Techniques of Differentiation

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′, dx dy ,

[

f(x)

]

dx d , 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) ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

(

_lim

_lim

_lim

cg x

x b x g b g c x b x cg b cg x b x f b f x f x b x b x b ′ = − − = − − = − − = ′ → → → (2) f(x)=g(x)±k(x) ′ = ₋− = ± ₋− ± = → → b x x k x g b k b g x b x f b f x f x b x b )] ( ) ( [ )] ( ) ( [ ) ( ) ( ) (

_lim

_lim

lim

( ) ( )

lim

( ) ( ) g (x) k (x) x b x k b k x b x g b g x b x b ′ ± ′ = − − ± − − → → (3) f(x)=g(x)k(x) ′ = ₋− = −₋ = → → b x x k x g b k b g x b x f b f x f x b x b ) ( ) ( ) ( ) ( ) ( ) ( ) (

_lim

_lim

= − − + − → b x x k x g x k b g x k b g b k b g x b ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

lim

=       − −       +       − −       → → → → b x x g b g x k x b x k b k b g x b x b x b x b ) ( ) ( ) ( ) ( ) ( )

(

_lim

_lim

_lim

lim

1 (4) = ⇒ ( ) ( )= ( )⇒ ′( )= ( ) ′( )+ ( ) ′( )⇒ ) ( ) ( ) ( f x k x g x g x f x k x k x f x x k x g x f

[

_{( )}

]

2 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( x k x k x g x g x k x k x k x k x g x g x k x k x f x g x f = ′ − ′ ′       − ′ = ′ − ′ = ′ .

This derivative rule is called the Quotient Rule. (5) f(x)=c

( )

lim

( ) ( )

lim

lim

0 =

lim

0 =0

− = − − = − − = ′ → → → →x b x b x b x b b x b x c c x b x f b f x f (6) f(x)=x

( )

lim

( ) ( )

lim

=

lim

1=1

− − = − − = ′ → → →x b x b x b b x x b x b x f b f x f (7) _f(x)=xn ′ = + − = + − = → → h x h x h x f h x f x f n n h h ) ( ) ( ) ( ) ( _{lim lim} 0 0 − ₌    _{+ − +} − _{− +} → h x h x n n h nx xn n n n h ... 2 ) 1 ( 2 2 1 0

lim

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

lim

− − − →  =           − ₊ + n n n h x nx n n h nx (Power Rule) 2

g(1)=7_, f′(1)=−2_{, and} g′(1)=4_{. Find (i)} (f +g)′(1)_{, (ii)} ) 1 ( ) (g − f ′ _, (iii) ( fg)′(1), (iv) (1) ′         f g _{, and} ) 1 ( ′         g f _. (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)

[

]

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

[

]

49 26 49 12 14 7 ) 4 ( 3 ) 2 ( 7 ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( ) 1 ( 2 2 − = − − = − − = ′ − ′ = ′     g g f f g g f Example 2: If f(x)=x4 −3x3 +5x2 −7x+11, find f ′(x). f′(x)=4x3 −3(3x2)+5(2x)−7(1)+0=4x3 −9x2 +10x−7 Example 3: If 3 2 5 7 5 3 4 ) ( x x x x x f = − + − _{, then find} f′(x). ( )=4 −_{3 2}3 +5− 7₅ =4x12−3x−23 +5x−1−7x−5⇒ x x x x x f ( ) 4 1₂ −12 3 2₃ −53 +5

(

−1 −2

) (

−7−5 −6

)

    − −     = ′ x x x x x f = 2 12 2 53 5 2 35 6 2 _{3 5}2 5₂ 35₆ x x x x x x x x− + − − − + − = + − + Example 4: If 4 3 3 2 ) ( 2 − − + = x x x x f , then find f ′(1). = − + − − − − = − − + − + − = ′ 2 2 2 2 2 ) 4 3 ( 9 6 3 8 2 6 ) 4 3 ( ) 3 )( 3 2 ( ) 2 2 )( 4 3 ( ) ( x x x x x x x x x x x f 3

[

]

1 4 4 4 ) 1 ( 3 1 ) 1 ( 8 ) 1 ( 3 ) 1 ( ) 4 3 ( 1 8 3 2 2 2 2 − = − = − + − = ′ ⇒ − + − _f x x x or

[

][

]

[

]

1 4 4 ) 1 ( ) 3 )( 0 ( ) 4 )( 1 ( 4 ) 1 ( 3 ) 3 ]( 3 ) 1 ( 2 1 [ 2 ) 1 ( 2 4 ) 1 ( 3 ) 1 ( 2_{2 2} =− =− − − − = − − + − + − = ′ f B. Trigonometric functions (1) f(x) =sin x = − + = − + = ′ → → h x h x h x f h x f x f h h sin ) sin( ) ( ) ( ) (

_lim

_lim

0 0 = + − = − + → → h x x h x x x h h sinh cos ) 1 (cosh sin sin sinh cos cosh sin

lim

lim

0 0 = + =       +       − → → (sin )(0) (cos )(1) sinh ) (cos 1 cosh )

(sin

_lim

_lim

0 0 x x h x h x h h

cos

x

(2) f(x) =cos x ′ = + − = + − = → → h x h x h x f h x f x f h h cos ) cos( ) ( ) ( ) (

_lim

_lim

0 0 = − − = − − → → h x x h x x x h h sinh sin ) 1 (cosh cos cos sinh sin cosh cos

lim

lim

0 0 = − =       −       − → → (cos )(0) (sin )(1) sinh ) (sin 1 cosh )

(cos

_lim

_lim

0 0 x x h x h x h h −sin x (3) x x x x f cos sin tan ) ( = = x x x x x x x x x x x f _{2 2} 2 2 2 2 _cos sec 1 cos sin cos ) (cos ) sin )( (sin ) )(cos (cos ) ( = − − = + = = ′ 4

(4) x x x f cos 1 sec ) ( = =

f (x) (cos x)(_(cos0) _x1₎(2 sinx) =_cossin2x_x =_cos1_x⋅_cossinx_x =sec xtan x

− − = ′ (5) x x x f sin 1 csc ) ( = = _{x x}x x x x x x x x x f csc cot sin cos sin 1 sin cos ) (sin ) (cos 1 ) 0 )( (sin ) ( 2 2 ⋅ =− − = − = − = ′ (6) x x x x f sin cos cot ) ( = = x x x x x x x x x x x f ₂ 2 ₂ 2 ₂ csc2 sin 1 sin sin cos ) (sin ) )(cos (cos ) )(sin (sin ) ( = − =− − = − =− ′

C. Composition and the generalized derivative rules (1) f(x) =(gk)(x) =g(k(x))

⋅

−

−

=

−

−

=

−

−

=

′

→

→

→

b

x

x

k

g

b

k

g

x

b

x

k

g

b

k

g

x

b

x

f

b

f

x

f

x

b

x

b

x

b

))

(

(

))

(

(

))

(

(

))

(

(

)

(

)

(

)

(

_lim

_lim

_lim

= − − ⋅ − − = − − → → b x x k b k x k b k x k g b k g x k b k x k b k x b x b ) ( ) ( ) ( ) ( )) ( ( )) ( ( ) ( ) ( ) ( ) (

lim

lim

lim

( (₍))_{) (}( ₎( ))

lim

( ) ( ) ( ( )) ( ) ) ( ) ( x k x k g x b x k b k x k b k x k g b k g x b x k b k ′ ⋅ ′ = − − ⋅ − − → → .

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 )] ( sin[ ) (x k x f = . )] ( cos[ )) ( ( cos ) ( sin ) (x x g x x g k x k x g = ⇒ ′ = ⇒ ′ = . Thus, f′(x)= ) ( )] ( cos[ ) ( )) ( (k x k x k x k x g′ ⋅ ′ = ⋅ ′ .

(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)=−csc2[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 g(9)=3

f′(1)=−2 f′(2)=−6 g′(1)=4 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} 2 )] ( [ 3 ) ( x f x l = _;

(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 (iii) = = ⇒ ′ = − ⋅ ′ = ′ ⇒ ) ( 2 ) ( ) ( )] ( [ 2 1 ) ( )] ( [ ) ( ) ( 12 12 x f x f x f x f x h x f x f x h h′(1) =₂f′_f(1₍)₁₎ =₂−2₉ =−₃1 (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 (v) = =3[ ( )]− ⇒ ′( )=−6[ ( )]− ⋅ ′( )⇒ ′(1)= )] ( [ 3 ) ( 2 3 2 f x l x f x f x l x f x l ₇₂₉12 ₂₄₃4 9 ) 2 ( 6 )] 1 ( [ ) 1 ( 6 3 3 = = − − = ′ − f f 6

(vi) 9 cos 2 ) 2 ( ) 9 cos( ) 1 ( )] 1 ( cos[ ) 1 ( ) ( )] ( cos[ ) ( = ⋅ ′ ⇒ ′ = ⋅ ′ = ⋅ − =− ′ x f x f x s f f s (vii) = ′ ⋅ = ′ ⇒ ′ ⋅ =

′(x) sec[g(x)]tan[g(x)] g (x) m(1) sec[g(1)]tan[g(1)] g (1) m

sec(2)tan(2)⋅4=4sec2tan2

Example 2: If _f(x)=3_2x4_−x2 _{+5x+2 , then find f} ′_(1).

= ′ ⇒ + + − = + + − = 2 5 2 (2 5 2) ( ) ) (_x 3 _x4 _x2 _{x x}4 _x2 _x 13 _{f x} f ⇒ + + − + − = + − + + − − 3 4 2 2 3 3 3 2 2 4 ) 2 5 2 ( 3 5 2 8 ) 5 2 8 ( ) 2 5 2 ( 3 1 x x x x x x x x x x 12 11 64 3 11 ) 2 5 1 2 ( 3 5 2 8 ) 1 ( 3 3 _{− + +} 2 = = + − = ′ f Example 3: If ₍ 3 ₄₎8 4 ) ( + = x x g _{, then find g′(x).} 9 3 2 2 9 3 8 3 8 3 _{( 4)} 96 ) 3 ( ) 4 ( 32 ) ( ) 4 ( 4 ) 4 ( 4 ) ( + − = + − = ′ ⇒ + = + = − − x x x x x g x x x g

Example 4: If h(x)=sin(cos x), then find h′(x). ) sin ( ) cos(cos ) (x x x h′ = ⋅ −

Example 5: If j(x)=tan( 2x2 −3x+1), then find j′(x).

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

Example 6: If k(x)=x2 3x+4, then find k′(x).

+       + = ′ ⇒ + = + = 3 4 (3 4) ( ) 1₂(3 4)− (3) ) (_{x x}2 _{x x}2 _x 12 _{k x x}2 _x 12 k = + + + = + + + = + 2 1 2 2 1 2 1 2 2 1 ) 4 3 ( 2 ) 4 3 ( 4 3 1 ) 4 3 ( 2 ) 4 3 ( 2 3 ) 2 ( ) 4 3 ( x x x x x x x x x x 2 1 2 ) 4 3 ( 2 16 15 + + x x x 7 Example 7: If 4 4 3 1 2 ) (       + − = x x x l , then find l′(x)_.

        + + − =         + − − +       + − = ′ 3 _{2 3}3 ₂ ) 4 3 ( 11 ) 4 3 ( ) 1 2 ( 4 ) 4 3 ( ) 3 )( 1 2 ( ) 2 )( 4 3 ( 4 3 1 2 4 ) ( x x x x x x x x x l ₌ 5 3 ) 4 3 ( ) 1 2 ( 44 + − x x . Example 8: If x x x k cos 1 sin ) ( + = , then find k′(x)_. = + + + = + − − + = ′ ₂ 2 ₂ 2 ) cos 1 ( sin cos cos ) cos 1 ( ) sin )( (sin ) )(cos cos 1 ( ) ( x x x x x x x x x x k x x x cos 1 1 ) cos 1 ( 1 cos 2 = ₊ + + .

Example 9: If s(x)=sin3(x2 −1), then find s′(x).

= ⋅ − ⋅ − = ′ ⇒ − = − = x x s x x x x x

s( ) sin3( 2 1) [sin( 2 1)]3 ( ) 3[sin( 2 1)]2 cos( 2 1) 2 ) 1 cos( ) 1 ( sin 6x 2 x2 − x2 − .

III. Implicit Differentiation

Example 1: Find the slope of the tangent line to the circle x2+y2 =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, _x2+_y2=₂₅⇒_y2 = 2

2

2 _{25 25}

25−x ⇒y=± −x ⇒y= −x is the equation of the upper

half circle and _{y=− 25−x}2 _{is the equation of the lower half circle.}

(

)

(

)

⇒ − − = − − = ′ ⇒ − = − − 2 2 1 2 2 1 2 2 25 ) 2 ( 25 2 1 ) ( 25 25 x x x x x f x x 4 3 16 3 9 25 3 3 25 3 ) 3 ( 3 =− − = − − = − − = ′ = f m _.

Sometimes, an equation [x2+y2 =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 dx dy

.

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

dx dy x

f′( ) = _{, you differentiate} with respect to x and substitute y for f(x) and

dx dy

for f ′(x)_{. Then you can} 9 solve for dx dy . Solution 2: 2+ 2 =25⇒ 2+[ ( )]2 =25⇒

(

x2+[f(x)]2 =25

)

⇒ dx d x f x y x

4

3

)]

(

[2

2

)

(

0

)

(

)]

(

[2

2

4

3

=

−

⋅

⇒

−

=

⇒

−

=

′

⇒

=

′

+

=

=

y

x

d x

d y

y

x

d x

d y

x

f

x

x

f

x

f

x

f

x

_.

Example 2: Suppose that the equation x y x + =

3 2

defines a function f with ) (x f y = . Find dx dy

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

Solution 1: Solve for y. ⇒

− = ⇒ = + ⇒ =     ₊ 2 3 3 2 ) ( 3 2 2 2 x x y y x x y x xy y x xy ₄18 ₂9 ) 2 ( 6 3 ) 2 ( ) 2 ( 3 ) 3 )( 2 ( 2 2 2 2 2 2 2 − = − = ⋅ ⇒ − − − = − − − = _x= dx dy x x x x x x dx dy

Solution 2: Clear fractions ⇒ + = ⇒

(

y+ x=x y

)

⇒ dx d y x x y 3 2 2 3 2 2

2

9

2

1 2

3

2

2

3

2

3

2

3

2

2

2

_⇒

_⋅

₌

−

₌

₋

−

−

=

⇒

+

=

+

=

=

y

x

d x

d y

x

x y

d x

d y

x y

d x

d y

x

d x

d y

Solution 3: _2+3 = _⇒

(

2 −1+3 −1=

)

⇒−2 −2−3 −2 =1⇒ dx dy y x x y x dx d x y x dx d − − = ⇒− 2− 2 = 2 2⇒ =− 2−₂ 2 2 ⇒ 2 2 ₃ 2 3 2 1 3 2 x y x y dx dy y x dx dy x y dx dy y x

2

9

1 2

5 4

1 2

3 6

1 8

3

2

=

−

−

=

−

=

−

⋅

=

=

y

x

d x

d y

Example 3: If cos( xy) =y_{, then find} dx dy

.

(

cos( )=

)

⇒−sin( )_ + (1)_= ⇒− sin( ) −ysin(xy)= dx dy xy x dx dy y dx dy x xy y xy dx d

sin( )

(

1 sin( )

)

₁ sin(_sin( )₎ xy x xy y dx dy xy x dx dy xy y dx dy + − = ⇒ + = − ⇒ 10 IV. Higher Order Derivatives

A. Notation

(1) 1st _{derivative (derivative of the original function y= f(x)): f (x)}

dx

dy _′

= (2) 2nd _{derivative (derivative of the 1}st _{derivative): ( )}

2 2 x f dx y d _′′ =

(3) 3rd _{derivative (derivative of the 2}nd _{derivative): ( )} 3 3 x f dx y d _′′′ = 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

) ( ) ( ) (t v t a t

s′′ = ′ = is an acceleration function with respect to time t. Example 1: If f(x)=x2sin x, then find f′(x)and f ′′(x).

x x x x x f′( )= 2cos +2 sin x x x x x x x x x x x x x

f ′′( )= 2(−sin )+2 cos +2 cos +2sin =− 2sin +4 cos +2sin

Example 2: If 5 4 3 2 ) ( − + = x x x

g _{, then find} g′(x) _and g′′(x)_.

2 2 2 2 _{(4 5)} 22(4 5) 22 ) 5 4 ( 12 8 10 8 ) 5 4 ( ) 4 )( 3 2 ( ) 2 )( 5 4 ( ) ( =− − − − − = − − − − = − + − − = ′ x x x x x x x x x g 3 3 3 ) 5 4 ( 176 ) 5 4 ( 176 ) 4 ( ) 5 4 ( 44 ) ( − = − = − = ′′ − − x x x x g

Example 3: If x2+y2 =25, then find

dx dy and 2₂ dx y d .

(

)

_yx y x dx dy dx dy y x y x dx d _{+ = ⇒ + = ⇒ =}− ₌− 2 2 0 2 2 25 2 2 3 2 2 2 2 2 2 ( 1) ( ) y x y y y x x y y dx dy x y y x dx d dx dy dx d dx y d ₌− −    − + − =       − − − =     − =       = = 3 3 2 2 _{) 25} ( y y y x + ₌− − 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)=x4 −5x3+9x2 −7x+5 (2) 9₂ 8₃ 2₄ x x x y= − + (3) ( ) 8 ₃ 6₂ x x x g = − ₍₄₎ 3 2 ₆ 3 x x x y= −

(5) 1 2 3 ) ( ₂ + + = x x x h ₍₆₎ y=x2cos x (7) x x x f( )=sin (8) _y=3 _x2_−3x+4 (9) g(x) =sin( x) (10) y=cos3 x (11) 4 3 1 2 ) ( − + = x x x h (12) x x y tan 1 sec + = (13) _{k(x)=x 9−x}2 _{(14) y} =_{sin(3x)cos( 4x)} (15) _f(x)=tan4

( )

_x3 ₍₁₆₎ 1 1 − + = x x y (17)       = x x x g( ) sec 1 ₍₁₈₎ y= 1+sin 2x II. Find dx dy

by implicit differentiation for each of the following: (1) −3xy −4y2 =2 (2) 8x2 =2y3+3xy2 (3) y y x+ = 1 2 3 (4) x y_y + − = 2 2 3 2 12 (5) x=tan y (6) y =cos(x−y)

(7) xsin y+ysin x=1 ₍₈₎ x=sec3(y2−1)

III. Find the slope of the tangent line at the given point on each curve defined by the given equation: (1) x2+3y2 =21; (3, – 2) (2) _x3₊3 _{y =}3_{; (1, 8)} (3) xy −y=−2_{; (1, 4) (4)} 3xy −2x4 =y3−23; (2, – 3) (5) x cos= y;       − 3 , 2 1 π (6) sin( xy)=x_;       2 , 1π IV. For each of the following functions f (x)_{, find} f ′(x) _and f ′′(x)_.

(1) f(x)=3x4 −4x2 +7x−11 (2) 1 2 1 3 ) ( − + = x x x f (3) f(x)=x3cos( 4x) (4) f(x)=sin4 x

V. Suppose the distance (in feet) that an object travels in t seconds is given by the formula s(t)=2t3 +4t−5. Find s(2), v(2), and a(2).

Solution Key for Techniques of Differentiation

I. (1) f′(x)=4x3−15x2−18x−7 (2) =9 −2 −8 −3+2 −4 ⇒ =−18x−3+24x−4−8x−5 dx dy x x x y (3) _g(x)=8x12 _−6x−23 _{⇒ g′(x)=4x}−12_+4x−53 13 (4) =3 −1−6 −2 ⇒ =−3x−2+12x−3 dx dy x x y (5) 2 _{2 2} ) 1 ( ) 2 )( 2 3 ( ) 3 )( 1 ( ) ( + + − + = ′ x x x x x h (6) x x x x dx dy 2 )( (cos ) sin ( 2 _{− +} = ) (7) ( ) (cos ) ₂(sin )(1) x x x x x f ′ = − (8) =

(

2−3 +4

)

13 ⇒ = 1₃

(

x2−3x+4

)

−23

(

2x−3

)

dx dy x x y (9) g′(x)=cos

( )

x ⋅_1₂x−12_

(10)

(

cos

)

3 3

(

cosx

)

2( sinx) dx dy x y= ⇒ = − (11)         − + − −       − + = ′ ⇒       − + = − 2 ₂ 1 2 1 ) 4 3 ( ) 3 )( 1 2 ( ) 2 )( 4 3 ( 4 3 1 2 2 1 ) ( 4 3 1 2 ) ( x x x x x x h x x x h (12) ₂ 2 ) tan 1 ( ) (sec sec ) tan )(sec tan 1 ( x x x x x x dx dy + − + =

(13) k(x) x

(

9 x2

)

12 k(x) x 1₂

(

9 x2

)

12( 2x) +

(

9−x2

)

12(1)         − − = ′ ⇒ − = −

(14) sin(3x)

[

sin(4x)(4)

]

cos( 4x)

[

cos(3x)(3)

]

dx

dy _{= − +}

(15) _f(x)=

[

_tan

( )

_x3

]

4 _⇒f′(x)=4

[

_tan

( )

_x3

]

3

[

_sec2

( )

_x3

]

( )

_3x2

(16)

(

)

(

)

(

)

2 1 2 1 1 2 1 1 −     + −     − = x x x x x dx dy

( )

_      ₌ x x dx d 2 1

(17) ( ) sec 1 tan 1 1₂ sec 1(1)      +          ₋             = ′ x x x x x x g

(18)

(

1 sin2

)

12 1₂

(

1 sin2x

) (

12 cos2x

)

(2) dx dy x y= + ⇒ = + − 14 II. (1) y y_dxdy _dxdy _x y_y dx dy x 8 3 3 0 8 3 3 + − = ⇒ = − − − (2) xy y y x dx dy y dx dy xy dx dy y x 6 6 3 16 3 6 6 16 ₂ 2 2 2 + − = ⇒ + + = (3) 3 4 2 2 2 4 2 3 2 2 3 1 2 3 2 2 2 − − = ⇒ + = + ⇒ = + ⇒ = + xy y dx dy y dx dy xy dx dy xy x y y y x (4) 1 3 6 12 6 3 12 2 3 6 2 2 2 ₂ + − − = ⇒ − = + + ⇒ − = + x xy x dx dy dx dy xy dx dy x x y y x x (5)

(

)

y y dx dy dx dy y ₂ 2 2 _cos sec 1 sec 1= ⇒ = = _{or 2 2} 1 1 tan 1 1 x y dx dy + = + = (6) ) sin( 1 ) sin( 1 ) sin( y x y x dx dy dx dy y x dx dy − − − − = ⇒     − − − = (7) y y x x dy_{dx dx}dy _x y_y y _xx dx dy y x sin cos cos sin 0 ) (sin cos sin ) (cos + − − = ⇒ = + + + (8)

[

]

⇒ =      − − − = dx dy dx dy y y y y 1)sec( 1)tan( 1) 2 ( sec 3 1 2 2 2 2

) 1 tan( 6 1 ) 1 tan( ) 1 ( sec 6 1 2 2 2 3 _{y − y −} = _{xy y −} y III. (1) 2 1 0 12 6 0 ) 2 ( 6 ) 3 ( 2 0 6 2 + = ⇒ + − = ⇒ − = ⇒ = dx dy dx dy dx dy dx dy y x (2) _ = ⇒       + ⇒ =         + ⇒ =     + − 0 8 3 1 ) 1 ( 3 0 3 1 3 0 3 1 3 3 2 2 3 2 2 3 2 2 dx dy dx dy y x dx dy y x 0 36 12 1 3+ = ⇒ =− dx dy dx dy (3) 3 4 4 4 1 ) 2 ( 2 ) 2 ( − 2 ⇒ + = − ⇒ + = ⇒ = = dx dy dx dy dx dy dx dy y y dx dy x y xy 15 (4) + − = ⇒ + − − = − ⇒ dx dy dx dy dx dy y x y dx dy x 3 8 3 3 2 3(2) 3( 3) 8(23) 3( 3)2 3 21 73 27 64 9 6 − − = ⇒ = − dx dy dx dy dx dy (5) 3 2 2 3 1 3 sin 1 ) sin ( 1 _ ⇒ = ⇒ =            − − = ⇒ − = dx dy dx dy dx dy dx dy y π (6)

=

























−

=

⋅

⇒

−

=

⇒

=









₊

=

=

2

c o s

1

2

c o s

2

1

)

c o s (

)

c o s (

1

1

)

c o s (

2

1

_π

π

π

π

y

x

d x

d y

x y

x

x y

y

d x

d y

y

d x

d y

x

x y

dx dy ⇒ 0 1

does not exist.

IV. (1) f′(x)=12x3−8x+7 and f′′(x)=36x2−8 (2) 2 _{(2 1)}2 5(2 1) 2 5 ) 1 2 ( ) 2 )( 1 3 ( ) 3 )( 1 2 ( ) ( =− − − − − = − + − − = ′ x x x x x x f _and 3 3 ) 1 2 ( 20 ) 2 ( ) 1 2 ( 10 ) ( − = − = ′′ − x x x f

(3) f′(x)=x3[−4sin( 4x)]+3x2cos( 4x)=−4x3sin( 4x)+3x2cos( 4x) and

−16x3cos( 4x)−24x2sin( 4x)+6xcos( 4x)

(4) f′(x)=4(sin x)3cos x and f ′′(x)=4

(

sin x

) (

3 −sin x

) (

+ cos x

)

(

12(sin x)2(cos x)

)

=

−4sin4 x+12sin2xcos2 x

V. s(2)=2

( )

23 +4(2)−5=16+8−5=19 ft

( )

2 4 28 6 ) 2 ( 4 6 ) ( ) (t =s′t = t2 + ⇒v = 2 + = v ft/sec 24 ) 2 ( 12 ) 2 ( 12 ) ( ) ( ) (t =v′t =s′′ t = t⇒a = = a _ft/sec 2 16