Chapter_2_-_Derivatives.

1. TANGENT LINES AND RATES OF CHANGES

1.1 Average rate of change

In general, if x and y are quantities related by an equationy=f (x ), we can consider the rate at which y changes with x. Let’s refer to Figure 1, geometrically; the average rate of change of y

with respect to x over the interval[ x 0 , x1 ] is the slope of the line joining the points ¿ and¿.

The slope of this line can be calculated as

m=∆ y∆ x

= f (x¿¿1)− f (x¿¿0)x1− x0

¿¿

Figure 1

Definition If y=f (x ), then the average rate of change of y with respect to x over the

interval [ x 0 , x1 ] israve=f (x¿¿1)−

f (x¿¿0)x1−x0

¿¿

Example1: Let y=x2+1, find the average rate of change of y with respect to x over the interval [3,5].Solution:

Let x0=3 and x1=5, then:

rave=f (x¿¿1)−f (x¿¿0)x1−x0

=f (5)−f (3)

5−3=26−10

2=8¿¿

Thus, on the average, y increases 8 units per unit increase in x over the interval [3,5].

1.2 Instantaneous rate of change (tangent of a curve)

Instantaneous rate of change is calculated with respect to a point and not an interval. Let’s refer to Figure 2. Geometrically, the instantaneous rate of change of y with respect to x at x=x0 is the slope of the line that touches the curve only once at ¿ x0 (This line is also called as tangent of y=f (x ) at ¿ x0 ).

Figure 2

To find the slope of this line, let’s employ the formula used in section 1.1 and approach to x0 form x1 (see Figure 3), and then we have

m= limx1→ x0

f (x¿¿1)− f (x¿¿0)x1−x0

¿¿

Figure 3

Definition If y=f (x ), then the instantaneous rate of change of y with respect to x when x=x0 is

rinst= limx1 → x0

f (x¿¿1)− f (x¿¿0)x1−x0

¿¿

Example2: Let y=x2+1, find the average rate of change of y with respect to x when x = -4.Solution:Let x0=−4, then

rinst= limx1 → x0

f (x¿¿1)−f (x¿¿0)x1−x0

= limx1 →−4

f (x¿¿1)−f (−4 )x1−(−4)

= limx1 →−4

( x12+1 )−17x1+4 = lim

x1 →−4

(x12−16)

x1+4 = limx1 →−4

( x1−4 )(x1+4)x1+4 = lim

x1 →−4(x¿¿1−4 )=−8¿¿¿¿

Thus, for a small change in x from x = -4, the value of y will change approximately 8 times as much in the opposite direction. That is, because the instantaneous rate of change is negative, the value of y decreases as the values of x move through x = -4 from left to right.

2. DERIVATIVE

Figure 4

As mentioned in section 1.2, at any point, tangent to a curve is a line that touches the curve only once at that point. Figure 4 shows another notation for finding the slope of the tangent line to the graph f(x) at x=x0. From figure 4, let x1−x0=h, then the slope is:

m= limx1→ x0

f (x¿¿1)− f (x¿¿0)x1−x0

=limh → 0

f ( x0+h )−f (x¿¿0)

h¿¿¿

And the equation of the tangent line: y−f ( xo )=m .( x−xo ).

The expression

f ( xo+h )−f (xo )h is called the difference quotient of f at xo with

increment h. If the difference quotient has a limit as h approaches zero, that limit is given a special name and notation. In other words, the slope of the tangent line to the graph f(x) is

also called derivative of f(x) and is usually denoted asf ' ( x ).

Definition The derivative of a function f at a point xo is denoted asf '( xo ) and its value is

f '( xo )=limh→ o

f ( xo+h)−f ( xo )h (Provided this limit exists)

Example 3:Find the slope of the graph of y=x2+1 at the point (2,5), and use it to find the equation of the tangent line to y=x2+1 at x=2.Solution:The slope of the graph at the point (2,5) is given by

f '(2 )= limh→o

f (2+h)− f (2)h

= limh→ o

(2+h )2+1−5h

=limh→o

h2+4 hh

=limh→o

( h+4 )=4

The tangent line is the line through the point (2,5) with slope 4,

y−f (2 )=4 ( x−2 )⇒ y−5=4( x−2)⇒ y=4 x−3

Note: The process of calculating a derivative is known as Differentiation.A function f is differentiable at a if f

'( a) exists. If f

'( x ) exist (the limit exist), => f is differentiable at x => f has a derivative at x.If f

'( x ) exists at every point in the domain of f, we call f differentiable.

Notation:The derivative of f ( x ) with respect to x:

f '( x )= y '= ddx

[ f (x )]= dydx

= dfdx

=Df ( x )=D x f (x ) if y=f ( x )

The derivative of f ( t ) with respect to t:

f '( t )= y '= ddt

[ f ( t ) ]=dydt

=Df ( t ) if y=f ( t )

The derivative of f at the point x=a :

dydx

|x=a=ddx

[ f ( x ) ]|x=a=f ' (a )

2.1 One-Sided Derivatives (Differentiable on an Interval)

A function y=f ( x ) is differentiable on an open interval if it has a derivative at each point of the interval.

It is differentiable on a closed interval [ a , b ] if it is differentiable on the interior (a , b ) and if

the limits limh→0+

f ( x+h)−f ( x )h and

limh→ 0−

f ( x+h)−f (x )h exists at the endpoints.

Note:If f

'( a)does not exist (the limit does not exist) => derivative of f does not exist at x=a .(occurs at corner points, cusp points, points of vertical tangency, points of discontinuity)

Theorem (Differentiation and continuity): If f has a derivative at x=c , then f is continuous at x=c .

3. TECHNIQUES OF DIFFERENTIATION

In the last sections, we defined the derivative of a function f as a limit, and we used that limit to calculate a few simple derivatives. In this section, we will develop some important theorems that will enable us to calculate derivatives more efficiently.

The graph of a constant function f ( x )=c is the horizontal line y = c and hence the tangent line to this graph has slope 0 at every value of x. thus, we should expect the derivative of a constant function to be 0 for all x.

Theorem (derivative of constant function): the derivative of a constant function is 0; that is, is c is any real number, then

f ' ( x )=0 or ddx

[ c ]=0

Theorem (The Power Rule): If n is an integer d

dx[ xn ]=nxn−1

In words, the derivative of x raised to a positive integer power is the product of the integer exponent and x raised to the next lower integer power.

Example4:d

dx[ x2 ]=2 x

,

ddx

[ x3 ]=3 x2

,

ddt

[ t12 ]=12t 11

,

Theorem ( The Constant Multiple Rule ): if f is differentiable at x and c is any real number, then cf is also differentiable at x and

ddx

[cf ( x ) ]=c . ddx

[ f ( x ) ]; (cf { )'=c f ' ¿

In words, a constant factor can be moved through a derivative sign.

Example5: ddx

[ 4 x8 ]=4 ddx

[ x8 ]=4 [8 x7 ]=32 x7

ddx

[−x12 ]=(−1 ) ddx

[ x12 ]=−12 x11

ddx

[ πx

]=π ddx

[ x−1 ]=π (−x−2)=− πx2

Theorem ( The Sum and Difference Rule): If f and g are both differentiable at x, then so are f + g and f - g

ddx

[ f ( x )±g (x )]= ddx

[ f ( x )]± ddx

[ g( x ) ]; ( f ±g)'= f '±g'

Example6:ddx

[ 2 x6+x−9 ]= ddx

[ 2x6 ]+ ddx

[ x−9 ]=12 x5+(−9 )x−10=12 x5−9 x−10

ddx

[ √x−2 x√x

= ddx

[1−2√x ]= ddx

[1 ]− ddx

[2√ x ]=0−2( 12√ x )=− 1

√ x

Example7:

At what points, if any, does the graph of y=x3−3 x+4 have a horizontal tangent line?

Solution:

Horizontal tangent line have slope zero, so we must find those values of x for which .y ' ( x )=0 . Differentiating yields

y '( x )= ddx

[ x3−3 x+4 ]=3 x2−3

Thus, horizontal tangent lines occur at

those values of x for which 3 x2−3=0 , that is, if x=−1or x=1 . The corresponding points on the curve y=x3−3 x+4 are (-1,6) and (1,2) (see Figure 5)

Figure 5

Theorem ( The Product Rule): If f and g are both differentiable at x, then

ddx

[ f ( x ) . g( x ) ]=f ( x ) ddx

[ g( x ) ]+g( x ) ddx

[ f ( x ) ];( f . g)'= f . g'+g , f '

Example8: find differentiation of y=(4 x2−1)(7 x3+x )Solution:There are two methods that can be used to find dy/dx. We can either use the product rule or we can multiply out the factors in y and then differentiate. We will give both methods.

Method 1.(using the product rule)dydx

=ddx

[(4 x2−1 )(7 x3+ x ) ]=(4 x2−1)ddx

[7 x3+x ]+(7 x3+x )ddx

[ 4 x2−1 ]

¿( 4 x2−1)(21 x2+1)+(7 x3+x )(8 x )=140 x4−9 x2−1

Method 2. (multiplying first)y=(4 x2−1)(7 x3+x )=28 x5−3 x3−xThus,ddx

[ 28 x5−3 x3−x ]=140 x4−9 x2−1

Which agrees with the results obtained using the product rule.

Example9:

Find ds/dt if s=(1+t )√t .Solution: Applying the product rule yields dsdt

= ddt

[(1+t )√ t ]=(1+t ) ddt

[√ t ]+√t ddt

[1+t ]=1+t2√t

+√t= 1+3t2√ t

Theorem ( The Quotient Rule) : If f and g are both differentiable at x, and g( x )≠0 , then

⇒ ddx [ f ( x )

g ( x ) ]=g ( x ) d

dx[ f ( x ) ]−f ( x ) d

dx[ g (x )]

[ g ( x ) ]2; ( f

g )'=g . f '− f . g'

g2

Example10: let f ( x )= x2−1

x4+1 , find f ' ( x )Solution:

dydx =

ddx [ x2−1

x4+1 ]=( x4+1 )d

dx[ x2−1]−( x2−1)d

dx[ x 4+1 ]

( x4+1 )2

¿( x4+1)(2 x )−( x2−1 )(4 x3 )( x4+1)2 =−2 x5+4 x3+2 x

( x4+1)2

Theorem ( The Reciprocal Rule) : If g is differentiable at x, andg( x )≠0 , then

ddx [ 1

g( x ) ]=− d

dx[ g( x ) ]

[ g( x ) ]2; ( 1

g )'=− g'

g2

4. HIGHER DERIVATIVES

First derivative of f: ( f )'=f 'dydx

= y '

Second derivative of f: ( f ' )'=f ' '

ddx [ dy

dx ]=d2 ydx2 = y ' '=D2 f ( x )

Third derivative of f: ( f ' ' )'=f ' ' '

ddx [ d2 y

dx2 ]=d3 ydx3 = y ' ' '=D3 f ( x )

Fourth derivative of f: ( f ' ' ' )'= f (4 )

.

.

.

nth derivative of f: f (n)dn ydxn = dn

dxn [ f ( x ) ]= y(n)=Dn f ( x )

5. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

We will assume in this section that the variable x in the trigonometric functions is measured in radians. The derivatives of the trigonometric functions are

ddx

[ sin x ]=cos x ddx

[ cos x ]=−sin x ddx

[ tan x ]=sec2 x

ddx

[ sec x ]=sec x tan x ddx

[ cot x ]=−csc2 x ddx

[ csc x ]=−csc x cot x

To find the derivatives that involves trigonometric functions, use the various differentiation formulas.

Example11: find dy/dx if y=x sin x .Solution: using derivatives of trigonometric functions and the product rule we obtain

dydx

= ddx

[ x sin x ]=x ddx

[sin x ]+sin x ddx

[ x ]=x cos x+sin x

Example12: find f' ' ( π /4 ) if f ( x )=sec x

Solution:f '( x )=sec x tan x

f ' ' ( x )=sec x ddx

[ tan x ]+ tan x ddx

[ sec x ]

=sec x sec2 x+tan x . sec x tan x=sec3 x+sec x tan2 xThus, f ' ' ( π /4 )=sec3 ( π /4 )+sec ( π /4 ) tan2 (π /4 )=(√2)3+(√2)(1)2=3√2

6. THE CHAIN RULE

Theorem (The Chain Rule): If g is differentiable at the point x and if f is differentiable at the point g(x), then the composite function ( f ∘g )is differentiable at the point x. moreover, if y=f [ g ( x ) ] and u=g ( x ) , then y=f (u )and

dydx

=dydu

⋅dudx

Example13: find dy/dx if y=cos( x3 )

Solution: let u=x3and express y as y=cos(u ). Applying chain rule yields

dydx

=dydu

⋅dudx

=ddu

[ cosu ]⋅ddx

[ x3 ]

¿(−sin u ).(3 x2 )¿(−sin (x3 )) .(3 x2)=−3 x2 sin( x3 )

Example14: find dw/dt if w=tan( x )and x=4 t3+t .Solution: in this case the chain rule computations take the form dwdt

=dwdx

⋅dxdt

=ddx

[ tan x ]⋅ddt

[ 4 t3+t ]

¿(sec2 x ) .(12t 2+1 )¿ [sec2( 4 t3+t ) ] .(12 t 2+1)=(12 t2+1)sec2(4 t3+ t )

3.3.1 An alternative version of the chain rule

dydx

=dydu

⋅dudx can be unwieldy in some problems because it involves so many variables. As

you become more comfortable with the chain rule, you may want to dispense with writing out the dependant variables by expressing the chain rule in the form

ddx

[ f ( g( x )) ]=( f ∘g )' ( x )=f ' (g ( x )) . g' ( x )

A convenient way to remember this formula is to callg( x ) is the “inside function” and f ( x ) is the “outside function” in the compositionf (g ( x )) . So, in words, the derivative of f (g ( x ))is the derivative of the outside function evaluated at the inside function times the derivative of the inside function.

Example15: find h' ( x ) if h( x )=cos (x3 ) .

Solution: we can this of h as a composition f (g ( x )) in which g( x )=x3is the inside function

and f ( x )=cos ( x )is the inside function. Thus, h' ( x )=f ' ( g( x )) .g '( x )=f '( x3 ). 3 x2

¿−sin( x3 ). 3 x2=−3 x2sin( x3 )

3.3.2 Generalized Derivative Formulas

There is a useful third variation of the chain rule that strikes a middle ground between the

previously discussed formulas in chain rule. If let u=g ( x ) in

ddx

[ f (g ( x )) ]=f ' (g ( x )) . g '( x ),

the we can rewrite that formula as ddx

[ f (u ) ]= f ' (u) . dudx

This result, called the generalized derivative formula for f, provides a way of using the derivative of f(x) to produce the derivative of f(u), where u is a function of xUsing this general formula, we can obtain the following theorem:

Theorem: If n is any real number and u=g ( x ) is differentiable, then

⇒ ddx

[ un ]=nun−1 . dudx

⇒ ddx

[ g ( x ) ]n=n[ g ( x ) ]n−1 . g '( x )

7. IMPLICIT DIFFERENTIATION

Up to now we have been concerned with differentiating functions that are given by equations of form y=f ( x ) . In this section, we will consider methods for differentiating functions for which it is inconvenient or impossible to express them in this form.

Functions defined explicitly and implicitly

An equation of the form y= f ( x ) is said to define y explicitly as a function of x because the variable y appears alone on one side of the equation and does not appear at all on the other side. However, sometimes functions are defined by equations in which y is not alone on one

side; for example, the equation yx+ y+1=x is not of the form y=f ( x ) , but still defines y as a function of x since it can be written as

y= x−1x+1

Thus, we say that yx+ y+1=x defines y implicitly as a function of x, the function being

f ( x )= x−1x+1

We can use the following steps to perform implicit differentiation.

1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x.

2. Collect the terms with

dydx on one side of the equation.

3. Solve for

dydx .

Example16: Use implicit differentiation to find dy/dx if 5 y2+sin y=x2.

Solution:ddx

[ 5 y2+sin y ]= ddx

[ x2 ]

5 ddx

[ y2 ]+ ddx

[ sin y ]=2 x

5(2 y dydx )+(cos y ) dy

dx=2 x

10 y dydx

+(cos y ) dydx

=2 x

dydx

= 2 x10 y+cos y

Note that this formula involves both x and y. in order to obtain a formula for dy/dx that involves x alone, we would have to solve the original equations for y in terms of x and the substitute in the final answer. However, it is impossible to do this, so we are forced to leave the formula for dy/dx in terms of x and y.

8. DERIVATIVES OF LOGARITHMIC FUNCTIONS

In this section we will obtain derivative formulas for logarithmic functions. Let’s start with f ( x )= ln x :

f '( x )= ddx

[ ln x ]=1x

x>0

A derivative formula for the general logarithmic function logb x can be obtained as below:

ddx

[ logb x ]= ddx

[ ln xln b

]= 1ln b

ddx

[ ln x ]

It follows from this thatddx

[ logb x ]= 1x ln b

x>0

Let u be a differentiable function of x, and if u(x) > 0, then applying the chain rule to the above formulas produces the following generalized derivative formulas:

ddx

[ ln u ]=1u

. dudx and

ddx

[ logb u ]= 1u ln b

. dudx

Example17: find

ddx

[ ln ( x2+1 )]

Solution:ddx

[ ln( x2+1 )]= 1x2+1

. ddx

[ x2+1 ]= 1x2+1

.2 x= 2 xx2+1

When possible, the properties of logarithms should be used to convert products, quotients, and exponents into sum, difference, and constant multiples before differentiating a function involving logarithms.

Example18:

ddx [ ln (x2 sin x

√1+x )]=ddx [2 ln x+ ln (sin x )−1

2ln (1+x )]

¿2x +

cos xsin x −

12(1+x )

=2x +cot x−1

2+2 x

Figure 6

Figure 6 shows the graph of f ( x )= ln|x|. This function is important because it “extends” the

domain of the natural logarithm function in the scenes that the values of ln|x| and ln x are the

same for x > 0, but ln|x|is defined for positive values of x

The derivative of ln|x|for x ≠ 0 can be obtained by considering the cases x > 0 and x < 0 separately:

Case x > 0. In this case |x|= x , soddx

[ ln|x|]= ddx

[ ln x ]=1x

Case x > 0. In this case |x|=−x , soddx

[ ln|x|]= ddx

[ ln (−x )]=1x

Since the same formula results in both cases, we have shown thatddx

[ ln|x|]=1x

if x≠0

Example19: using the above formula and the chain rule: ddx

[ ln|sin x|]= 1sin x

ddx

[ sin x ]=cos xsin x

=cot x

Example20: The derivative of y= x2 3√7 x−14

(1+x2 )4is messy to calculate directly. However, if we

first take the natural logarithm of both sides and then use its properties, we can write

ln y=2 ln x+ 13

ln (7 x−14 )−4 ln(1+x2)

Differentiating both sides with respect to x yields1y

dydx

=2x+ 7/3

7 x−14− 8x

1+x2

Thus, on solving for dy/dx and using y= x2 3√7x−14

(1+x2 )4, we obtain

dydx =

x2 3√7 x−14(1+x2)4 [ 2

x +7 /3

7 x−14−8 x

1+ x2 ]9. DERIVATIVES OF EXPONENTIAL FUNCTIONS

To obtain a derivative formula of bx

, we rewrite y=bxas x= logb y and differentiate

implicitly to obtain:

1= 1y ln b

. dydx

Solving for dy/dx and replacing y by bx

we havedydx

= y ln b=bx ln b

Thus, we have shown that

ddx

[ bx ]=bx ln b

In special cases where b = e we have ln e = 1, and we have

ddx

[ ex ]=ex

Moreover, if u is a differentiable function of x, then using the chain formula, we can obtain

⇒ ddx

[ bu ]=bu ln b . dudx and

ddx

[ eu ]=eu . dudx

10. DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS

Definition The inverse sine function, denoted by sin−1 , is defined to be the inverse of the restricted sine function

sin x , −π /2≤x≤π /2Definition The inverse cosine function, denoted by cos−1

, is defined to be the inverse of the restricted cosine function

cos x , 0≤x≤πDefinition The inverse tangent function, denoted bytan−1

, is defined to be the inverse of the restricted tangent function

tan x , −π /2<x<π /2Definition The inverse secant function, denoted by sec−1

, is defined to be the inverse of the restricted secant function

sec x 0≤x≤π with x≠π /2

Let u be a differential function of x, then, derivatives of inverse trigonometric functions are:

ddx

[sin−1 u ]= 1√1−u2

dudx

(−1<u<1)

ddx

[cos−1 u ]= 1√1−u2

dudx

(−1<u<1 )

ddx

[ tan−1 u ]= 11+u2

dudx

(−∞<u<+∞)

ddx

[cot−1 u ]=− 11+u2

dudx

(−∞<u <+∞)

ddx

[sec−1u ]= 1|u|√u2−1

dudx

(1<|u|)

ddx

[csc−1 u ]=− 1|u|√u2−1

dudx

Example21:

Find

dydx if

(a) y=sin−1 ( x3) (b) y=sec−1 ( ex )

a) Solution dydx

= 1

√1−( x3 )2(3 x2)= 3 x2

√1−x6

b) Solutiondydx

= 1

ex √( ex )2−1(ex )= 1

√e2 x−1

11. L'HÔPITAL'S RULE

L'Hôpital's rule is a general method of using derivatives to find limits. This method enables us to resolve indeterminate forms of various type such as 0/0, ∞/∞, and so on. In chapter 1, we used numerical and algebraic methods to resolve indeterminate forms in limits.

11.1 L'Hôpital's rule for form 0/0

Suppose that f and g are differentiable functions on an open interval containing x = a, except possibly at x = a, and that

limx→a

f ( x )=0 and

limx→a

g( x )=0

If limx→a

[ f '( x ) /g' ( x )]exists, or if this limit is +∞ or -∞, then

limx→a

f ( x )g ( x )

=limx→a

f '( x )g' ( x )

Moreover, this statement is also true in the case of a limit as x→a−, x→a+

, x→−∞ or as x→+∞ .

Applying L'Hôpital's rule:Step1: check that the limit of f ( x )/g ( x )is an indeterminate form of 0/0.Step2: differentiate f and g separately.Step3: find the limit of f

'( x )/g' ( x ) . If this limit is finite, +∞ or -∞,then it is equal to the limit of f ( x )/g ( x )

Example22: find the limit limx→2

x2−4x−2 using L'Hôpital's rule.

Solution: Try substitution first:

limx→2

x2−4x−2

=22−42−2

=00

Applying L'Hôpital's rule:

limx→2

x2−4x−2

= limx→2

ddx

[ x2−4 ]

ddx

[ x−2 ]=lim

x→2

2 x1

=4

This agrees with the computation

limx→2

x2−4x−2

= limx→2

( x−2 )(x+2 )( x−2 )

=limx→2

( x+2)=4

11.2 L'Hôpital's rule for indeterminate for of ∞/∞

Suppose that f and g are differentiable functions on an open interval containing x = a, except possibly at x = a, and that

limx→a

f ( x )=∞ and

limx→a

g( x )=∞

If limx→a

[ f '( x ) /g' ( x ) ]exists, or if this limit is +∞ or -∞, then

limx→a

f ( x )g ( x )

=limx→a

f '( x )g' ( x )

Moreover, this statement is also true in the case of a limit as x→a−, x→a+

, x→−∞ or as x→+∞ .

Example22: find the limit lim

x→+∞

xex

using the L'Hôpital's ruleSolution:

Try substitution first:lim

x→+∞

xex

= ∞e∞ =∞

∞, this is indeterminate form

Applying L'Hôpital's rule: lim

x→+∞

xex

= limx→+∞

1ex

= 0∞=0

11.3 L'Hôpital's rule for indeterminate form of 0.∞, ∞-∞

Example: Evaluate the following limits using hospitals rule

a) limx→0+

x ln x b)

limx→π /4

(1−tan x )sec 2 xc)

limx→0+( 1

x− 1

sin x )Solution:

a) try substitution: limx→0+

x ln x=0 . ln0=0 .∞

Re-writing the limit:limx→0+

x ln x=limx→0+

ln x1/ x now, we have indeterminate form of ∞/∞.

Applying L'Hôpital's rule: limx→0+

x ln x= limx→ 0+

ln x1/ x

= 1 /x−1/ x2 =lim

x→0+(−x )=0

b) try substitution: lim

x→π /4(1−tan π /4 )sec2 π /4=0 .∞

Re-writing the limit: lim

x→π /4(1− tan x )sec 2 x= lim

x→ π /4

1− tan xcos2 x now, we have indeterminate

form of 0/0.

Applying L'Hôpital's rule:

limx→π /4

(1− tan x )sec 2x= limx→ π /4

1− tan xcos2 x

= limx→π /4

−sec2x−2sin 2 x

=−2−2

=1

c) try substitution: limx→0+( 1

x− 1

sin x )=( 10− 1

sin 0 )=∞−∞

combining the two terms: limx→0+( 1

x− 1

sin x )=limx →0+(sin x−x

xsin x ) now, we have indeterminate

form of 0/0.Applying L'Hôpital's rule twice:

limx→0+( 1

x− 1

sin x )=limx →0+(sin x−x

x sin x )= limx →0+(cos x−1

sin x+x cos x )= lim

x→0+(cos x−1sin x+xcos x )= lim

x→0+( −sin xcos x+cos x−x sin x )=0

2=0