# Techniques of Differentiation

November 13, 2017 | Author: Raheel's Manners | Category: Derivative, Equations, Mathematical Concepts, Space, Mathematical Analysis

#### Description

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 D x [ 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 (4) f ( x ) =

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) = = = k ( x) k ( x) [ k ( x)] 2

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 h+ x h + ... − x n  x + nx 2   = lim h h →0  n −1  n( n −1) n −2  h +h 2  x +...  nx 2   = lim  h  h →0     

lim nx

h →0 

n −1

 n(n − 1) n − 2  + h x + ...  = nx n −1 (Power Rule)  2 

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) ,

′    (1) , f 

g (iii) ( fg ) ′(1) , (iv)  

′ 

f and     (1) . 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 (iv) (v)

′  f (1) g ′(1) − g (1) f ′(1) 3( 4) −7( −2) 12 +14 26  = = =  (1) = 2 2 9 9 3 [ f (1)]  ′ f  g (1) f ′(1) − f (1) g ′(1) 7( −2) −3( 4) −14 −12 − 26   = = =  g  (1) = 2 2 49 49 7 [ g (1)]   g  f 

Example 2: If

f ( x ) = x 4 −3 x 3 + 5 x 2 −7 x +11 ,

find f ′(x) .

f ′( x ) = 4 x 3 −3(3 x 2 ) +5( 2 x ) −7(1) +0 = 4 x 3 −9 x 2 +10 x −7

Example 3: If f ( x) = 4 x − 3 f ( x) = 4 x −

3 3 2

+

x

3 x

2

+

5 7 − f ′(x ) . x x 5 , then find

1 −2 5 7 − = 4 x 2 − 3 x 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 − 5 x − 6 = 2 3    

2x

−1

2

+ 2x

−5

3

− 5 x − 2 + 35x − 6 =

2 x

+

2 3 5

x

5 x

2

+

35 x6

2 Example 4: If f ( x) = x + 2 x − 3 , then find f ′(1) .

3x − 4

f ′( x) =

(3 x − 4)( 2 x + 2) − ( x 2 + 2 x − 3)( 3) (3 x − 4) 2

=

6 x 2 − 2 x − 8 − 3x 2 − 6 x + 9 (3 x − 4) 2

3 2

3x − 8 x +1 (3 x − 4)

2

⇒ f ′(1) =

3(1) 2 − 8(1) +1

[3(1) − 4]

2

=

−4 = −4 or 1

=

f ′(1) =

[3(1) − 4][ 2(1) + 2] −[12 + 2(1) − 3]( 3) [3(1) − 4] 2

=

( −1)( 4) − (0)( 3) (−1) 2

=

−4 = −4 1

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  h →0 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  h →0 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

4

=

cos 2 x + sin 2 x cos

2

x

=

1 cos

2

x

= sec 2 x

(4) f ( x) = sec x = f ′( x) =

(cos x)( 0) −1( −sin x) (cos x)

(5) f ( x) = csc x = f ′( x ) =

2

=

sin x cos

2

x

=

1 sin x ⋅ = sec x tan x cos x cos x

1 sin x

(sin x )( 0) −1(cos x) (sin x)

(6) f ( x) = cot x = f ′( x ) =

1 cos 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 sin

2

x

=

−1 sin

2

x

= −csc 2 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 ( 2) = −5

g (1) = 2

f ′(1) = −2 f ′( 2) = −6

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 h( x ) = f ( x ) ; (iv) j ′(1) if j ( x ) =[ g ( x )] 5 ; (v) l ′(1) if l ( x) =

3

;

[ 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 )] 2 ⇒ h ′( x ) = 1 [ f ( x)] 2

(iii) h( x) = h ′(1) =

(iv)

f ′(1) 2

f (1)

=

−2 2 9

=−

−1

2 ⋅ f ′( x) =

f ′( x ) 2 f ( x)

1 3

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

(v) l ( x) =

3 [ f ( x)]

− 6 f ′(1) [ f (1)]

3

=

2

= 3[ f ( x)] −2 ⇒ l ′( x ) = −6[ f ( x )] −3 ⋅ f ′( x ) ⇒ l ′(1) =

− 6( −2) 9

3

=

12 4 = 729 243

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 + 5 x + 2

, then find f ′(1) . 1

3 f ( x) = 2 x 4 − x 2 + 5 x + 2 = (2 x 4 − x 2 + 5 x + 2) 3 ⇒ f ′( x) = 2 8x3 − 2x + 5 1 (2 x 4 − x 2 + 5 x + 2) − 3 (8 x 3 − 2 x + 5) = ⇒ 3 33 ( 2 x 4 − x 2 + 5 x + 2) 2 8 − 2 +5 11 11 f ′(1) = = = 3 12 3 64 33 (2 −1 + 5 + 2) 2

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 (3 x 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 −3 x +1) ,

then find j ′(x ) .

j ′( x ) = sec 2 (2 x 2 −3 x +1) ⋅ ( 4 x −3)

Example 6: If

k ( x) = x 2

3 x +4 ,

then find k ′(x ) .

1  −1  k ( x) = x 2 3x + 4 = x 2 (3 x + 4) 2 ⇒ k ′( x) = x 2  1 (3 x + 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(3x + 4) 2 2(3x + 4) 2 1

15 x 2 + 16 x 1

2(3x + 4) 2

7 4

2 x −1   , then find l ′(x ) .  3x + 4 

Example 7: If l ( x) = 

3 3  11  2 x −1  (3 x + 4)( 2) − (2 x −1)( 3)  4(2 x −1)  l ′( x) = 4 =      = 2 3 2  3x + 4   (3 x + 4) (3 x + 4)      (3 x + 4) 

44 ( 2 x −1) 3 (3 x + 4) 5

.

Example 8: If k ( x ) = k ′( x ) =

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 = 2 1 + cos x. (1 + cos x )

Example 9: If

s ( x ) =sin 3 ( x 2 −1) ,

=

cos x + cos 2 x + sin 2 x (1 + cos x ) 2

=

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 point (3, 4). y (0, 5)

x 2 + y 2 = 25

at the

(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 25 − x 2 = 25 − x 2 2 ⇒ f ′( x) = 1 25 − x 2 2 ( −2 x) = 2

m = f ′(3) =

−3 25 − 33

=

−3 25 − 9

=

−3

−x 25 − x 2

3 =− . 4 16

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. (3) Solve for

dy . 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 generalized forms of the Chain Rule. Since f ′( x) = with respect to x and substitute y for f(x) and

dy , you differentiate dx

dy for f ′(x) . Then you can dx

9 dy solve for . dx

Solution 2: x 2 + y 2 = 25 ⇒ x 2 + [ f ( x)]2 = 25 ⇒

2 x + 2[ f ( x) ]f ′ ( x) = 0 ⇒ f ′ ( x) =

(

)

d 2 x + [ f ( x)]2 = 25 ⇒ dx

− 2x dy − x dy 3 ⇒ = ⇒ ⋅ x= 3 = − . 2[ f ( x)] d x y d x y = 4 4

2

3

Example 2: Suppose that the equation x + y = x defines a function f with y = f (x) . Find

dy and the slope of the tangent line at the point (2, 3). dx 2

3

3x

2 ⇒ Solution 1: Solve for y. xy  +  = xy ( x) ⇒ 2 y + 3x = x y ⇒ y = 2 x −2 x y

dy ( x 2 − 2)( 3) − 3 x( 2 x) − 3x 2 − 6 dy −18 9 = = ⇒ ⋅ x =2 = =− 2 2 2 2 dx dx 4 2 ( x − 2) ( x − 2)

Solution 2: Clear fractions ⇒ 2 y + 3 x = x 2 y ⇒

(

)

d 2 y + 3x = x 2 y ⇒ dx

dy dy d y 3 − 2x y d y 3− 1 2 9 2 + 3 = x 2 + 2 x y⇒ = ⇒ ⋅ x= 2 = =− 2 dx dx d x x − 2 d x y= 3 2 2 Solution 3: −2 x2

(

)

 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 − 3 x 2 = x2 y2 ⇒ = ⇒ dx dx y 2 dx 3x 2

dy − 1 8− 3 6 − 5 4 9 ⋅ x= 2 = = =− d x 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 dy − y sin( xy ) ⇒ − y sin( xy ) = (1 + x sin( xy ) ) ⇒ = dx dx dx 1 + x sin( xy )

10 IV. Higher Order Derivatives A. Notation (1) 1st derivative (derivative of the original function y = f (x) ): (2) 2nd derivative (derivative of the 1st derivative):

d2y dx 2

= f ′′( x )

dy = f ′(x ) dx

d3y

(3) 3rd derivative (derivative of the 2nd derivative):

= f ′′′( x)

dx 3

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

f ( x ) = x 2 sin x

, then find f ′(x) and f ′′(x ) .

cos x +2 x sin x

2

f ′′( x ) = x ( −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)

=

8 x −10 − 8 x −12

=

Example 3: If

x 2 + y 2 = 25

(

, then find

2

− 22

(4 x − 5) (4 x − 5) 2 176 g ′′( x) = 44 ( 4 x − 5) −3 (4) = 176 (4 x − 5) −3 = (4 x − 5) 3 ( 4 x − 5)

2

= −22 (4 x − 5) −2

dy d2y 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 2 2 d y d  dy  d  − x   dx  =  y  = −y −x =   = = =     dx  dx  dx  y  dx 2 y2 y2 y3 2

−(x2 + y 2 ) y

3

=

− 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 −5 x 3 + 9 x 2 − 7 x + 5

(3) g ( x) = 8 x − 3

6 x2

(2) y =

(4) y =

9 x

2

8 x

3

+

3x 2 − 6 x x3

2 x4

(5) h( x) =

3x + 2

(7) f ( x ) = (9)

(6)

x 2 +1

sin x x

g ( x) =sin(

(11) h( x) =

(8) (10)

x)

2 x +1 3x − 4

k ( x) = x 9 −x 2

(15)

f ( x ) = tan 4 x 3

x 2 −3 x + 4

y = cos 3 x

sec x 1 + tan x

(14) y = sin( 3 x) cos( 4 x)

( )

(16) y =

1  x

(18)

(17) g ( x) = x sec  

(1)

3

(12) y =

(13)

II. Find

y=

y = x 2 cos x

x +1 x −1

y = 1 +sin 2 x

dy by implicit differentiation for each of the following: dx

(2)

−3 xy −4 y 2 = 2

3

8 x 2 = 2 y 3 +3 xy 2

2−y

1

(3) 2 x + y = y

2 (4) 3x = 2 + y

(5) x = tan y

12 (6) y = cos( x − y )

(7) x sin y + y sin x =1

(8)

x =sec 3 ( y 2 −1)

III. Find the slope of the tangent line at the given point on each curve defined by the given equation: (1)

x 2 +3 y 2 = 21 ;

(3)

xy −y = −2 ;

(3, – 2)

(2)

x 3 +3 y = 3 ;

(1, 4)

(4)

3 xy −2 x 4 = y 3 −23

 1 −π   (5) x = cos y ;  , 2

3 

(1, 8) ; (2, – 3)

 π (6) sin( xy ) = x ; 1,  

2

IV. For each of the following functions f ( x ) , find f ′(x) and f ′′(x) .

(1)

f ( x ) = 3 x 4 −4 x 2 +7 x −11

(2) f ( x) =

3 x +1 2 x −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

dy = −18 x − 3 + 24 x − 4 − 8 x − 5 dx

(2) y = 9 x − 2 − 8 x − 3 + 2 x − 4 ⇒ 2

1

5

1

(3) g ( x) = 8 x 2 − 6 x − 3 ⇒ g ′( x) = 4 x − 2 + 4 x − 3 13 (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 2 (8) y = x 2 − 3x + 4 3 ⇒ = x − 3x + 4 3 ( 2 x − 3) 3 dx

( )

(9) g ′( x) = cos x ⋅  1 2 x  3 (10) y = ( cos x ) ⇒

−1

  

2

dy = 3( cos x ) 2 (− sin x) dx 1

(11)

1

−  2 x +1  2  2 x +1  2 (3 x − 4)( 2) − ( 2 x +1)( 3)  h( x ) =  ⇒ h ′( x) = 1      2  3x − 4   3x − 4  (3 x − 4) 2    

dy

(12) dx =

(1 + tan x )(sec x tan x ) − sec x (sec 2 x ) (1 + tan x ) 2

(

2 (13) k ( x) = x 9 − x

) 12 ⇒ k ′( x) = x  12 (9 − x 2 )− 12 (−2 x) + (9 − x 2 ) 12 (1)

  dy = sin( 3 x )[ − sin( 4 x)( 4)] + cos( 4 x)[ cos( 3 x)( 3)] (14) dx

[ ( )]

(15) f ( x) = tan x 3

(16) dy =

(

4

[ ( )] [sec (x )](3x )

)

 1 x −1  2 x

(

dx

  − 

(

)

 1 x + 1  2 x

)2

2

  

 1   x 

1 x 1

(18) y = (1 + sin 2 x ) 2 ⇒

3

d  dx 

x −1

(17) g ′( x ) = x sec   tan   − 

3

⇒ f ′( x) = 4 tan x 3

2

( x ) = 2 1 x  

1  1  + sec  (1) 2  x x 

1 dy 1 = (1 + sin 2 x ) − 2 ( cos2 x ) (2) 2 dx

14

dy dy dy −3y II. (1) − 3 x dx − 3 y − 8 y dx = 0 ⇒ dx = 3 x + 8 y dy dy dy 16 x − 3 y 2 + 6 xy +3y 2 ⇒ = dx dx dx 6 y 2 + 6 xy

(2)

16 x = 6 y 2

(3)

3 1 dy dy dy 2 −2y2 + = y ⇒ 3 y + 2 x = 2 xy 2 ⇒ 3 + 2 = 4 xy +2y2 ⇒ = 2x y dx dx dx 4 xy − 3

dy

dy

dy

2 2 2 (4) 6 x + 3x y = 2 − y ⇒ 12 x + 3x dx + 6 xy = − dx ⇒ dx =

(

) dy

dy

1

dy

− 12 x − 6 xy

1

3x 2 + 1 1

2 2 = (5) 1 = sec y dx ⇒ dx = 2 = cos y or dx = sec y 1 + tan 2 y 1 + x 2

(6)

dy dy − sin( x − y )  dy  = − sin( x − y ) 1 −  ⇒ = dx dx 1 − sin( x − y )  dx  dy

dy

dy

− sin y − y cos x

(7) x(cos y ) dx + sin y + y cos x + (sin x) dx = 0 ⇒ dx = x cos y + sin x

[

]

2 2 2 2 (8) 1 = 3 sec ( y − 1) sec( y − 1) tan( y − 1)  2 y

dy  dy = ⇒ dx  dx

1 3

6 y sec ( y

III. (1) 2 x + 6 y

(2)

2

−1) tan( y

2

−1)

=

1 6 xy tan( y 2 −1)

dy dy dy dy 1 = 0 ⇒ 2(3) + 6(−2) = 0 ⇒ 6 − 12 =0⇒ = dx dx dx dx 2

  − 2  dy  1 3 x 2 +  1 y 3  = 0 ⇒ 3x 2 +  3  33 y 2   dx 

3+

   dy 2  1 = 0 ⇒ 3 ( 1 ) +  dx  3 2 3 8 

 dy  =0⇒  dx 

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 (4) 3 x 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 )

(6)

 dy  −π ⇒ 1 = − sin  dx  3 

 dy c o sx( y) x +  dx

3 dy dy 2  dy ⇒1= ⇒ =  2 dx dx 3  dx

d y 1 − y c o sx( y) d y  y = 1 ⇒ = ⇒ ⋅ x= 1 d x x c o sx( y) dx π  y=

2

π π  1 − c o s  2  2 = = π  1c o s   2

1 dy ⇒ does not exist. 0 dx

IV. (1)

f ′( x ) =12 x 3 −8 x +7

(2) f ′( x) =

and

( 2 x −1)( 3) − (3 x +1)( 2) ( 2 x −1)

2

f ′′( x ) = 10 ( 2 x −1) −3 (2) =

(3)

f ′′( x ) =36 x 2 −8

=

−5 ( 2 x −1)

2

= −5( 2 x −1) −2 and

20 ( 2 x −1) 3

f ′( x ) = x 3 [− 4 sin( 4 x )] +3 x 2 cos( 4 x ) = −4 x 3 sin( 4 x ) +3 x 2 cos( 4 x )

and

f ′′( x ) = −4 x 3 [4 cos( 4 x )] −12 x 2 sin( 4 x ) +3 x 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 2 3 + 4( 2) −5 =16 +8 −5 =19 2

ft

( ) + 4 = 28 2

ft/sec a (t ) = v ′(t ) = s ′′(t ) = 12t ⇒ a (2) = 12(2) = 24 ft / sec

v (t ) = s ′(t ) = 6t

+ 4 ⇒ v ( 2) = 6 2

16

2

)