# lecture6 ```Lecture 6
More On The Derivative Rules
Derivative Rules
1. If f is a function and c a constant then
(cf) ‘ = c (f ‘)
2. If f and g are functions then
(f+g) ‘ = f ‘ + g ‘
3. (Leibnitz Rule) If f and g are functions
then (fg) ’ = f ’ g + f g ’
Where Does Product Rule Come From?
Recall key idea:
For the purposes of studying a function f “near a” we can
replace f by the function whose graph is the tangent line
to f at x = a.
If we have f and g we replace each by the tangent at x=a
f -&gt; f(a) + f ‘ (a)*(x-a) and g -&gt; g(a) + g ‘(a)*(x-a)
Then fg gets replaced by their product
f(a)g(a)  (f '(a)g(a)+f(a)g '(a))
( x  a ) f '(a)g'(a) ( x  a )
The derivative of this is
(f '(a)g(a)+f(a)g '(a))
Which when x = a is
 2 f '(a)g'(a) ( x  a )
f '(a)g(a)+f(a)g '(a)
2
More Derivative Rules
f
4. Quotient Rule
5. Power Rule
(
g
)‘ =
(f )‘
n
f ' g  f g'
g
2
( n  1 )
nf
f
‘
Power Rule Most often
Encountered in Following Form
5
 
6
Find f ‘ (x) if
5
f ‘( x) =
3
f ( x )  (  x  3 x )
 5

  1 
 6

3
(  x  3 x )
'
3
(  x  3 x )
6
f ‘( x) =
5
6
 5

  1 
 6

3
(  x  3 x )
(   9 x
2
)
Chain Rule
6.
f(g) '  f ' ( g ) g '
Power rule is a special case where
f( x ) x
n
There are really only 3 rules
2. Product rule
3. Chain rule
The others come from these –
Doesn’t hurt to remember the six.
h 
Quotient Rule Comes
from the Product Rule
f
g
f g h
(g h ) '  g ' h  g h '
f '  g '
f
g
 g h '
f ' g  g ' f g
2
h'
f
(
g
) ' 
f ' g  g ' f
g
2
Power Rule Comes from Chain Rule
Want
Then
(f
n
)'
Let
n
f  p( f )
p( x ) x
n
so we are looking for (p(f)) ’
By chain rule (p(f)) ’ = p ‘ (f)* f ‘
p '(x)  n x
( n  1 )
(f )‘
n
so
p '(f)  f
( n  1 )
nf
f
‘
( n  1 )
Rules Apply to More than Formulas
• The (red) graph of f(x) is
given at right and the blue
lines are tangent lines to
the graph. Using
information from the graph
and the derivative rules
estimate h’(3) where
h( x ) x f(x)
Solution: h ‘ (x) = (x ‘) f(x) + x f ‘ (x) = f(x) + x f ’(x)
h ‘(3) = f(3) + 3 f ‘ (3). From the graph we estimate that
f(3) = -3 and f ‘ (3) = -3/2 so we estimate
h ‘ (3) = -3 + 3 (-3/2) = -15/2
Example
Suppose f and g are functions such that f(5) = 4, f ‘ (5) = 7, g(5) = -3,
g ‘ (5) = 2. What is the equation of the tangent line to the graph of
h(x) = (fg) (x) at x = 5?
Solution:
(fg)(5) = f(5)(g(5) = (4)(-3) = -12.
(fg)’ (x) = f ‘(x)g(x) + f(x)g ‘(x)
so (fg)’(5) = f ‘ (5)g(5) + f(5)g ‘ (5) = 7(-3) + 4(2) = -13.
y = (fg)(5) + (fg)’(5)(x-5)
y = -12 -13(x-5)
Example
Suppose f is a function such that f(8) = -3 and
f ‘ (8) = 5. What is the slope of the tangent line
to the graph of
h( x ) ( 2 x  f( x ) )
9
at x = 8?
Sln:
8
h ' (x)  9 ( 2 x  f( x ) ) ( 2  f ' ( x ) )
8
h ' (8)  9 ( 2*8  3 ) ( 2  5 )
Example
Suppose f and g are functions such that f(7) = 2, f ‘(7) =11
g(3) = 7, and g ‘ (3) = 4.What is the equation of the
tangent line to the graph of h(x) = f(g)(x) at x = 3?
y – h(a) = h ‘ (a) ( x –a)
y – h(3) = h ‘ (3) ( x –3)
y – f(g(3)) = f ‘ (g(3))g ’(3) ( x –a)
y – f(7) = f ‘ (7)*4* ( x –a)
y – 2 = 11*4* ( x –a)
or y = 2 +44 ( x –a)
Example
Suppose f is a function such that f(4) = 5 and f ‘(4) = 2.
What is the slope of the tangent line at x = 2 to the graph
of
x
h( x ) 
2
f ( x )  x
2
x ' ( f ( x )  x )  x ( f ' ( x )  2 x )
h ' ( x ) 
2
2
( f ( x )  x )
h ' ( 4 ) 
2
1 ( f ( 4 )  4 )  4 ( f ' ( 4 )  2*4 )
2
2
( f ( 4 )  4 )
h ' ( 4 ) 
13
121
```