2.3: Product and Quotient Rules and Higher Order Derivatives

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AP CALCULUS
Section Number:
LECTURE NOTES
Topics: Differential Technique 1
- Product Rule and Quotient Rule
- Higher Order Derivatives
- Absolute Value and the Derivative
2.3
MR. RECORD
Day: 13
Although the derivative of the sum (and difference) of two functions is simply the sum (or difference) of
their derivatives, the product and quotient of two functions are handled quite differently when
differentiating.
1. The Product Rule
The Product Rule
The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first
function times the derivative of the second, plus the second function times the derivative of the first function.
d
 f ( x ) g ( x )   f  ( x ) g ( x )  f ( x ) g ( x )
dx
or
f ( x) g ( x)  g ( x) f ( x)
Proof can be seen on the Power Point Presentation.
Example 1: Find the derivative of h( x)  (3x  2 x 2 )(5  4 x) .
Did we have to use the product rule for Example 1? What other way could we have found h( x) ?
Example 2: Find the derivative of each of the following
a.
y  x sin x
b.
y  2 x cos x  2sin x
2. The Quotient Rule
The Quotient Rule
The quotient, f/g, of two differentiable functions f and g is itself differentiable at all values for x for which g ( x )  0 .
Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the
numerator times the derivative of the denominator, all divided by the square of the denominator.
d  f ( x )  f  ( x ) g ( x )  f ( x ) g ( x )

2
dx  g ( x) 
 g ( x) 
The proof for the Quotient Rule is similar to that of the Product Rule.
Example 3: Find the derivative of y 
5x  2
x2  1
Example 4: Find the derivative of y 
3  (1/ x)
x5
Not every function that contains a quotient requires use of the Quotient Rule.
Example 5: Complete the table.
Original Function
a. y 
b.
y
x 2  3x
6
3(3x  2 x 2 )
7x
Rewrite
Differentiate
Simplify
3. Higher Order Derivatives
Just as you can obtain a velocity function from differentiating a position function, you can obtain an
acceleration function by differentiating a velocity function.
s (t ) Position function
v(t )  s(t ) Velocity function
a(t )  v(t )  s(t ) Acceleration function
Common Higher Order Derivative Notation
First derivative
y
f ( x )
Example 6:
Second derivative
y
f ( x )
Third derivative
y 
f ( x)
Fourth derivative
y (4)
f (4) ( x)
nth derivative
y (n)
f ( n ) ( x)
dy
dx
d2y
dx 2
d3y
dx 3
d4y
dx 4
dny
dx n
Because the moon has no atmosphere, a falling object on the moon encounters no air
resistance. In 1971, astronaut David Scott demonstrated that a feather and a hammer fall
at the same rate on the moon. The position function for each of these falling objects is
given by
s(t )  0.81t 2  2
where s (t ) is the height in meters and t is the time in seconds. What is the ratio of the
earth’s gravitational force to the moon’s?
Definition of Absolute Value of x
 x, x  0
x 
  x, x  0
Example 7: Find the derivative of
f ( x)  2 x  5 .
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