Techniques of Differentiation
I. Notations for the Derivative
The derivative of y  f (x) may be written in any of the following ways:
f (x) ,
y ,
dy
,
dx
d
 f (x) ,
dx
or Dx  f (x) .
II. Basic Differentiation Rules
A. Suppose c and n are constants, and f and g are differentiable functions.
(1) f ( x)  cg ( x)
f (b)  f ( x)
cg (b)  cg ( x)
g (b)  g ( x)
 lim
 c lim
 cg ( x)
b

x
b

x
b

x
b x
b x
b x
f ( x)  lim
(2) f ( x)  g ( x)  k ( x)
f (b)  f ( x)
[ g (b)  k (b)]  [ g ( x)  k ( x)]
 lim

bx
bx
b x
b x
f ( x)  lim
g (b)  g ( x)
k (b)  k ( x)
 lim
 g ( x)  k ( x)
bx
bx
b x
b x
lim
(3) f ( x)  g ( x)k ( x)
f (b)  f ( x)
g (b)k (b)  g ( x)k ( x)
 lim

bx
bx
b x
b x
f ( x)  lim
g (b)k (b)  g (b)k ( x)  g (b)k ( x)  g ( x)k ( x)

bx
b x
lim



k (b)  k ( x)  
g (b)  g ( x) 
 lim g (b)  lim
   lim k ( x)  lim

bx 
b  x
 b  x b  x  b  x
 b  x
g ( x)k ( x)  k ( x) g ( x) (Product Rule)
1
g ( x)
 f ( x)k ( x)  g ( x)  g ( x)  f ( x)k ( x)  k ( x) f ( x) 
k ( x)
 g ( x) 
g ( x)  
k ( x)
k ( x) 
g ( x)  f ( x)k ( x)
k ( x) g ( x)  g ( x)k ( x)

.
f ( x) 


2
k ( x)
k ( x)
k ( x)
(4) f ( x) 
This derivative rule is called the Quotient Rule.
(5) f ( x)  c
f (b)  f ( x)
cc
0
 lim
 lim
 lim 0  0
bx
b x
b x b  x b x b  x b x
f ( x)  lim
(6) f ( x)  x
f (b)  f ( x)
bx
 lim
 lim 1  1
bx
b x
b x b  x b x
f ( x)  lim
(7) f ( x)  x n
f ( x  h)  f ( x )
( x  h) n  x n
 lim

h
h
h 0
h 0
f ( x)  lim
n(n  1) n  2 2
 n

n 1
x
h  ...  x n
 x  nx h 
2



lim
h
h 0
 n 1

2  n(n  1) n  2
 ...  
 nx h  h  2 x

 
lim 
h

h0 





n(n  1) n  2

x
 ...  nx n 1 (Power Rule)
 2


n 1
lim nx  h
h 0 
2
Example 1: Suppose f and g are differentiable functions such that f (1)  3 ,
g (1)  7 , f (1)  2 , and g (1)  4 . Find (i) ( f  g )(1) , (ii) ( g  f )(1) ,


g
f

(iii) ( fg ) (1) , (iv)   (1) , and   (1) .
f
g
(i) ( f  g )(1)  f (1)  g (1)  2  4  2
(ii) ( g  f )(1)  g (1)  f (1)  4  (2)  6
(iii) ( fg )(1)  f (1) g (1)  g (1) f (1)  3(4)  7(2)  12  (14)  2

g
f (1) g (1)  g (1) f (1) 3(4)  7(2) 12  14 26



(iv)   (1) 
9
9
32
 f (1)2
f

f 
g (1) f (1)  f (1) g (1) 7(2)  3(4)  14  12  26



(v)   (1) 
49
49
72
g (1)2
g
Example 2: If f ( x)  x 4  3x 3  5 x 2  7 x  11, find f (x) .
f ( x)  4 x 3  3(3x 2 )  5(2 x)  7(1)  0  4 x 3  9 x 2  10 x  7
Example 3: If f ( x)  4 x 
f ( x)  4 x 
3
3
x2

3
3
x2

5 7

, then find f (x) .
x x5
1
2
5 7

 4 x 2  3x 3  5 x 1  7 x  5 
x x5

 


5 

1 
f ( x)  4 1 x 2   3  2 x 3   5  1x  2  7  5x  6 =
2
3

 

2x
1
2
 2x
Example 4: If f ( x) 
f ( x) 
5
3
 5 x  2  35 x  6 
2
x

2
3 5
x

5
x2

35
x6
x 2  2x  3
, then find f (1) .
3x  4
(3x  4)( 2 x  2)  ( x 2  2 x  3)(3)
(3 x  4) 2
3

6 x 2  2 x  8  3x 2  6 x  9
(3 x  4) 2

3x 2  8 x  1
(3x  4) 2
f (1) 
 f (1) 
3(1) 2  8(1)  1
3(1)  42

4
 4 or
1
3(1)  42(1)  2  [12  2(1)  3](3)  (1)(4)  (0)(3)   4  4
1
(1) 2
3(1)  42
B. Trigonometric functions
(1) f ( x)  sin x
f ( x  h)  f ( x )
sin( x  h)  sin x
 lim

h
h
h0
h0
f ( x)  lim
sin x cosh  cos x sinh  sin x
sin x(cosh  1)  cos x sinh
 lim

h
h
h 0
h 0
lim


cosh  1
sinh 
(sin x)  lim
  (cos x)  lim
  (sin x)(0)  (cos x)(1) 
h  0 h 
h  0 h 
cos x
(2) f ( x)  cos x
f ( x  h)  f ( x )
cos( x  h)  cos x
 lim

h
h
h 0
h 0
f ( x)  lim
cos x cosh  sin x sinh  cos x
cos x(cosh  1)  sin x sinh
 lim

h
h
h 0
h0
lim


cosh  1
sinh 
(cos x)  lim
  (sin x)  lim
  (cos x)(0)  (sin x)(1) 
h  0 h 
h  0 h 
 sin x
(3) f ( x)  tan x 
f ( x) 
sin x
cos x
(cos x)(cos x)  (sin x)(  sin x)
(cos x)
2

cos 2 x  sin 2 x
2
cos x
4

1
2
cos x
 sec 2 x
(4) f ( x)  sec x 
f ( x) 
1
cos x
(cos x)(0)  1( sin x)
(cos x)
(5) f ( x)  csc x 
f ( x) 

sin x

2
cos x
1 sin x

 sec x tan x
cos x cos x
1
sin x
(sin x)(0)  1(cos x)
(sin x) 2
(6) f ( x)  cot x 
f ( x) 
2

 cos x
sin 2 x

 1 cos x

  csc x cot x
sin x sin x
cos x
sin x
(sin x)(sin x)  (cos x)(cos x)
(sin x)
2

 cos 2 x  sin 2 x
2
sin x

1
2
  csc 2 x
sin x
C. Composition and the generalized derivative rules
(1)
f ( x)  ( g  k )( x)  g (k ( x))
f (b)  f ( x)
g (k (b))  g (k ( x))
g (k (b))  g (k ( x))
 lim
 lim

bx
bx
bx
b x
b x
b x
f ( x)  lim
k (b)  k ( x)
g (k (b))  g (k ( x))
k (b)  k ( x)
 lim
 lim

k (b)  k ( x) b  x
k (b)  k ( x)
bx
b x
g (k (b))  g (k ( x))
k (b)  k ( x)
 lim
 g (k ( x))  k ( x) .
k
(
b
)

k
(
x
)
b

x
k (b )  k ( x )
b x
lim
This derivative rule for the composition of functions is called the Chain Rule.
(2) Suppose that f ( x)  g (k ( x)) where g ( x)  x n . Then f ( x)  [k ( x)] n .
g ( x)  x n  g ( x)  nx n 1  g (k ( x))  nk ( x)n 1 . Thus, f (x) 
g (k ( x))  k ( x)  nk ( x)n 1  k ( x) . This derivative rule for the power of a
function is called the Generalized Power Rule.
5
(3) Suppose that f ( x)  g (k ( x)) where g ( x)  sin x . Then f ( x)  sin[ k ( x)] .
g ( x)  sin x  g ( x)  cos x  g (k ( x))  cos[ k ( x)] . Thus, f (x) 
g (k ( x))  k ( x)  cos[ k ( x)]  k ( x) .
(4) Similarly, if f ( x)  cos[ k ( x)] , then f ( x)   sin[ k ( x)]  k ( x) .
(5) If f ( x)  tan[ k ( x)] , then f ( x)  sec 2 [k ( x)]  k ( x) .
(6) If f ( x)  sec[ k ( x)] , then f ( x)  sec[ k ( x)] tan[ k ( x)]  k ( x) .
(7) If f ( x)  cot[ k ( x)] , then f ( x)   csc 2 [k ( x)]  k ( x) .
(8) If f ( x)  csc[ k ( x)] , then f ( x)   csc[ k ( x)] cot[ k ( x)]  k ( x) .
Example 1: Suppose f and g are differentiable functions such that:
f (1)  9
f (1)  2
f (2)  5
f (2)  6
g (1)  2
g (1)  4
g (9)  3
g (9)  7
Find each of the following: (i) ( f  g )(1); (ii) ( g  f )(1); (iii) h(1) if
3
;
h( x)  f ( x) ; (iv) j (1) if j ( x)  [ g ( x)]5 ; (v) l (1) if l ( x) 
[ f ( x)] 2
(vi) s (1) if s ( x)  sin[ f ( x)] ; and (vii) m(1) if m( x)  sec[ g ( x)] .
(i) ( f  g )(1)  f ( g (1))  g (1)  f (2)  g (1)  (6)( 4)  24
(ii) ( g  f )(1)  g ( f (1))  f (1)  g (9)  f (1)  7(2)  14
1
f ( x)
f ( x)  [ f ( x)] 2  h( x)  1 [ f ( x)] 2  f ( x) 

2
2 f ( x)
f (1)
2
1
h(1) 


3
2 f (1) 2 9
(iii) h( x) 
1
(iv) j ( x)  [ g ( x)]5  j ( x)  5[ g ( x)] 4  g ( x)  j (1)  5[ g (1)] 4  g (1) 
5(2) 4 (4)  320
3
(v) l ( x) 
 3[ f ( x)]  2  l ( x)  6[ f ( x)]  3  f ( x)  l (1) 
2
[ f ( x)]
 6 f (1)  6(2) 12
4



3
3
729 243
[ f (1)]
9
6
(vi) s ( x)  cos[ f ( x)]  f ( x)  s (1)  cos[ f (1)]  f (1)  cos(9)  (2)  2 cos 9
(vii) m( x)  sec[ g ( x)] tan[ g ( x)]  g ( x)  m(1)  sec[ g (1)] tan[ g (1)]  g (1) 
sec( 2) tan( 2)  4  4 sec 2 tan 2
3
Example 2: If f ( x)  2 x 4  x 2  5x  2 , then find f (1) .
1
3
f ( x)  2 x 4  x 2  5x  2  (2 x 4  x 2  5x  2) 3  f ( x) 
2
8x 3  2 x  5
1 (2 x 4  x 2  5 x  2)  3 (8 x 3  2 x  5) 

3
4
2
2
3
3 (2 x  x  5 x  2)
825
11
11
f (1) 


3
33 (2  1  5  2) 2 3 64 12
Example 3: If g ( x) 
g ( x) 
4
( x 3  4) 8
4
3
( x  4) 8
, then find g (x ) .
 4( x 3  4)  8  g ( x)  32( x 3  4)  9 (3x 2 ) 
 96 x 2
( x 3  4) 9
Example 4: If h( x)  sin(cos x) , then find h(x) .
h( x)  cos(cos x)  ( sin x)
Example 5: If j ( x)  tan( 2 x 2  3x  1) , then find j (x) .
j ( x)  sec 2 (2 x 2  3x  1)  (4 x  3)
Example 6: If k ( x)  x 2 3x  4 , then find k (x) .
1


1
k ( x)  x 2 3x  4  x 2 (3x  4) 2  k ( x)  x 2  1 (3x  4) 2 (3) 
2


1
3x 2
2 x(3x  4) 2 3x 2  4 x(3x  4)
2
(3x  4) (2 x) 



1
1
1
2
2
2(3x  4)
2(3x  4)
1
15 x 2  16 x
1
2(3x  4) 2
7
4
 2x  1 
Example 7: If l ( x)  
 , then find l (x) .
 3x  4 
3
3
 2 x  1   (3x  4)( 2)  (2 x  1)(3)  4(2 x  1)
l ( x)  4

 

3
 3x  4  
(3x  4) 2
 (3x  4)
44(2 x  1) 3
(3 x  4) 5
.
Example 8: If k ( x) 
k ( x) 
 11 

 =
2
(
3
x

4
)


sin x
, then find k (x) .
1  cos x
(1  cos x)(cos x)  (sin x)(  sin x)
(1  cos x) 2
cos x  1
1
.

(1  cos x) 2 1  cos x

cos x  cos 2 x  sin 2 x
(1  cos x) 2

Example 9: If s( x)  sin 3 ( x 2  1) , then find s (x) .
s( x)  sin 3 ( x 2  1)  [sin( x 2  1)]3  s ( x)  3[sin( x 2  1)] 2  cos( x 2  1)  2 x 
6 x sin 2 ( x 2  1) cos( x 2  1) .
III. Implicit Differentiation
Example 1: Find the slope of the tangent line to the circle x 2  y 2  25 at the
point (3, 4).
y
(0, 5)
(3, 4)
(– 5, 0)
x
(5, 0)
m=?
(0, – 5)
8
Solution 1 : A circle is not a function. However, x 2  y 2  25  y 2 
25  x 2  y   25  x 2  y  25  x 2 is the equation of the upper
half circle and y   25  x 2 is the equation of the lower half circle.
Since the point (3, 4) is on the upper half circle, use the function f (x) =
1
1
x
2 ( 2 x) 
25  x 2  25  x 2 2  f ( x)  1 25  x 2

2
25  x 2
3
3
3
3
m  f (3) 


 .
4
25  9
16
25  33




Sometimes, an equation [ x 2  y 2  25 ] in two variables, say x and y, is given, but it
is not in the form of y  f (x) . In this case, for each value of one of the variables,
one or more values of the other variable may exist. Thus, such an equation may
describe one or more functions [ y  25  x 2 and y   25  x 2 ]. Any function
defined in this manner is said to be defined implicitly. For such equations, we may
not be able to solve for y explicitly in terms of x [in the example, I was able to solve
for y explicitly in terms of x]. In fact, there are applications where it is not essential
to obtain a formula for y in terms of x. Instead, the value of the derivative at certain
points must be obtained. It is possible to accomplish this goal by using a technique
called implicit differentiation. Suppose an equation in two variables, say x and y, is
given and we are told that this equation defines a differentiable function f with y =
f(x). Use the following steps to differentiate implicitly:
(1) Simplify the equation if possible. That is, get rid of parentheses by
multiplying using the distributive property or by redefining subtraction, and
clear fractions by multiplying every term of the equation by a common
denominator for all the fractions; simplify and combine like terms.
(2) Differentiate both sides of the equation with respect to x. Use all the relevant
differentiation rules, being careful to use the Chain Rule when differentiating
expressions involving y.
dy
(3) Solve for
.
dx
Note: It might be helpful to substitute f (x) into the equation for y before
differentiating with respect to x. This will remind you when you must use the
dy
generalized forms of the Chain Rule. Since f ( x) 
, you differentiate
dx
dy
with respect to x and substitute y for f(x) and
for f (x) . Then you can
dx
9
solve for
dy
.
dx


d 2
x  [ f ( x)] 2  25 
Solution 2: x 2  y 2  25  x 2  [ f ( x)] 2  25 
dx
 2x
dy  x
dy
3
2 x  2[ f ( x)] f ( x)  0  f ( x) 



x 3   .
2[ f ( x)]
dx
y
dx y  4
4
2 3
  x defines a function f with y  f (x) .
x y
Example 2: Suppose that the equation
Find
dy
and the slope of the tangent line at the point (2, 3).
dx
2 3
3x
Solution 1: Solve for y. xy    xy( x)  2 y  3x  x 2 y  y 

2
x 2
x y
dy ( x 2  2)(3)  3x(2 x)  3x 2  6
dy
 18
9



x2 

dx
dx
4
2
( x 2  2) 2
( x 2  2) 2


d
2 y  3x  x 2 y 
Solution 2: Clear fractions  2 y  3x  x 2 y 
dx
dy
dy
dy 3  2 xy
dy
3  12
9
2  3  x2
 2 xy 


x 2 

2
dx
dx
dx x  2
dx y  3
2
2
Solution 3:
2




d 2 3
d
dy
   x  
2 x 1  3 y 1  x  2 x  2  3 y  2
1
dx  x y
dx
dx

3 dy
dy
dy  2 y 2  x 2 y 2
 1  2 y 2  3x 2
 x2 y2 


dx
dx
y 2 dx
3x 2
x2
dy
 18  36  54
9
x2 


dx y  3
12
12
2
Example 3: If cos( xy)  y , then find
dy
.
dx
d
cos( xy)  y    sin( xy) x dy  y(1)  dy   x sin( xy) dy  y sin( xy) 
dx
dx
 dx
 dx
dy
dy
1  x sin( xy)  dy   y sin( xy)
  y sin( xy) 
dx
dx
dx 1  x sin( xy)
10
IV. Higher Order Derivatives
A. Notation
dy
 f (x )
dx
(1) 1st derivative (derivative of the original function y  f (x) ):
d2y
(2) 2nd derivative (derivative of the 1st derivative):
dx 2
d3y
(3) 3rd derivative (derivative of the 2nd derivative):
dx 3
 f ( x)
 f ( x)
B. Distance functions
Suppose s (t ) is a distance function with respect to time t. Then s (t )  v(t )
is an instantaneous velocity (or velocity) function with respect to time t, and
s (t )  v (t )  a(t ) is an acceleration function with respect to time t.
Example 1: If f ( x)  x 2 sin x , then find f (x) and f (x) .
f ( x)  x 2 cos x  2 x sin x
f ( x)  x 2 ( sin x)  2 x cos x  2 x cos x  2 sin x   x 2 sin x  4 x cos x  2 sin x
Example 2: If g ( x) 
g ( x) 
2x  3
, then find g (x ) and g (x) .
4x  5
(4 x  5)( 2)  (2 x  3)( 4)
(4 x  5)
2

8 x  10  8 x  12
(4 x  5)
g ( x)  44(4 x  5)  3 (4)  176(4 x  5)  3 
2

 22
(4 x  5)
2
 22(4 x  5)  2
176
(4 x  5) 3
dy
d2y
Example 3: If x 2  y 2  25 , then find
and
.
dx
dx 2


d 2
dy
dy  2 x  x
x  y 2  25  2 x  2 y
0


dx
dx
dx
2y
y
 x
 dy 

y (1)  ( x)   y  x
y   y 2  x 2
d y d  dy  d   x 
dx 


 
=
    


dx 2 dx  dx  dx  y 
y2
y2
y3
2
 (x2  y 2 )
y3

 25
y3
11
Practice Sheet – Techniques of Differentiation
I. Find the derivative of each function defined as follows; there is no need to simplify
your answers.
(1) f ( x)  x 4  5x 3  9 x 2  7 x  5
(3) g ( x)  8 x 
(5) h( x) 
(7) f ( x) 
6
3
x2
3x  2
2
x 1
sin x
x
x2

8
x3

3x 2  6 x
(4) y 
x3
(6) y  x 2 cos x
2x  1
3x  4
 
(10) y  cos 3 x
(12) y 
sec x
1  tan x
(14) y  sin( 3x) cos( 4 x)
x 1
(15) f ( x)  tan 4 x 3
(16) y 
1
(17) g ( x)  x sec 
 x
(18) y  1  sin 2 x
x 1
dy
by implicit differentiation for each of the following:
dx
(1)  3xy  4 y 2  2
(3)
x4
3
(13) k ( x)  x 9  x 2
II. Find
2
(8) y  x 2  3x  4
(9) g ( x)  sin( x )
(11) h( x) 
9
(2) y 
3 1
 y
2x y
(2) 8 x 2  2 y 3  3xy2
(4) 3x 2 
12
2 y
2 y
(5) x  tan y
(6) y  cos( x  y )
(7) x sin y  y sin x  1
(8) x  sec3 ( y 2  1)
III. Find the slope of the tangent line at the given point on each curve defined by the
given equation:
(2) x 3  3 y  3 ; (1, 8)
(1) x 2  3 y 2  21 ; (3, – 2)
(3)
(4) 3xy  2 x 4  y 3  23 ; (2, – 3)
xy  y  2 ; (1, 4)
 1  
(5) x  cos y ;  ,

2 3 
 
(6) sin( xy)  x ; 1, 
 2
IV. For each of the following functions f (x) , find f (x) and f (x) .
(1) f ( x)  3x 4  4 x 2  7 x  11
(2) f ( x) 
3x  1
2x  1
(3) f ( x)  x 3 cos(4 x)
(4) f ( x)  sin 4 x
V. Suppose the distance (in feet) that an object travels in t seconds is given by the
formula s(t )  2t 3  4t  5 . Find s ( 2) , v ( 2) , and a(2) .
Solution Key for Techniques of Differentiation
I. (1) f ( x)  4 x 3  15 x 2  18 x  7
(2) y  9 x  2  8 x  3  2 x  4 
dy
 18 x  3  24 x  4  8 x  5
dx



(3) g ( x)  8x 2  6 x 3  g ( x)  4 x 2  4 x 3
1
2
1
13
5
(4) y  3x 1  6 x  2 
(5) h ( x) 
(6)
(7)
dy
 3x  2  12 x  3
dx
( x 2  1)(3)  (3 x  2)( 2 x)
( x 2  1) 2
dy
 x 2 ( sin x)  (cos x)( 2 x )
dx
f ( x) 
x(cos x)  (sin x)(1)


x2


1
2
dy 1

x 2  3x  4 3 2 x  3
(8) y  x 2  3x  4 3 
3
dx

1 
(9) g ( x)  cos x   1 x 2 
2


dy
 3cos x 2 ( sin x)
(10) y  cos x 3 
dx
 

 2x  1  2
 2 x  1  2  (3x  4)( 2)  (2 x  1)(3) 
1

(11) h( x)  

h
(
x
)




2  3x  4 
 3x  4 
(3x  4) 2


1
1
dy (1  tan x)(sec x tan x)  sec x(sec 2 x)

(12)
dx
(1  tan x) 2






1
1


1
(13) k ( x)  x 9  x 2 2  k ( x)  x  1 9  x 2 2 (2 x)  9  x 2 2 (1)
2


dy
 sin( 3x) sin( 4 x)( 4)  cos( 4 x)cos(3x)(3)
(14)
dx
  
      
4
3
(15) f ( x)  tan x 3  f ( x)  4 tan x 3 sec 2 x 3 3x 2
dy
(16)

dx


 1 
x  1 
 
2 x 



 1 
x  1 

2 x 
2
x 1
d
 dx

 x   2 1 x 
  1   1  1 
1
  sec (1)
(17) g ( x)  x sec  tan   
2
 x
  x   x  x 
(18) y  1  sin 2 x  2 
1
1
dy 1


1  sin 2 x  2 cos 2 x (2)
2
dx
14

dy
dy
dy
 3y
 3y  8y
0

dx
dx
dx 3x  8 y
II. (1)  3x
(2) 16 x  6 y 2
(3)
dy
dy
dy 16 x  3 y 2
 6 xy  3 y 2 

dx
dx
dx 6 y 2  6 xy
3 1
dy
dy
dy 2  2 y 2
  y  3 y  2 x  2 xy 2  3  2  4 xy  2 y 2 

2x y
dx
dx
dx 4 xy  3
(4) 6 x 2  3x 2 y  2  y  12 x  3x 2

(5) 1  sec 2 y
(6)
dy
dy
dy  12 x  6 xy
 6 xy  


dx
dx
dx
3x 2  1
dydx  dydx  sec12 y  cos 2 y or dydx  1  tan1 2 y  1 1x 2
dy
dy
 sin( x  y )
 dy 
  sin( x  y ) 1   

dx
dx 1  sin( x  y )
 dx 
(7) x(cos y )
dy
dy
dy  sin y  y cos x
 sin y  y cos x  (sin x)
0

dx
dx
dx
x cos y  sin x


dy
 dy 

(8) 1  3 sec 2 ( y 2  1) sec( y 2  1) tan( y 2  1)  2 y  
dx
 dx 
1
1

3 2
2
6 y sec ( y  1) tan( y  1) 6 xy tan( y 2  1)
III. (1) 2 x  6 y
dy
dy
dy
dy 1
 0  2(3)  6(2)
 0  6  12
0

dx
dx
dx
dx 2



 1  2 3  dy
2  1  dy
2  1


(2) 3x  
y

0

3
x


0

3
(
1
)

 dx
 3 2  dx
3
 3 2


3 8
3 y 
2
3
1 dy
dy
0
 36
12 dx
dx
(3) xy  ( y  2) 2  x
dy
dy
dy
dy
dy 4
 y  2( y  2)
1  4  4


dx
dx
dx
dx
dx 3
15
 dy

0
 dx

(4) 3x
6
dy
dy
dy
dy
 3 y  8x 3  3 y 2
 3(2)  3(3)  8(2 3 )  3(3) 2

dx
dx
dx
dx
dy
dy
dy  73
 9  64  27


dx
dx
dx
21
(5) 1  ( sin y )

dy
 
 1   sin 
dx
 3

3 dy
dy
2
 dy
1



2 dx
dx
3
 dx
dy 1  y cos( xy)
dy
 dy



 x 1 
(6) cos( xy)  x  y   1 
dx
x cos( xy)
dx
 dx


y
2
 
cos 
2
2 
 
1 cos 
2
1

1
dy

does not exist.
0
dx
IV. (1) f ( x)  12 x 3  8 x  7 and f ( x)  36 x 2  8
(2) f ( x) 
(2 x  1)(3)  (3x  1)( 2)
(2 x  1)
2
f ( x)  10(2 x  1)  3 (2) 

5
(2 x  1)
2
 5(2 x  1)  2 and
20
(2 x  1) 3
(3) f ( x)  x 3  4 sin( 4 x)  3x 2 cos( 4 x)  4 x 3 sin( 4 x)  3x 2 cos( 4 x) and
f ( x)  4 x 3 4 cos( 4 x)  12 x 2 sin( 4 x)  3x 2  4 sin( 4 x)  6 x cos( 4 x) =
 16 x 3 cos( 4 x)  24 x 2 sin( 4 x)  6 x cos( 4 x)


(4) f ( x)  4sin x 3 cos x and f ( x)  4sin x 3  sin x   cos x  12(sin x) 2 (cos x) =
 4 sin 4 x  12 sin 2 x cos 2 x
 
V. s(2)  2 23  4(2)  5  16  8  5  19 ft
 
v(t )  s (t )  6t 2  4  v(2)  6 2 2  4  28 ft/sec
a(t )  v (t )  s (t )  12t  a(2)  12(2)  24 ft / sec 2
16