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Calculus

http://matrix.skku.ac.kr/Cal-Book1/Ch1/

http://matrix.skku.ac.kr/Cal-Book1/Ch2/

Chapter 3.

3.1 Definition of Derivatives, Differentiation   http://youtu.be/A-vDsF9ulTs

3.2 Derivatives of Functions, The Product and Quotient Rules

http://youtu.be/XXMnCESesfQ

3.3 The Chain Rule and Inverse Functions      http://youtu.be/HfScHEsPfKI

3.4 Approximation and Related Rates  http://youtu.be/ViRwEJ0Wfkw

3.1 Definition of Derivatives, Differentiation

DEFINITION 1

Let be a function defined on an open interval .

The function defined by the formula

is called the derivative with respect to of the function

assigns the number to each , so that we may regard as a new function. Also we note that the domain of is the set .

EXAMPLE 1

Find if .

Solution.   From the definition, we have

.

Some common notation for the derivative is given below

The operator , which transforms to its derivative, is called the differentiation operator.

The value of a derivative at a specific number , is denoted by or .

DEFINITION 2

A function is differentiable at if exists.

A function is differentiable on if it is differentiable at every number in . One of the typical examples of a non-differentiable function is at

There is a relationship between continuity and differentiability. The relationship is indicated in the following theorem.

THEOREM 3 Differentiability Implies Continuity

If is differentiable at , then is continuous at .

Proof  Assume is differentiable at a point , which implies that

exists.

Note This follows from Definition 1(with , )

To show that is continuous at , we must show that . The key is the identity

,   .

Taking the limit as approaches on both sides of this formula and simplifying, we have

.

Therefore, , which means that is continuous at .

The converse of Theorem 3 is not true. For example, is continuous at but it is not differentiable at .

3.1 EXERCISES (Definition of Derivatives, Differentiation)

1-7. Differentiate each of the following functions using Definition 3.1.1, if the derivative exists.

1.

Solution.

2.

Solution.

3.

Solution.

4.

Solution.

5.

Solution.

6.

Solution.

7.

Solution.

8.Is the function

differentiable at ?

proof.

Hence is differentiable at and .

9. Discuss a geometric and physical interpretation of .

Solution.  1. Geometric interpretation: The slope of the tangent to the curve at a point is equal to the derivative of the function at that point.

2. Physical interpretation: The average velocity is the ratio between

distance travelled () and the time elapsed ().

3.2 Derivatives of Functions, The Product and Quotient Rules

Figure 1

In this section, we shall consider derivatives of some of the basic functions such as polynomials, exponential, trigonometric and inverse trigonometric functions. The product rule and quotient rule are helpful in derivations of derivatives of functions which are expressed as the product and/or quotient of other functions. We also consider applications of the derivative to other sciences.

For the constant function , the graph is the horizontal line , which has slope . So we expect . From the definition of a derivative,

.

Thus, exists for every and its value is zero, that is, we have the following fact and theorem.

THEOREM 1  The Power Rule

If is a positive integer, then

.

proof  Let . Using the formula

we have

.

Hence .                   ■

EXAMPLE 1

Find where

.

Solution.  Let us define and . Then and .

Notethat . This implies

and

.

Hence is differentiable and

since .

Note that might not be differentiable at .

THEOREM 2  The General Power Rule

If is any nonzero real number, then

.

Theorem 1, , is a special case of Theorem 2.

EXAMPLE 2

For the curve find an equation of the tangent line at the point .

Solution. Since , the slope of the tangent line at (1, 1) is . Thus the equation of the tangent line is

or  .

THEOREM 3 The Constant Multiple Rule

If is a constant and is a differentiable function at , then is differentiable at and

.

That is, the derivative of a constant times a function is the constant times the derivative of the function.

proof Let   for some constant . By the limit definition of the derivative: and

To prove the proposition, it suffices to show that .

.

THEOREM 4  The Sum Rule

If and are both differentiable at , then is differentiable at and

.

That is, the derivative of a sum of functions is the sum of the derivatives.

proof.  Left to the reader as an exercise.

The sum rule can be extended to the sum of a finite number of functions, for example

.

In general .

THEOREM 5  The Difference Rule

If and are both differentiable  at , then  is differentiable at and

.

This simple proof is omitted.

EXAMPLE 3

EXAMPLE 4

Find the points on the curve where the tangent line is horizontal.

Solution. The tangent line is horizontal where the derivative is zero. We

have

Thus, if or The corresponding

points are and

Exponential Function

Consider the exponential function where , . From the definition of the derivative:

.

Thus,

.

That is, the rate of change of any exponential function is proportional to the

function itself.

Definition of the Number

The number is an important mathematical constant, it is an irrational

umber which approximately equals to 2.71828. This number arises in the study of

compound interest, and can also be calculated as the sum of the infinite series

.

The constant may be defined in many ways; for example, is the unique real number

such that the value of the derivative (slope of the tangent line) of the function

at the point is equal to 1. The number is defined so that

as when .

Derivative of the Natural Exponential Function

Figure 2

Figure 2

There is a very important exponential function that arises naturally in many places.

This function is called the natural exponential function. However, for most people this

is simply the exponential function.

Then for , implies , and from the general derivative

above we have .

Thus the slope of a tangent line to the curve is equal to the -coordinate of

the point.

EXAMPLE 5

Find the point on the curve where the tangent line is parallel to the line .

Solution. We have . If the tangent line is parallel to , then the slope

of the tangent line is 2. If then The point is .

The Product Rule

In calculus, the product rule is a formula used to find the derivatives of products of two

or more differentiable functions. It may be stated as:

THEOREM 6 The Product Rule

If and are both differentiable at , then is differentiable at and

or

.

That is, the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

proof. Let . We use the continuity of in the proof.

Then

.

EXAMPLE 6

Differentiate the function .

proof. Using the Product Rule, we have

The Quotient Rule

THEOREM 7  The Quotient Rule

If and are differentiable at and , then is differentiable at and

or

That is, the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

proof. Let . Then

.

EXAMPLE 7

For the curve , find the equation of the tangent line at the point .

Solution.    By the Quotient Rule, we have

.

Thus the slope at the given point is . The equation of the tangent line is .

Table of Differentiation Formulas

Derivatives of Trigonometric Functions

In this section, we make use of the following limits.

THEOREM 8  Trigonometric Limits

,

Figure 3

Using the Squeeze Theorem, we have

(Note that if in Figure 3)

Now,

Vibrations, waves, elastic motion and other quantities that vary in a periodic manner can be described using trigonometric functions.

Consider . From the definition of a derivative, we have

.

Thus we have  .

We note that . Similarly, we can show

.

To find the derivative of tan , we will use the quotient rule.

We have

.

Figure 4

We may deduce derivatives of the remaining three trigonometric functions in a similar manner.

So in summary, we have the following.

Derivatives of Trigonometric Functions

For the “cofunctions” namely cosine, cosecant, and cotangent, notice the appearance of the minus signs in their derivatives.

EXAMPLE 8

If find .

Solution.  Differentiation gives us that .

EXAMPLE 9

Find the values of where the graph of has a horizontal tangent.

http://matrix.skku.ac.kr/cal-lab/cal-3-2-Exm10.html

Figure 5

Solution.  Using the Quotient Rule, we have

.

When , (and is not zero for all ). Thus, for the curve has horizontal tangents.

Use http://sage.skku.edu/ or http://www.sagemath.org/eval.html

f(x) = csc(x)/(1 + cot(x))

plot(f(x),-3*pi, pi, ymax=10, ymin= -10,ticks=[[n*pi+pi/4 for n in [-3,3]],[-10,-5,0,5,10]])

Applications

Figure 6

In this section, we consider applications of the rate of change of with respect to , , in physics, chemistry, biology, economics, and other sciences. We have mentioned the following, rates of change that are important in physics are velocity, density, current, power (the rate at which work is done), the rate of heat flow, temperature gradient (the rate of change of temperature with respect to position) and the rate of decay of a radioactive substance.

If and changes from to , then the change in is written as and the corresponding change in is .

So, the average rate of change of with respect to over the interval is given as

which is the slope of the secant line through the points and . Its limit as is the derivative which can therefore be interpreted as the instantaneous rate of change of with respect to , or the slope of the tangent line at .

Physics: Let the position function of a particle that is moving in a straight line be denoted . Then, the average velocity over a time period is the average rate of change and the instantaneous velocity is (the rate of change of displacement with respect to time).

EXAMPLE 10

The position function of a particle is given by the function where time is measured in seconds and position in meters.

(a) Find the velocity at time .

(b)When is the particle moving in the positive direction?

(c) Find the total distance traveled by the particle during the time .

(d) Find the total displacement traveled by the particle during the time .

Solution.  (a)The velocity function is the derivative of the position function:

.

(b)When , the particle is moving in the positive direction. Thus,                         or .

(c) The total distance is

.

(d) The total displacement is

.

Biology: Let be the number of individuals in an animal or plant population at time . The change in the population size between the times and is . So, the average rate of growth during the time period is .

The Instantaneous Rate of Growth is

The instantaneous rate of change is important in many sciences. Consider a population of bacteria in a homogeneous nutrient medium. In the ideal situation, the bacteria’s population has an exponential rate of growth. For example, if the bacteria’s population is , , then we have .

We note that the derivatives of the demand, revenue, and profit functions are known as marginal demand, marginal revenue, and marginal profit, these are used in economics.

In psychology, if represents the performance of someone learning a skill as a function of the training time , then the rate at which  performance improves as time passes is, . In sociology, if denotes the proportion of a population that knows a rumor (innovations, fads or fashions) by time , then the derivative represents the rate at which the rumor is spreading.

In other disciplines, a meteorologist is concerned with the rate of change of atmospheric pressure with respect to height, an urban geographer is interested in the rate of change of the population density in a city as the distance from the city center increases, a geologist may be interested in knowing the rate at which an intruded body of molten rock cools by conducting heat to surrounding rocks, and an engineer wants to determine the rate at which water flows into or out of a reservoir.

CAS EXAMPLE 11

Consider the function

.

Find and plot the graph of and together. Also plot the tangent at

http://matrix.skku.ac.kr/cal-lab/cal-3-2-Exm12.html

[Practice Math & Code in http://sage.skku.edu/

Solution.

Figure 7

var('t')

f(t)=t*(1-t)*exp(-t/2)

f1(t)=f.diff()  # derivative

p1=plot(t*(1-t)*exp(-t/2),-2,3)

p2=plot(f1(t),-2,3,ymin=-10,ymax=5,color='red')

T(t)=f(-1)+f1(-1)*(t+1) # tangent line at

pt=plot(T(t),-1.5,0.2,color='green')

p1+p2+pt

Derivative:

<Curve> Math Love Exhibition

3.2 EXERCISES (Derivatives of Functions, The Product and Quotient Rule)

CAS 1. Find where .

http://matrix.skku.ac.kr/cal-lab/cal-3-1-6.html

[Practice Math & Code in http://sage.skku.edu/

Solution.

var('x');
diff(x^(15/14)+5*e^x,x)

2-5. [You may have to use a Chain Rule see Section 3.3] Find the derivative where is

2.

Solution.

3.

Solution.

4.

Solution.

5.

Solution.

6. Find the equation of the tangent line to the curve at .

Solution. .

So the slope of the tangent line is 20.

(Since passes through

[Practice Math & Code in http://sage.skku.edu/

f(x)=x^2*sqrt(x)

df(x)=diff(f(x),x)

y(x)=df(4)*(x-4)+32

y(x)

p1=plot(f(x),x,0,10, color='blue')

p2=plot(y(x),x,0,10, color='red')

show(p1+p2,ymax=50,ymin=-10)

7. The normal line to a curve at a point is the line that passes through

and is perpendicular to the tangent line to at . Find an equation of

the normal line to the curve at the point .

Solution.

So the slope of the tangent line is . (Since )

Then the slope of the normal line is .

Thus, .

Therefore, the normal line is .

[Practice Math & Code in http://sage.skku.edu/

f(x)=1+e^x

df(x)=diff(f(x),x)

y(x)=-1/df(0)*x+2

y(x)

p1=plot(f(x),x,-5,5, color='blue');

p2=plot(y(x),x,-5,5, color='red');

show(p1+p2,ymax=5,ymin=-5)

8.Where is the function differentiable? Give a formula for

.

Solution. ,

then .So is continuous on .

,

then is not differentiable at .

Therefore, is differentiable on .

9. Let . Find the values of and that make

differentiable everywhere.

Solution. Note that is differentiable everywhere except . We must check at . For to be differentiable at 2,

, so . And also have to be continuous at .

,

since , . Therefore, .

CAS 10. Evaluate .

http://matrix.skku.ac.kr/cal-lab/cal-3-1-11.html

Solution. Method (1) Let , and  .

Then by the definition of derivative,  .

Method (2) Note that   .

So

lim((x^2020-1)/(x-1),x=1)

11-12. Differentiate the following functions.

11.

Solution.

12.

Solution.

13. Show that if is differentiable and satisfies the identity for all

and and , then satisfies for all .

Solution.

Thus, either or .

Since for all and . Now,

Since , then

.

Therefore, .

14-16. Find derivatives of the following functions.

14.

Solution.

f(x)= (csc(x))^2  * cot(x)

df(x)=diff(f(x),x)

print df(x)

15.

Solution.

16.

Solution.

17. Show that the curve has no tangent line with slope .

Solution.  .

Since is always positive, there is no such that .

So has no tangent line with slope 0.

18. A stone is thrown into a pond, creating a wave whose radius increases at the rate of meter per second. In square meters per second, how fast is the area of the circular ripple increasing seconds after the stone hits the water?

Solution. Let radius time. . when .  Area

, . Therefore, .

19.A particle moves along a straight line with equation of motion .

(a) When is the particle moving forward?

(b) When is (the acceleration) zero?

(c)When is the particle speeding up? Slowing down?

Solution. (a) , , when .

When , the particle moves forward.

(b) .

Therefore the acceleration is 0, when .

(c)When the particle is speeding up, the acceleration is positive and when it is

slowing down, the acceleration is negative. when and when . Therefore the particle speeds up when

, and the particle slows down when .

20. A particle moves in a straight line with equation of motion , where is measured in seconds and in meters.

(a)What is the position of the particle at and ?

(b) Find the velocity of the particle at time .

(c) When is the particle moving forward?

(d) Find the total distance traveled by particle on the time interval .

(e) Find the acceleration of the particle at time .

Solution. (a)

(b)

(c) When , .

Therefore, .

(d)

(e)

21. The population of a bacteria colony after hours is . Find the growth rate when .

Solution.  . . .

22.A cost function is given by .

(a) Find the marginal cost function.

(b) Find .

Solution. (a)

(b)

23.If a stone is thrown vertically upward with a velocity , then its height after seconds

is

(a) What is the maximum height reached by the stone?

(b) What is the velocity of the stone when it is above the ground on its way up?

On its way down?

Solution.  (a)

and

(b)

The time is when the height of the stone is 5.

The velocity of the stone is on its way up. The velocity of the stone is on

its way down.

24. is the total value of the production when there are workers in a plant, then the

average productivity is  .

Find . Explain why the company wants to hire more workers if ?

Solution.    Since , implies .

This means .  This means the rate of productivity is larger than

the average productivity .

Therefore, if the company hires more workers, then they can expect to have a better

productivity.

25. Let be the population of a bacteria colony at time hours. Find the

growth rate of the bacteria after 10 hours.

Solution.

26. The angular displacement of a simple pendulum is given by with an

angular amplitude , an angular frequency and a phase constant . Find .

Solution.

27. Show that .

Solution.  We note . Let . Then , and as , .

Therefore,   .

3.3 The Chain Rule and Inverse Functions

In this section, we consider the derivative of the composition of two functions, derivatives of implicitly defined functions, higher order derivatives, the inverse trigonometric functions, logarithmic functions and the hyperbolic functions.

THEOREM 1   The Chain Rule

If and are both differentiable and is the composite function defined by , then is differentiable and

.

Differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function.

If and are both differentiable functions, then the rate of change of with respect to is

(the rate of change of with respect to ) (rate of change of with respect to ).

EXAMPLE 1

Find if .

Solution.

.

Suppose the outer function is a power function. If , then we can write

, where . By using the Chain Rule and the Power Rule, we get

.

THEOREM 2 The Power Rule Combined with the Chain Rule

Let be any real number and be differentiable. If , then .

EXAMPLE 2

Find if .

Solution. .

EXAMPLE 3

Find the derivative of the function .

Solution.  Combining the Power Rule, Chain Rule, and Quotient Rule, we get

EXAMPLE 4

Find the derivative of the function    using .

Solution.

Suppose that , , and , where , , and are differentiable functions.

Then, to compute the derivative of with respect to , we use the Chain Rule twice:

EXAMPLE 5

Find if

.

Solution. Apply the chain rule twice,

.

EXAMPLE 6

Differentiate .

Solution.  .

Tangents to Parametric Curves

Now, consider the particle which is moving along the curve in the plane. Then, the trajectory of the particle can be described by the parametric equations

.

The chain rule allows us to find , in particular, to find equations of the tangent lines to the

parametric curve.

Since , we have when .

When and , we have a vertical tangent line. When and ,

we have a horizontal tangent line.

EXAMPLE 7

Figure 1

Consider the cycloid    , .

Find the equation of the tangent line which is tangent at .

Solution.  At , the point is .

From the above formula, we have the slope of the tangent line

The slope at is . The equation of the tangent is

or      .

CAS EXAMPLE 8

Find an equation of the tangent line to the curve at the given point.

Solution. var('t, x, y, u')

x=sin(t)

y=cos(2*t)

dxdt=diff(x,t)

dydt=diff(y,t)

dydx=dydt/dxdt

yu=y(pi/4) + dydx(pi/4)*(u-x(pi/4))

p1=parametric_plot((x,y),(t,0,pi/2), color='blue');

p2=plot(yu, u, 0, 1.5, color='red');

show(p1+p2)

[Practice Math & Code in http://sage.skku.edu/

Figure 2

Higher Order Derivatives

The second order derivative of , denoted by , or is the derivative of the

erivative of . Thus, .

If is the position function of an object that moves in a straight line, then its first

derivative represents the velocity of the object as a function of time:

.

The instantaneous rate of change of velocity with respect to time is called the acceleration

of the object. Thus, the acceleration function is the derivative of the velocity function and is

therefore the second derivative of the position function:

or

.

In general, a second derivative is the rate of change of a rate of change.

The third order derivative is the derivative of the second order derivative: So can be interpreted as the slope of the curve or as the rate of change of .

Similarly the fourth order derivative is usually denoted by . In general, the th order derivative of is denoted by and is obtained from by differentiating times. If , we write

.

EXAMPLE 9

If , find .

Solution.  We have , ,

,   repeatedly.

Thus using induction on , we obtain a formula,

.

EXAMPLE 10

The position of a particle is given by the equation , where is measured in seconds and in meters.

(a) Find the velocity and acceleration at time .

(b) Draw the graph of .

Solution.  (a) The velocity function is the derivative of the position function,

.

The acceleration is the derivative of the velocity function.

.

(b)

Figure 3 Position

Leibnizs Rule

Figure 4 Acceleration

When Gottfried Wilhelm von Leibniz discovered his version of the calculus he derived the rules governing the process of differentiation (and introduced the notation that we use today). Among those rules we find a theorem concerning multiple differentiations of a product of two functions.

If and are (smooth, differentiable) functions of on , we have

where , , .

EXAMPLE 11

Find the th derivative of

Solution.

CAS EXAMPLE 12

Find  of the function

Solution.

Similarly,     .

Implicit Differentiation

So far we are dealing with explicit functions, that is, with functions of the form . We have noted that this function has the property that each value of is associated with one and only one value . But in practice, sometimes the relation between two variables and is expressed in a more general form, we can regard as an implicit function of , that is, regard as independent variable and as the dependent variable. Of course, we can also regard as the independent variable and as a function of . Whether one determines or as independent variable depends either on the problem or on the concrete application.

Note that, while explicit functions are a particular case of implicit functions, since equation can be written as (So it has form , with ), not every implicit function can be written as explicit function, that is, it is not always possible to solve the equation in order to write . Also, note that, for implicit functions , one value of can be associated with more than one value of .

EXAMPLE 13

The equation shows is as an implicit function of , but it is not an explicit function, since it is impossible to find an expression for in terms of (that is, it is impossible to solve this equation for ).

EXAMPLE 14

Figure 5 The folium of Descartes

The equation is another example of an implicit function. This curve is called the folium of Descartes shown in Figure 5.

In order to find the derivative of it is not necessary to solve an equation for in terms of . Instead we may use the method of implicit differentiation which consists of differentiating both sides of the equation with respect to and then solving the resulting equation for . Assume that the given equation determines implicitly as a differentiable function of so that the method of implicit differentiation may be applied.

EXAMPLE 15

If , find .

Solution.  Differentiate both sides with respect to :

Solving for , we obtain .

EXAMPLE 16

Given , find the tangent line at the point . Also find at .

Solution.  Differentiate both sides with respect to :

Solving for , we obtain .

At , the slope of the tangent line is .

Thus the equation of the tangent line is .

Thus, .

Orthogonal Trajectories

Two curves are called orthogonal if at each point of intersection their tangent lines are perpendicular. Two families of curves are orthogonal trajectories of each other if every curve in one family is orthogonal to every curve in the other family. For example in physics, the lines of force in an electrostatic field are orthogonal to the lines of constant potential. In thermodynamics, the isotherms (curves of equal temperature) are orthogonal to the flow lines of heat. In aerodynamics, the streamlines (curves of direction of airflow) are orthogonal trajectories of the velocity-equipotential curves.

EXAMPLE 17

Show that the family of curves ( is an arbitrary constant) and ( is an arbitrary constant) are orthogonal trajectories.

Solution.   Differentiating gives

.

Differentiating gives .

So at any of the intersection points, the slopes of the tangents are negative reciprocals of each other. Therefore, the two families are orthogonal families.

p=circle((0,0), 0.1) for i in srange(0.2, 2, 0.2):   p=p+circle((0, 0), I)

q=plot(x, 0, 2, color='red')  r=plot(-x, -2, 0) for j in srange(-4, 4, 0.4):

q=q+plot(j*x,0,2,color='red')r=r+plot(-j*x,-2,0,color='red')(p+q+r).show(ymin=-2, ymax=2)

Figure 6

[Practice Math & Code in http://sage.skku.edu/

var('x,y,k')  p=plot(x^2, -2, 2)  for i in srange(-4, 4, 0.4):

p=p+plot(i*x^2, -2, 2) q=implicit_plot(x^2+2*y^2==1, (x,-2,2), (y,-2,2),

color='red') for j in srange(-4, 4, 0.4): q=q+implicit_plot(x^2+2*y^2==j,(x, -2, 2), (y, -2, 2), color='red')  p+q

Derivatives of Inverse Functions

DEFINITION 3

A function with domain is said to be one-to-one, if then .

DEFINITION 4

Suppose is one-to-one function on . Then its inverse, , is a function which is defined on the range of by

,

that is, if , then .

Using the Chain Rule, we can find the derivative of the inverse function. Namely, from

it follows

so that

.

Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions are found using implicit differentiation. If is any one-to-one differentiable function, it can be proved that its inverse function is also differentiable, except where its tangents are vertical.

The definition of the inverse sine(arcsine) function:

For , if and .

(Implicit Differentiation) Differentiating implicitly with respect to , we obtain

.

Now , , so

.   Therefore .

,

The formula for the derivative of the inverse tangent function(arctangent) is derived in a similar way. If , then tan . Differentiating the latter equation implicitly with respect to , we have

.

Similarly, the derivative of any inverse function may be found using the above procedure.

EXAMPLE 18

Differentiate.

Solution.

Derivatives of Inverse Trigonometric Functions

Derivatives of Logarithmic Functions

The derivatives of the logarithmic functions and the natural logarithmic function are obtained using implicit differentiation.

1.

proof. Let . Then, .

Differentiating this equation implicitly with respect to , we get

and so .

If we put in Formula 1. then the factor on the right side becomes and we get the formula for the derivative of the natural logarithmic function :

2.

In general, we have

3.      or

EXAMPLE 19

Find if .

Solution.  Since

It follows that

Thus, for all .  Hence,

4.

EXAMPLE 20

Differentiate .

Solution.      .

EXAMPLE 21

Differentiate .

Solution.  Applying the logarithm to both sides, we get

Differentiating with respect to , we obtain

.

EXAMPLE 22

Let be implicitly defined by Compute in terms of and .

Solution.   Taking logarithms gives .

Differentiating implicitly, we have

,

and solving for yields

Apply the definition to the derivative to the logarithmic function, , we have

.

So, . Thus,   .

5.

Formula 5 can be rewritten in the following way, by substituting with ,

6.    .

Hyperbolic Functions

Figure 7  Catenary curve

Hyperbolic functions which arise frequently in mathematics and its applications, are certain combinations of the exponential functions and . In many ways they are analogous to the trigonometric functions, and they have the same relationship to the hyperbola that the trigonometric functions have to the circle. For this reason they are called hyperbolic functions.

Definition of the Hyperbolic Functions

The graphs of the hyperbolic sine and cosine can be sketched using graphical addition as in Figures 8-9. The remaining hyperbolic functions are shown in Figures 10-13.

Figure 8     Figure 9    Figure 10

Figure 11                  Figure 12                Figure 13

Hyperbolic Identities

The hyperbolic functions satisfy various identities similar to the identities for the trigonometric functions. The most fundamental of these is

7.      If we divide 7 by , we obtain

8.      and if we divide 7. by , we obtain

9.    The addition formulas for and are

10a.

10b.    For example, to prove 10a. we use the relations

11a.

11b.    From  11a. and 11b. it follows that

which proves 10a. . The proof of 10b is similar.

The derivative of hyperbolic sine function,

follows immediately from the definition of the hyperbolic functions. By similar techniques we can deduce the derivatives of the other five hyperbolic function.

Derivatives of the Hyperbolic Functions

If we use 10a and 10b , we obtain the following analogs of the trigonometric double-angle formulas

12a

12b

To obtain subtraction formulas for the hyperbolic function we shall use the addition formulas and the relations

13a

13b

13c

If we replace by in 10a and 10b and then apply 13a and 13b, we obtain

EXAMPLE 23

Differentiate .

Solution.  .

EXAMPLE 24

Differentiate .

Solution.  .

EXAMPLE 25

Evaluate the derivative of at .

Solution.  Using the Quotient Rule

.

Because and , we obtain

.

CAS EXAMPLE 26

Find .

Solution.  var('x')     diff((1+ cos(x))^(1/x),x)

Inverse Hyperbolic Functions

Because the hyperbolic functions are expressed in terms of combinations of , it should not be surprising that the inverse hyperbolic function are expressed in terms of the natural logarithms. For example,is equivalent to , so that

Hence, .

Figure 14           Figure 15                Figure 16

Figure 17       Figure 18                Figure 19

The Inverse Hyperbolic Functions

The proofs of some important identities, relative to derivatives and the inverse

hyperbolic functions will be presented in the exercises.

Derivatives of the Inverse Hyperbolic Functions

The derivatives of inverse hyperbolic functions may be obtained by differentiating the logarithmic expression for the inverse hyperbolic functions. For example,

.

Derivatives of the Inverse Hyperbolic Functions

EXAMPLE 27

Find derivative of the function .

Solution.

EXAMPLE 28

Find derivative of the function .

Solution.  .

EXAMPLE 29

Find derivative of the function .

http://matrix.skku.ac.kr/cal-lab/cal-3-3-Exm-27.html

Solution.   By the Product Rule, we obtain

.

3.3 EXERCISES (The Chain Rule and Inverse Functions)

1-3. Find the derivative .

1.

Solution.

2.

Solution.

3.

Solution.

4-5. Find of these functions.

4.

Solution.        , so

Therefore, .

5.

Solution.

6. Find derivative of the function .

Solution.

7. Find where .

Solution.

Using induction on , we get   .

8. Find for an integer if .

Solution.

Therefore, using induction   .

9.Show that the curves and are orthogonal.

Solution.  (ⅰ)    (ⅱ)

At a point of intersection,

.

Thus, the intersections of the two curves are and .

At ,

Therefore, the curves are orthogonal.

At the derivative for the equation does not exist (or, in other words, it is equal to infinity). Hence the tangent line to the curve at is vertical. At the same point, the derivative for the equation is equal 0, hence the tangent line for the curve at is horizontal.

Thus, the two curves are orthogonal at .

10-13. Find

10. ; Note

Solution.  ,   ,

, ...,

11.

Solution.

12.

Solution.

13.

Solution.

We observe that the denominator is , and the numerator is .

Therefore .

14. Find the th derivative of

Solution.

15. Differentiate .

Solution.  By differentiating implicitly, we have

Thus, .

16. Use differentiation to show that

Solution.

for some constant

Therefore

17-20.  In problems below, find .

17.

Solution.

18.

Solution.

CAS 19.

http://matrix.skku.ac.kr/cal-lab/cal-3-2-12.html

Solution. var('x')  diff((sin(x))^(sqrt(x)))

[Practice Math & Code in http://sage.skku.edu/

CAS 20.

http://matrix.skku.ac.kr/cal-lab/cal-3-2-13.html

Solution.  var('x')  diff((arccos(x))^arctan(x))

[Practice Math & Code in http://sage.skku.edu/

*arccos(x)^arctan(x)

CAS 21. Find if .

http://matrix.skku.ac.kr/cal-lab/cal-3-2-14.html

Solution.

[Practice Math & Code in http://sage.skku.edu/

var('x')     df(x)=diff(x^(arctan(x)),x);   df(x)

Answer : (log(x)/(x^2 + 1) + arctan(x)/x)*x^arctan(x)

ddf(x)=diff(df(x),x)        ddf(x)

Answer : (log(x)/(x^2 + 1) + arctan(x)/x)^2*x^arctan(x) - (2*x*log(x)/(x^2+1)^2 +

arctan(x)/x^2- 2/((x^2 + 1)*x))*x^arctan(x)

22-23. Find and of the following parametric equations ( and are constants).

22. .

Solution.

.

23. .

Solution.  ,

,

24. Given , show that it satisfies the following identity.

Using this identity, find .

Solution.

Therefore,  ,

Therefore, for is even,  for is odd.

25. Given , show that .

Solution.  Since so is continuous on .

And ,   .

Therefore, .

26-28. Find for the following implicit functions.

26.

Solution.

.

27.

Solution.  . . .

28.

Solution.

29. If , where and are three times differentiable, find expressions for and .

Solution.

30.Given , find at the point .

Solution.

Therefore,.

31. Determine the points on the curve at which the tangent is either vertical or horizontal.

32. Find an equation of the tangent line to the curve  at for an arbitrary value .

Solution.  , so we have to find the

tangent line at

Because .

33. Establish the following derivative rules.

(a)  b)

Solution.  (a)

(b)

34-46. Prove the following identities.

34.

Solution.

35.

Solution.

or

36.

Solution.

or

37.

Solution.

38.

Solution.

39.

Solution.  Use Mathematical Induction. If , trivial.

Let

So,

40.

Solution.

41.

Solution.  Let . Then , so .

Note that , but   because .

Thus, .  So, .

42.

Solution.  Let . Then , so .

Note that , but   because .

Thus, for So .

43.

Solution.  Let . Then

44. Find , where .

Solution.

45. Find , where .

Solution.

46. Find , where .

Solution.

47. Find , where .

Solution.

3.4 Approximation and Related Rates

http://youtu.be/ViRwEJ0Wfkw

In this section, we deal with approximating a given function by means of a class of functions which are simple and easy to deal with. One such class of functions is polynomials.

Linear Approximation

Given a function , sometimes it is desirable to choose another function from a class of functions which are convenient to work with as an approximation of . For the approximation to be a good approximation of the function , the difference which is called the error of the approximation must be small in some definite sense.

In this section, we consider linear approximations, that is, the class of functions from which we choose an approximation consists of linear functions. If we require to be a linear function, such that its value of the its first derivative at a given point coincides with the value of and at , respectively, that is, , , then we have

.

This function is the linear approximation of the function at the point , and is denoted by (often we simply write instead of , if it is clear at which point we consider the linear approximation). Thus,

.

Figure 1

Note that from the Mean Value Theorem, we have

,

where is some number between and .

Under the assumption that is continuous (on an interval contain and ), then is small for all values which are in some small interval containing , hence the error of the linear approximation is small for all values of which are close to . This fact is illustrated in Figure 1. For values which are far from , since can be large, the error of the linear approximation can be large. Thus, linear approximation does not give a good

approximation for which are far from .

EXAMPLE 1

Determine the linear approximation for at .

Solution.  All we need to do is find the tangent line to at .

The linear approximation is,

So, as long as remains small we can say that .

Differentials

In this section, we introduce some notation that we will use frequently the next chapter.

Given a function , if differential (or ) is defined by

Let denote an actual change in the variable , and denote the corresponding actual change in the variable , that is,

The idea of using differential is to give an approximation of the value , so that we can use it to approximate (since ).

If the function is continuous, as in our case, then a small change in produces a small change in .

Geometrically, the differential may be explained as follows:

Let and be two points on the curve . Let . Since the derivative at is the same as the slope of the tangent line , we have

.

Figure 2

In other words, is the differential of the coordinates to the tangent line while is the increment of the coordinates to the curve.

Therefore, the differential is an approximation for the increment .

More  formally

.

In other words, can be used for the approximation of the increment near . We now write . From the above formula, we can find the approximating value of if we know and .

EXAMPLE 2

Use differentials to approximate .

Solution.  Let . Set and .

Since , , we have

Hence approximately,

.

EXAMPLE 3

Let . Use differentials to approximate the value of at .

Solution.

Thus, we set and Since we have

, the differential is

As an approximation at , . When (and ),

(In this case, we can compute 24.9491078380810 and the true error, which is . Hence, the approximating value is accurate within an error of 0.0002, .)

Sometimes, a better description of the error can be given by the relative error, which is computed as the error divided by the total quantity , i.e., the relative error is . Also, sometimes percentage error is more suitable. The percentage error is the (relative error) (%).

EXAMPLE 4

The radius of the sphere was measured and found to be 5cm with a possible error in measurement of at most 0.02cm. What is the maximum error and the relative error in using this value of radius in computing the volume of the sphere?

Solution.  If the radius of the sphere is , then its volume is . If the error in is denoted by , then the corresponding error in will be denoted by . Since , ,

we have, Hence, the maximum error is and the relative error is .

Figure 3

In the above, we considered the linear (tangent line) approximation for a curve. The graph of the cosine function is a very nice looking curve. But it is just that, a curve. The best linear approximation to the cosine function near 0 is quite unexciting; you can check that for , the best linear approximation near 0 is given by . One reason that linear approximations are popular is because linear functions are easy to work with. But as we see from the cosine function, linear approximations have their limitations. After linear functions, the next simplest functions are the quadratics. For a better approximation, we approximate a curve by using a parabola instead of a straight line. We will look for a second-degree polynomial in the following form:

,

this is a better approximation of near than the linear approximation. Similarly, as with the linear approximation, we require , and . Then, and .

Hence if is at least twice differentiable, then

is the “quadratic approximation” for near .

We note that linear approximations at given point , while quadratic approximation at , at first and second derivative at .

EXAMPLE 5

Figure 4

Consider the cosine function and let . It is not hard to verify that if we let , then and have the same first two derivatives at 0. In addition, the graph of is very close to the graph of (very nicely) near 0.

EXAMPLE 6

Find the quadratic approximation for near . Use it to approximate .

Solution.  and .

The quadratic approximation near is  .

A more accurate approximation is so we have found agreement in 6 decimal places.

Related Rates

In the real world, many quantities are related with each other. In turn, their derivatives (rates of change) will be related with each other. This is also a useful approach to the solution of real-world problems. Let us consider some simple examples.

EXAMPLE 7

Figure 5

A balloon is released at a point 100m away from an observer, who is on level ground. (See Figure 5.) If the balloon goes straight up at a rate of 5 meters per second, how fast is the distance from the observer to the balloon increasing when the balloon is 50 meters high?

Solution.  We have a relationship between and , i.e., .

Implicit differentiation gives us that Note that and . Since , we have At , the distance between the balloon and the observer is increasing at a rate of .

EXAMPLE 8

A rotating camera is focusing on a car driving on the straight road as depicted below. The camera lies 100 meters away from the road. Right now, the car is 200 meters away from the camera, and the angle is increasing at a rate of 1/2 radian per second. How fast is the car moving?

Solution.  We have Differentiating, with respect to we have

Figure 6

If then

Thus, we have . So, the car is moving at the speed of 200.

CAS EXAMPLE 9

Use differential to approximate

Solution.  is and

var('x,dx');

f(x)=sin(x)

f(pi/6 + pi/60).n()  # Numerical approximation

dy(x)=diff(f(x),x)*dx

f(pi/6)+dy(x=pi/6,dx=pi/60).n()

# Approximation by derivative

3.4 EXERCISES (Approximations and Related Rates)

1-3.Use differential to approximate the following quantities.

1.

Solution.  var('a,b,x,dx');

f(x)=(x+25)^(1/2);

dy(x)=diff(f(x),x)*dx;

dy(x)

[Practice Math & Code in http://sage.skku.edu/

(f(0)+dy(x=0, dx=2)).n()

2. (and )

Solution.  In order to find an approximate of , we define a function .

Then

Now we use differentials to approximate. An approximate of can be found by using

.  Set and at

.  Therefore .

var('a,b,x,dx')

f(x)=(x+60)^(1/3);

f(x);

f(1.05)

[Practice Math & Code in http://sage.skku.edu/

dy(x)=diff(f(x),x)*dx

(f(1)+dy(x=1, dx=0)).n()

3.

Solution.  In order to find an approximation of , we define a function .

Then   .

var('a,b,x,dx');

f(x)=(x+32)^(1/5);

f(x);

Now we use differentials to approximate. Set and at . Find an approximation of by using

[Practice Math & Code in http://sage.skku.edu/

dy(x)=diff(f(x),x)*dx;

dy(x)

f(1)+dy(x=1,dx=0.05)

There are Sage built in command  .n()  for finding such approximations.

(32.05)^(1/5).n()

4.  The height of a circular cone is the same as the radius of its circular bottom. The height and radius were measured and found to be 4cm with a possible error in measurement of at most 0.04cm. What is the relative error in using this value to compute the volume?

Solution.    (Because )

5. Find an approximation of the difference between the surface areas of two spheres whose radii are 5cm and 5.05cm, respectively.

Solution. . Thus.

Since we have

6. The period of a pendulum is given by the formula , where is the length of the pendulum measured in meters and is the gravitational constant. If the length of the pendulum is measured to be 3m with a possible error in mea-surement 1cm. What is the approximate percentage error in calculating the period ?

Solution.  ,

The approximate percentage error(relative error)100% .

7. A ladder 10 meters long is leaning against a wall. If the foot of the ladder is being pulled away from the wall at 4m/s, how fast is the top of the ladder sliding down the wall when the foot of the ladder is 8 meter away from the wall?

Solution.  Let be the distance between the foot of the ladder to the wall and be the distance between the top of the ladder to the bottom.

Then from the Pythagorean Theorem, , ,.

Since , , so .

Therefore, the speed of the ladder sliding down from the top is .

8. A ladder 10 meters long is leaning against a wall. If the top of the ladder is sliding down the wall at 3m/s, how fast is the foot of the ladder being pulled away from the wall when the foot of the ladder is 6 meter away from the wall?

Solution.

Since , .

9.  A ladder 10m long is leaning against a wall. If the top of the ladder is sliding down the wall at 1m/s, how fast is the angle between the top of the ladder and the wall changing when the foot of the ladder is 5m away from the wall?

Solution.  when .

, .

Therefore, when the foot of ladder is 5m away, angle at the top of the ladder changes is .

10. Two cars start moving from the same point. One travels south at km/hr and the other travels west at km/hr. How fast is the distance changing between the two cars?

Solution.  Distance of travel south is , distance of travel west is , and the distance between the two travelers is at time

11. Water is being pumped at a rate of 20 liters per minute into a tank shaped like a frustrum of a right circular cone. The tank has an altitude of 8 meters and lower and upper radii of 2 and 4 meters, respectively. How fast is the water level rising when the depth of the water is 3 meters?

Solution.  .

Let

If

Therefore the water level rises when the depth of the water is 3 meters.

12. Water is being pumped at a rate of 20 liters per minute into a tank shaped like a hemisphere. The tank has a radius of 8 meters. How fast is the water level rising when the depth of the water is 3 meters?

Solution.  .

13. A snowball melts at a rate proportional to its surface area. Does the radius shrink at a constant rate? If it melts to 1/4 its original volume in one hour, how long does it take to melt completely?

Solution.  The rate of melting is described by . Since the snowball melts at a rate proportional to its surface area. Notice that

so is constant. Let the volume of first time

be , after one hour, .

.

The snowball melts completely at time ().

Calculus