Math 1431

Section  14839  M  TH  4:00  PM-­‐5:30  PM    Online  

Susan Wheeler

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Math 1431 Page 1 of 19 Section 2.3 !!

Section 2.3 – Differentiation Rules

We know how to take the derivative of polynomials and basic trigonometric functions. What if

the function is more complicated? How do we differentiate ( ) 25 1

4x

f xx

+=+

, ( ) ( )53g x x x= + , or

h x( ) = x2 sin x( ) ? In this section, we study some rules that will help us with these functions.

We will begin with the product rule. If a function is defined as the product of other functions, then we must use the product rule to differentiate that function.

Theorem 2.3.1: The Product Rule

If f and g are differentiable at x , then so is the product f ⋅ g . Moreover,

f ⋅ g( )' x( ) = f ' x( )g x( ) + f x( )g' x( )

This formula may be written as:

( )uv ' u' v uv'= + or ( )d du dvu v v u

dx dx dx⋅ = ⋅ + ⋅ .

This rule can be extended to the product of more than two functions:

( )uvw ' u' vw uv' w uvw'= + + or ( )d du dv dwu v w v w u w u v

dx dx dx dx⋅ ⋅ = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ .

Example 1: Find the derivative of h x( ) = x2 sin x( ) .

Solution: Since this function is the product of the functions ( ) 2f x x= and ( )g x sin x= , we

need to use the product rule:

ddx

h x( )⎡⎣ ⎤⎦ =ddx

x2( ) ⋅sin x + x2 ⋅ ddx

sin x( ) = 2x sin x + x2 cos x .

Math 1431 Page 2 of 19 Section 2.3 !!

Example 2: Find ( )2

2d

sec xdx

⎡ ⎤⎣ ⎦ .

Solution: Start with finding the first order derivative:

( ) ( ) ( )dsec x sec x tan x

dx=⎡ ⎤⎣ ⎦ .

Notice that the first derivative is the product of two functions; we need to use the product rule while differentiating it.

d2

dx2sec x( )⎡⎣ ⎤⎦ =

ddx

sec x( )tan x( )⎡⎣ ⎤⎦

= sec x( )tan x( )derivative of sec x( )! "## $##

⋅ tan x( ) + sec x( ) ⋅ sec2 x( )derivative of tan x( )!"# $#

= sec x( )tan2 x( ) + sec3 x( ) .

That is,

( ) ( ) ( ) ( )2

2 32

dsec x sec x tan x sec x

dx= +⎡ ⎤⎣ ⎦ .

Theorem 2.3.2: The Reciprocal Rule

If f is differentiable at x and ( ) 0f x ≠ , then so is the reciprocal 1f

. Moreover,

( )( )( ) 2

1 f ' xddx f x f x

⎡ ⎤= −⎢ ⎥

⎣ ⎦ ⎡ ⎤⎣ ⎦

Example 3: For ( ) 21

g xx x

=+

, find ( )3g' .

                   

     

Math 1431 Page 4 of 19 Section 2.3 !!

Now, think about differentiating a rational function. If a function is the quotient of two functions, we need to use the quotient rule to differentiate that function.

Theorem 2.3.3: The Quotient Rule

If f and g are differentiable at x and ( ) 0g x ≠ , then the quotient f / g is differentiable at x

and

fg

⎛⎝⎜

⎞⎠⎟′

x( ) = f ' x( )g x( )− f x( )g' x( )g x( )⎡⎣ ⎤⎦

2

This formula may be written as:

2

'u u' v u v'v v

⋅ − ⋅⎛ ⎞ =⎜ ⎟⎝ ⎠ .

Since 1f fg g= ⋅ , the quotient rule can be obtained by using a combination of the product rule

and the reciprocal rule.

Example 5: Find the derivative of ( ) 25 1

4x

f xx

+=+

.

Solution: Since this function is the quotient of 5 1x + and 2 4x + , we need to use the quotient rule:

f ' x( ) =5x +1( )' x2 + 4( )− 5x +1( ) x2 + 4( )'

x2 + 4( )2

=5 x2 + 4( )− 5x +1( ) 2x( )

x2 + 4( )2

Math 1431 Page 6 of 19 Section 2.3 !!

Finally, how do we differentiate the composition of two functions? For instance, what is the

derivative of ( ) ( )52h x x x= + ? We see that expanding this expression and then taking the

derivative will be very tedious. Since this function is the composition of two functions:

( ) 5f x x= , ( ) 2g x x x= + and ( ) ( )( )h x f g x= ,

we can use the following rule.

Theorem 2.3.4: The Chain Rule

If g is differentiable at x and f is differentiable at g x( ) , then the composition f ! g is

differentiable at x . Moreover,

f ! g( )' x( ) = f ' g x( )( ) ⋅ g' x( ) .

This rule is one of the most important rules of differentiation. It helps with many complicated functions.

Example 7: Find the derivative of ( ) ( )52h x x x= + .

Solution: As stated earlier, h is the composition of two functions:

( ) 5f x x= , ( ) 2g x x x= + and h x( ) = f g x( )( ) = f ! g( ) x( ) .

That is, using the chain rule,

h' x( ) = f ! g( )' x( ) = f ' g x( )( ) ⋅ g' x( ) .

Here, ( ) 45f ' x x= , ( ) 2 1g' x x= + and ( )( ) ( )425f ' g x x x= + . Hence,

( ) ( )( ) ( ) ( ) ( )425 2 1h' x f ' g x g' x x x x= ⋅ = + + .

       

 

Math 1431 Page 7 of 19 Section 2.3 !!

Now that we know the derivative, we can evaluate it at given points. For example,

( ) ( ) ( )421 5 1 1 2 1 1 240h' = + ⋅ + = and ( ) ( ) ( )

420 5 0 0 2 0 1 0h' = ⋅ + ⋅ + = .

As seen in the previous example, the chain rule gives a somewhat “generalized” power rule:

Fact: If u is a differentiable function at x , and n is any real number other than 0, then

( ) 1n nd duu n u

dx dx−= ⋅ ⋅ .

Note that if n is a negative real number, this fact holds true for all x such that u x( ) ≠ 0 .

Example 8: For ( ) ( )506 11f x x= − , find ( )2f ' .

Solution: One can expand this function into a polynomial using the binomial theorem. However, this is not the best approach to solve this problem as it will be very difficult to keep track of those 51 terms. Using the chain rule is a much more efficient way.

( ) ( ) ( ) ( ) ( )49 49 4950 6 11 6 11 50 6 11 6 300 6 11df ' x x x x x

dx= ⋅ − ⋅ − = ⋅ − ⋅ = − .

Direct substitution gives:

( ) ( )49 492 300 6 2 11 300 1 300f ' = ⋅ − = ⋅ = .

Example 9: Find the derivative of ( )41

1x

h xx−⎛ ⎞= ⎜ ⎟+⎝ ⎠

at 0x = .

Solution: Use the chain rule:

( )31 14

1 1x d x

h' xx dx x− −⎛ ⎞ ⎛ ⎞= ⋅ ⋅⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

.

To find the derivative of 11

xx−+

, use the quotient rule: Math 1431 Page 10 of 19 Section 2.3 !

Hence,

f ' x( ) = g ! h( )' x( ) = g' h x( )( ) ⋅h' x( ) .

Here,

( ) 23 1h' x x= − , ( )g' x sin x= − and ( )( ) ( )3g' h x sin x x= − − .

The derivative is:

f ' x( ) = g' h x( )( ) ⋅h' x( ) = − sin x3 − x( ) ⋅ 3x2 −1( ) = − 3x2 −1( )sin x3 − x( ) ⋅ .

We can generalize this approach to all basic trigonometric functions.

The Chain Rule Applied to Six Trigonometric Functions

If u is a differentiable function of x , then

ddx

sinu( ) = cosu ⋅ dudx

ddx

cosu( ) = − sinu ⋅ dudx

ddx

tanu( ) = sec2 u ⋅ dudx

ddx

cot u( ) = −csc2 u ⋅ dudx

ddx

secu( ) = secu ⋅ tanu ⋅ dudx

ddx

cscu( ) = −cscu ⋅cot u ⋅ dudx

More  Examples:  1. Find the derivative:

a. f x( ) = − 1

x2

         

b. f x( ) = x2 +2( )

x3

         

 

c. f x( ) = x3 + 3x

x2 −1

                         

2. Find ( )0f ' given that h (0) = 3 and h’ (0) = 2 if f x( ) = h x( ) + x

h x( )

             

3. Find ddx

3 f (x)+ x2( )3( )            

4. Find an equation for the tangent line to the graph of f x( ) = x2 − 10

x

at the point (-2, f(-2)).                                

5. Find x such that a)        f  ’(x)  =  0,          b)      f  ’  (x)  >  0,       c)    f  ’  (x)  <  0    

Given that f x( ) = 3x 4 − 4x3 − 2

                           

6.        5. Find an equation for the normal line to the graph of f (x) = sin2 x at the

point π4, f π

4⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟ .

MORE  examples:    

7.      Given   f (x) = x3      and     g(x) = sin(x)  

                 If,    h(x) = f ⋅g( ) x( ), find ′h x( ) .  

         

           8.        If,          h(x) = fg( ) x( ), find ′h x( ) find   h '(x) .  

           

 9.        If,    h(x) = f g(x)( ), find ′h x( ) .  

                               

10. Given:    f (2) = 7, f '(2) =1, f (5) = 4, f '(5) = 4,g(2) = 5, g '(2) = 3, g(5) =10, g'(5) = 6.  

 

a) If   ( ) ( )( )h x fg x= ,  find   '(2)h .    

             

b) If   h(x) = f

g⎛⎝⎜

⎞⎠⎟

x( ) ,  find   h '(2) .            

Given:  f (2) = 7, f '(2) =1, f (5) = 4, f '(5) = 4,g(2) = 5, g '(2) = 3, g(5) =10, g'(5) = 6.

 

 

c) If   ( )3

( )h x f x= ⎡ ⎤⎣ ⎦ ,  find   '(2)h .            d) If   h(x) = ( f ! g)(x) ,  find   '(2)h .  

             

   

Implicit  Differentiation    

Section  2.4        If  we  can’t  solve,  or  can’t  solve  easily,  for  y  as  a  function  of  x,  use  implicit  differentiation  to  find  the  derivative  of  y.  Discuss:       + =2 2x y 1                  

 “Rules”  for  implicit  differentiation    1)   Differentiate  both  sides  of  the  equation  with  respect  to  x.    2)   Collect  all   dy

dx  (or  y  ’  )  terms.  

   3)   Factor  out   dy

dx  (or  y  ’  )    .  

   4)   Solve  for   dy

dx  (or  y  ’  )  .  

 If  solving  in  terms  of  x,  take  the  derivative  as  usual.    If  solving  in  terms  of  y,  use  the  chain  rule.  

For  the  following,  use  implicit  differentiation  to  find  the  derivative.    1.   = 4y x        2.   ( )3d x ydx

 

     3.   =2x y 5          

 4.   + =3 22x y 8          5.   ( )sin 2d ydx

 

       6.   ( )sin2d ydx

 

       

 7.   + − + =3 2y 2y 3y x 2                                

 8.                              9.   sin cos =2 x y 1          

2 3x 3xy y 10+ + =

10.     Find  the  slope  of  the  graph  at  the  given  point,  then  find  the  equation  of  the  tangent  line  at  (1,  1)  for   + =3 3x y 2xy.                                

11.     Find  the  second  derivative  of   − =2 2x y 16    in  terms  of  y.