ( x)
 lim g ( x)  lim
h 0
h 0
f ( x  h)  f ( x )
h
(D) Power Rule - Examples
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ex 1. Find the derivative of f(x) = 3x4(5x3 + 5x - 7)
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ex 2. Find the derivative of f(x) = (x4-4x3–2x2+5x+2)2
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ex 3. Find the equation of the tangent to the function
f(x) = (2x + 4)(3x3 – 3x2 + x - 2) at (1,-6)
(E) Derivatives of Rational
Functions – The Quotient Rule
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Since the derivative of a product does not equal the
product of the derivatives, what about a quotient?
Would the derivative of a quotient equal the quotient of
the derivatives?
Since quotients are in one sense nothing more than
products of a function and a reciprocal  we would
guess that the derivative of a quotient is not equal to the
quotient of the derivatives
(F) Quotient Rule - Derivation
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First, set up a division
and then rearrange
the division to
produce a
multiplication so that
we can apply the
product rule
developed earlier
H ( x) 
f ( x)
g ( x)
f ( x)  H ( x)  g ( x)
f ( x )  H ( x )  g ( x )  H ( x )  g ( x )
H ( x ) 
f ( x )  H ( x )  g ( x )
g ( x)
f ( x ) 
H ( x ) 
f ( x)
 g ( x )
g ( x)
g ( x)
H ( x ) 
f ( x ) g ( x)  f ( x ) g ( x )
[ g ( x )]
2
(G) Examples Using the Quotient
Law
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Differentiate each of the following rational functions
ex 1.
f ( x) 
2x  5
3x  1
x 3
3
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ex 2.
ex 3.
g ( x) 
h( x ) 
1 4x
2
x
2
( x  2)( x  x )
4
2
(H) Internet Links
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Calculus I (Math 2413) - Derivatives Product and Quotient Rule
Visual Calculus - Calculus@UTK 3.2
solving derivatives step-by-step from
Calc101
(I) Homework
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IB Math HL/SL  Stewart, 1989, Chap
2.4, Q2eol,3eol,5,6,9b as well as Chap
2.5, Q2eol,3eol,4-7
MCB4U  Nelson text, Chap 4.4, p301
Q5,7,9,10,13 and Chap 5.4,
Q2,4,7,8,15,17