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Chapter 2 DIFFERENTIATION
Contents
[1.] Tangent Lines and Their Slopes
[2.] Derivatives
[3.] Differentiation Rules[4.] Rates of Change in Natural and Social Sciences
[5.] Derivatives of Trigonometric Functions
[6.] Chain Rule and Implicit Differentiation
[7.] Derivatives of Inverse Functions
[8.] Linear Approximation and Applications
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2.1 TANGENT LINES AND THEIR SLOPES
Differential calculus is concerned with how one quantity changes inrelation to another quantity.
Its central concept of differential calculus is the derivative.
Questions: How can we define the tangent line to a
graph and how can we compute its slope?
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2.1 TANGENT LINES AND THEIR SLOPES
Let C be the graph of y = f(x) and let P be the point (a, f(a)) on
C. Consider a nearby point Q(x, f(x)), where x = a, and the linethrough P and Q, which is called a secant line to the curve. Thisline rotates around P as Q moves along the curve.
Slope of secant line through P and Q = fx
= f(x) f(a)x a
where f = f(x) f(a) and x = x a.
The expressionf(x) f(a)
x ais called the difference quotient.
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2.1 TANGENT LINES AND THEIR SLOPES
Let Q approach P along the curve C by letting x approach a.Then the tangent line is the limiting position of the secant line PQas Q approaches P.
Definition 1.1 The tangent line to the curve y = f(x)at the point P = (a, f(a)) is the line through P with slope
m = limxa
f(x) f(a)x a
provided that this limit exists.
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2.1 TANGENT LINES AND THEIR SLOPES
Let h = x a. Then x = a + h and the slope of the secant line PQis
mPQ =f(a + h)
f(a)
h .
As x a, h 0 and so the slope of the tangent line is
m = limh0
f(a + h) f(a)h
Example 1.1 Consider the graph of f(x) = 3
x = x1/3. Let ustry to calculate the limit of the difference quotient for f at x = 0:
limh0
f(0 + h) f(0)h
= limh0
h1/3
h= lim
h01
h2/3=
Hence although the limit does not exist, the slope of the secantline joining the origin to another point Q on the curve approaches
infinity as Q approaches the origin from either side.Dr. Nguyen Ngoc Hai CALCULUS I
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2.1 TANGENT LINES AND THEIR SLOPES
Definition 1.2 If f(x) is continuous at the pointP = (a, f(a)) and if either
limh0
f(a + h) f(a)h
= or limh0
f(a + h) f(a)h
= ,
then the vertical line x = a is tangent to the graphy = f(x) at P.
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2.1 TANGENT LINES AND THEIR SLOPES
Definition 1.3 The slope of a curve C at a point P isthe slope of the tangent line to C at P if such a tangent lineexists. In particular, the slope of the graph of y = f(x) at the
point (a, f(a)) is
limh0
f(a + h) f(a)h
.
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2.1 TANGENT LINES AND THEIR SLOPES
Definition 1.4 If a curve C has a tangent line L at pointP, then the straight line N through P perpendicular to L iscalled the normal to C at P.
If L is horizontal, then N is vertical; if L is vertical, then N is
horizontal. If L is neither horizontal nor vertical, then the slope ofN is the negative reciprocal of the slope of L:
slope of the normal =1
slope of the tangent
Example 1.2 Find equations of the straight lines that aretangent and normal to the curve y =
x at the point (4, 2).
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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE
Definition 2.1 The derivative of a function f at a
number a, denoted by f(a), is the limit of the differencequotients
f(a) = limh0
f(a + h) f(a)h
if this limit exists. When the limit exists, we say that f isdifferentiable at a. The process of computing the derivativeis called differentiation.
If we writex
=a
+h
, thenh
=x
a
andh
approaches 0 if andonly if x approaches a. Therefore, an equivalent definition of thederivative is
f(a) = limxa
f(x) f(a)x a
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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE
Finding f(a) from the definition of derivative
The algebraic steps to calculate f(a):
Step 1. Find f(a) and f(a + h).
Step 2. Find and simplify f(a + h) f(a) .Step 3. Divide by h to get f(a+h)f(a)
h.
Step 4. Take the limit as h 0.
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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE
Geometrical Meaning of the Derivative
The tangent line to the curve y = f(x) at the point P(a, f(a)) hasslope
m = limh0
f(a + h) f(a)h
= f(a)
So
The tangent line to y = f(x) at (a, f(a)) is the line through(a, f(a)) whose slope is equal to f(a), the derivative of f at a.
Therefore, an equation of the tangent line to the curve y = f(x)at the point (a, f(a)) is
y = f(a) + f(a)(x a)
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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE
Suppose y is a quantity that depends on another quantity x. Thus,y is a function of x: y = f(x).
If x changes from x1 to x2, then the change in x (also called theincrement of x) is x = x2 x1.
The corresponding change in y is y = f(x2) f(x1).
The difference quotient
y
x =
f(x2)
f(x1)
x2 x1is the average rate of change of y with respect to x over theinterval [x1, x2].
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2.2 DERIVATIVES2.2.1 DERIVATIVES. RATES OF CHANGE
The limit of average rates of change as x 0 is called the(instantaneous) rate of change of y with respect to x at x = x1:
limx0y
x = limx2x1f(x2)
f(x1)
x2 x1
Thus,
The derivative f(a) is the instantaneous rate of change of
y = f(x) with respect to x when x = a.
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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION
Definition 2.2 The derivative of a function f is thefunction f whose value at x is
f(x) = limh
0
f(x + h) f(x)h
if this limit exists.
The domain of f is the set
{x| f(x) exists}and may be smaller than the domain of f.
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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION
We extend the definition to allow for a right derivative at x = a
and a left derivative at x = b:
f+(a) = limh0+
f(a + h) f(a)h
, f(b) = limh0
f(b+ h) f(b)h
.
We say that f is differentiable on [a, b] if f(x) exists for all x in(a, b) and f+(a) and f
(b) both exist.
In general,
f is differentiable on an interval I if it is differentiable atevery number in the interval.
If f has a derivative at every point of its domain, we call fdifferentiable.
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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION
Right and left derivatives may be defined at any point of a
functions domain.
Remark:
f is differentiable at x if and only if the one-sided derivatives
f+
(x) and f
(x) exist and are equal.
f(x) f+(x), f(x) and f+(x) = f(x)
Example 2.1 Find the derivative of f(x) = |x|.
Example 2.2 Find the derivative of f(x) =
x. State thedomain of f.
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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION
Solution
f(x) = limh0 x + h xh
= limh0
(
x + h x)(x + h + x)h(
x + h +
x)
= limh0(x + h)
x
h(x + h + x) = limh0h
h(x + h + x)= lim
h01
x + h +
x=
1x +
x
=1
2
x.
At x = 0,
limh0+
0 + h 0
h= lim
h0+
h
h= lim
h0+1
h= .
Thus f is not differentiable at x = 0 and the domain of f is
(0, ). This is smaller than the domain of f, which is [0, ).Dr. Nguyen Ngoc Hai CALCULUS I
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2.2 DERIVATIVES2.2.2 DERIVATIVE AS A FUNCTION
Other Notations
The most common notations for the derivative of a functiony = f(x), besides f(x), are
y =dy
dx =df
dx = Dx(f) =d
dx(f)
If we want to indicate the value of a derivative at a specificnumber a, we use the notations
Dxyx=a
= dydx
x=a
= dfdx
x=a
= ddx
f(x)x=a
which are synonyms for f(a).
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2.2 DERIVATIVES2.2.3 CONTINUITY AND DIFFERENTIABILITY
Theorem 2.1 If f is differentiable at a, then f iscontinuous at a.
The converse of this theorem is false. That is, there are
functions that are continuous but not differentiable.
It follows from Theorem 2.1 thatA function can never have a derivative at a point ofdiscontinuity.
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2.2 DERIVATIVES2.2.3 CONTINUITY AND DIFFERENTIABILITY
When does a Function Fail to be Differentiable?
At any discontinuity a function fails to be differentiable.
If the graph of a function f has a corner in it, then the graphof f has no tangent at this point and f is not differentiablethere.
If f is continuous at a but the curve has a vertical tangent line
x = a then f is not differentiable at this point becauselimxa |f(x)| = .
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2 3 DIFFERENTIATION RULES
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2.3 DIFFERENTIATION RULES
Constant Rule
The derivative of a constant function is 0:
d
dx(c) = 0
Constant Multiple Rule
If u is a differentiable function of x, and c is a constant, then
d
dx(cu) = c
du
dx
The derivative of a constant times a function is the constant
times the derivative of the function.
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2.3 DIFFERENTIATION RULES
Sum and Difference Rule
If u and v are differentiable functions at x, then
(u+ v)(x) = u(x) + v(x)
(u
v)(x) = u(x)
v(x)
The derivative of a sum is the sum of the derivatives. The
derivative of a difference is the difference of the derivatives.
Example 3.1 Does the curve y = x4 2x2 + 2 have anyhorizontal tangents? If so, where?
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2.3 DIFFERENTIATION RULES
Product RuleIf u and v are differentiable functions at x, then
(uv)(x) = u(x)v(x) + v(x)u(x)
The derivative of a product of two functions is the first
function times the derivative of the second, plus the second
function times the derivative of the first.
Example 3.2 Find the derivative of (x2 + 1)(x3 5).
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3 O U S
Power Rule
If n is a positive integer, then
d
dx(xn) = nxn1
d
dx
xexponent
= exponent xexponent1
Example 3.3
d
dxx3 = 3x2,
d
drr2 = 2r,
d
dtt48 = 48t47.
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Quotient Rule
If u and v are differentiable functions at x, and if v(x)
= 0,
then uv
(x) =
v(x)u(x) u(x)v(x)[v(x)]2
The derivative of a quotient is the denominator times the
derivative of the numerator, minus the numerator times thederivative of the denominator, all divided by the square of the
denominator.
In particular,1
v
(x) = v
(x)[v(x)]2
Example 3.4 Find the derivative of y = x21
x2+1.
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Power Rule
If n is a negative integer and x = 0, thend
dx(xn) = nxn1.
More generally,
For any real number ,
d
dx(x) = x1
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Second and Higher Order Derivatives
If f is a differentiable function, then its derivative f may alsohave a derivative of its own. If so, (f) is called the secondderivative of f and denoted by f. We refer to f as the zerothderivative and f as the first derivative. We also write thesecond derivative of y = f(x) as
y =dy
dx=
d
dx
dydx
=
d2y
dx2
If f is differentiable, its derivativef = y =
dy
dx=
d3y
dx3
is the third derivative of f.
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In general, the nth derivative of f is the derivative of the
(n 1)st derivative and denoted by
y(n) = f(n)(x) =dny
dxn=
d
dxy(n1)
Example 3.5 Find the nth derivative of the functiony = x3 4x2 + 10.
Example 3.6 Calculate the first four derivatives of y = x1.Then find the pattern and determine a general formula for y(n).
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Suppose we have a body moving along a coordinate line and weknow that its position at time t is s = f(t). In the interval fromany time t to the slightly later time t + t, the body moves fromposition s = f(t) to position s + s = f(t + t). The bodys netchange in position, or displacement, for this short time interval is
s = f(t + t) f(t).
Definition 4.1
If a body moves along a line from position s = f(t) to
position s + s = f(t + t), the the bodys averagevelocity for the time interval from t to t + t is
vav =displacement
travel time=
s
t=
f(t + t) f(t)t
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Definition 4.2
Instantaneous velocity (velocity) is the derivative ofposition with respect to time. If the position function of thebody moving along a line is s = f(t), the bodysinstantaneous velocity at time t is
v =ds
dt
= lim
t0
f(t + t) f(t)
t
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Motion under the Influence of Gravity
Galileo discovered that the height s(t) and velocity v(t) of anobject tossed vertically in the air are given as function of time bythe formulas
s(t) = s0 + v0t1
2 gt2
, v(t) =ds
dt = v0 gtThe constants s0 and v0 are the initial values:
s0 = s(0) is the position at time t = 0.
v0 = v(0) is the velocity at time t = 0.
g is the acceleration due to gravity on the surface of theearth, with value
g = 9.80 m/s2 = 32 ft/s2
The maximum height is attained when v(t) = 0.
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Example 4.1 A dynamite blast blows a heavy rock straight upwith a launch velocity of 160 ft/sec. It reaches a height ofs = 160t 16t2 ft after t sec.(a) How high does the rock go?(b) How fast is the rock traveling when it is 256 ft above theground on the way up? on the way down?
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Speed
Definition 4.3 Speed is the magnitude of the velocity.
speed = |velocity|
Example 4.2 Suppose a person standing at the top of abuilding 112 feet high throws a ball vertically upward with aninitial velocity of 90 ft/sec.
(a) Find the balls height and velocity at time t.
(b) When does the ball hit the ground and what is its impactspeed?
(c) When is the velocity 0? What is the significance of this time?
(b) How far does the ball travel during its flight?
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AND SOCIAL SCIENCES2.4.1 VELOCITY
Acceleration
In studies of motion, we usually assume that the bodys positionfunction s = f(t) has a second derivative as well as a first. Thefirst derivative gives the bodys velocity as a function of time; the
second derivative gives the bodys acceleration.
Definition 4.4 Acceleration is the derivative of velocity:
a =
dv
dt =
d2s
dt2 .
So
acceleration is the second derivative of the position.
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AND SOCIAL SCIENCES2.4.1 VELOCITY
If a(t) > 0, the velocity is increasing. This does not necessarymean that the speed is increasing. The object is speeding up onlywhen the velocity and acceleration have the same sign.
If velocity is and acceleration is then object is and speed is
positive positive moving forward increasingpositive negative moving forward decreasingnegative positive moving backward decreasing
negative negative moving backward increasing
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
Definition 4.5 The average rate of change of a functionf(x) over the interval from x to x + h is
Average rate of change =f(x + h) f(x)
h
The (instantaneous) rate of change of f at x is thederivative:
Rate of change of f at x = f(x) = limh0 f(x + h) f(x)h
provided that the limit exists.
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
Example 4.3 Let A = r2 be the area of a circle of radius r.
(a) Calculate the rate of change of area with respect to radius.(b) Compute dA/dr for r = 2 and r = 5, and explain why dA/dr islarger at r = 5.
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
The Effect of a One-Unit Change
For small values of the increment h, the difference quotient is closeto the derivative itself:
f(a) = limh0f(a + h)
f(a)
h f(a + h)
f(a)
h .
In some applications, the approximation is already useful withh = 1:
f(a)
f(a + 1)
f(a)
In other words,
f(a) is approximately equal to the change in f caused byone-unit change in x when x = a .
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
Example 4.4 Researchers have determined that the body massof yearling bighorn sheep on Ram Mountain in Alberta, Canada,can be estimated by
M(t) = 27.5 + 0.3t
0.001t2,
where M(t) is the mass of the sheep (in kg) and t is the numberof days since May 25.
(a) Find the average rate of change of the weight of a bighorn
yearling between 70 and 75 days after May 25.(b) Find the (instantaneous) rate of change of weight for a
yearling sheep whose age is 70 days past May 25.
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
We now consider some other rates of change in Physics.
A current exists whenever electric charges move. If Q is the netcharge that passes through a surface during a time period t,then the average current during this time interval is defined as:
Average Current = Qt
= Q Q0t t0
If we take the limit of this average current over smaller and smallertime intervals, we get what is called the current I at a given time
t0:I = lim
tt0Q Q0
t t0 =dQ
dt
Thus, the current is the rate at which charge flows through asurface.
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
If a given substance is kept at a constant temperature, then its
volume V depends on its pressure P. The rate of change of volumewith respect to pressure is the derivative dV/dP. As P increases,V decreases, so dV/dP < 0. The compressibility is defined by
Isothermal Compressibility = = 1
V
dV
dP
Velocity, compressibility, and current are not the only rates ofchange important in physics. Others include:
Density Power (the rate at which work is done) Rate of heat flow Temperature gradient (the rate of change of temperature with
respect to position)
Rate of decay of a radioactive substance in nuclear physics
Dr. Nguyen Ngoc Hai CALCULUS I
2.4 RATE F HAN E IN THE NATURALAND SOCIAL SCIENCES
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
Derivatives in Economics
Marginal Cost Suppose c(x) is the total cost of producing xunits of a particular commodity. It costs more to produce x + hunits, and the cost difference, divided by h, is the average increasein cost per unit:
c(x + h) c(x)h
= average increase in cost per unit
The limit of the ratio as h
0 is the marginal value when x units
of commodity are produced:
c(x) = limh0
c(x + h) c(x)h
= marginal cost
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AND SOCIAL SCIENCES2.4.2 OTHER RATES OF CHANGE
If the company, currently producing x units, increases itsproduction by one unit, then the additional cost c of producingthat one unit is
c = c(x + 1) c(x) c(x) 1 = c(x).
Thus,
Marginal cost estimates the cost of producing one unit beyond
the present production level.
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2.4.2 OTHER RATES OF CHANGE
Marginal Revenue and Marginal Profit Suppose r(x) is therevenue generated when x units of a particular commodity areproduced, and p(x) is the corresponding profit. When x = a unitsare being produced, then
The marginal revenue is r(a). It approximates r(a + 1) r(a),the additional revenue generated by producing one more unit.
The marginal profit is p(a). It approximates p(a + 1) p(a),the additional profit generated by producing one more unit.
Note Revenue = (number of units sold)(price per unit)
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2.4.2 OTHER RATES OF CHANGE
Example 4.5 Marginal Cost. Suppose it costs
c(x) = x3 6x2 + 15xdollars to produce x stoves and your shop is currently producing 10stoves a day. About how much extra will it cost to produce one
more stove a day?
Solution The cost of producing one more stove a day when 10are produced is about c(10). Since
c(x) = (x3
6x2
+ 15x) = 3x2
12x + 15,c(10) = 3(100) 12(10) + 15 = 195.
Thus the additional cost will be about $195 if you produce 11stoves a day.
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2.4.2 OTHER RATES OF CHANGE
Example 4.6 Marginal Revenue. If
r(x) = x3 3x2 + 12xgives the dollar revenue from selling x thousand candy bars, themarginal revenue when x thousand are sold is
r(x) = (x3 3x2 + 12x) = 3x2 6x + 12.The marginal revenue function estimates the increase in revenuethat will result from selling one additional unit. If you currently sell
10 thousand candy bars a week, you can expect your revenue toincrease by about
r(10) = 3(100) 6(10) + 12 = $252if you increase sales to 11 thousand bars a week.
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2.4.2 OTHER RATES OF CHANGE
Example 4.7 Marginal Revenue and Profit. A manufacturerestimates that when x units of a particular commodity areproduced, the total cost will be c(x) = 18 x
2 + 3x + 98 dollars, andfurthermore, that all x units will be sold when the price isp(x) = 13 (75
x) dollars per units.
(a) Find the marginal cost and the marginal revenue.
(b) Use marginal cost to estimate the cost of producing the ninthunit.
(c) What is the actual cost of producing the ninth unit?
(d) Use marginal revenue to estimate the revenue derived fromthe sale of the ninth unit.
(e) What is the actual revenue from the sale of the ninth unit?
Dr. Nguyen Ngoc Hai CALCULUS I
2.5 DERIVATIVE FTRIGONOMETRIC FUNCTIONS
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Derivative of the Sine
ddx
(sin x) = cos x
Derivative of the Cosine
ddx
(cos x) = sin x
Example 5.1 A body hanging from a spring is stretched 5 unitsbeyond its rest position and released at time t = 0 to bob up anddown. Its position at any latter time t is
s = 5 cos t.
What are its velocity and acceleration at time t?Dr. Nguyen Ngoc Hai CALCULUS I
2.5 DERIVATIVE FTRIGONOMETRIC FUNCTIONS
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Because sin x and cos x are differentiable functions of x, therelated functions
tan x =sin x
cos xsec x =
1
cos x
cot x =cos x
sin xcsc x =
1
sin x
are differential functions. Their derivatives are
d
dx(tan x) =
1
cos2 x= sec2 x
d
dx(sec x) =
sin x
cos2 x= sec xtan x
ddx
(cot x) = 1sin2 x
ddx
(csc x) = csc xcot x
Example 5.2 Differentiate(a) y = x2 sin x (b) y = 3
xcot x.
Dr. Nguyen Ngoc Hai CALCULUS I
2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.1 THE CHAIN RULE
Theorem 6.1 If g is differentiable at x and f is
differentiable at g(x), the composite function y = f g isdifferentiable at x and (f g)(x) is given by the product:
(f g)(x) = f(g(x)) g(x)
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2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.1 THE CHAIN RULE
In Leibniz notation,
If y = f(u) is a differentiable function of u and u = g(x) is adifferentiable function of x, then
dy
dx=
dy
du du
dx
ord
dxf(u) = f(u) du
dx
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2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.1 THE CHAIN RULE
Example 6.1 Find (f
g)(x) at x = 2 iff and g are
differentiable functions with g(2) = 3, g(2) = 4, and f(3) = 15.
Example 6.2 Find F(x) if F(x) = sin(x2 + 1)
Example 6.3 Find dy/dx at x = 0 ify = cos2 3x
.
Example 6.4 Find dy/dx for
y = 1 x
1 + x2
2.
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2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.2 IMPLICIT DIFFERENTIATION
The functions that we have met so far can be described by
expressing one variable explicitly in terms of another variable. Afunction of this form is said to be in explicit form.
Sometimes practical problems will lead to equations in which thefunction y is not written explicitly in terms of the independent
variables x; for example, equations such as:
x2 + y2 = 25, (1)
x3 + y3 = 6xy. (2)
cos(ty) =t2
y (3)
Since it has not been solved for y, such an equation is said todefine y implicitly as a function of x and the function y is said tobe in implicit form.
Dr. Nguyen Ngoc Hai CALCULUS I
2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
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2.6.2 IMPLICIT DIFFERENTIATION
There is a simple technique based on the Chain Rule that youcan use to find dy/dx without first solving for y explicitly.
This technique is called implicit differentiation.
Example 6.5 Find an equation of the tangent line to the curvehaving equation
x2 + y2
3xy + 4 = 0
at the point (2, 4).
Dr. Nguyen Ngoc Hai CALCULUS I
2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
2 6 2 IMPLICIT DIFFERENTIATION
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2.6.2 IMPLICIT DIFFERENTIATION
IMPLICIT DIFFERENTIATION
Suppose an equation defines y implicitly as a differentiablefunction of x. To find dydx
:
1. Differentiate both sides of the equation with respect to x.
2. Solve the differentiated equation algebraically for dydx
.
Dr. Nguyen Ngoc Hai CALCULUS I
2.6 HAIN RULE ANDIMPLICIT DIFFERENTIATION
2 6 2 IMPLICIT DIFFERENTIATION
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2.6.2 IMPLICIT DIFFERENTIATION
Example 6.6 Find dy/dt where cos(ty) = t2
y.
Example 6.7 Suppose that the output at an certain factory isQ = 2x3 + x2y + y3 units, where x is the number of hours ofskilled labor used and y is the number of hours of unskilled labor.The current labor force consists of 30 hours of skilled labor and 20hours of unskilled labor. Use calculus to estimate the change in
unskilled labor y that should be made to offset a 1-hour increase inskilled labor x so that output will be maintained at its current level.
Dr. Nguyen Ngoc Hai CALCULUS I
2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.1 GENERAL FORMULA
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Theorem 7.1 Assume that f(x) is differentiable andone-to-one with inverse g(x) = f1(x). If a belongs to thedomain of g(x) and f
g(a)
= 0, then g(a) exists and
g(a) = 1f
g(a)
In Leibniz notation,
df
1
dx =1dfdx
Dr. Nguyen Ngoc Hai CALCULUS I
2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.2 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
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For 1 < x < 1,
ddx
sin1 x = 11 x2
d
dxcos1 x = 1
1 x2
Example 7.1 Calculate f( 12 ), where f(x) = arcsin(x2).
d
dx tan1(x) =
1
1 + x2d
dx cot1(x) =
1
1 + x2
Example 7.2 Differentiate f(x) = xarctan
x.
Dr. Nguyen Ngoc Hai CALCULUS I
2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.3 DERIVATIVES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
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dy
dx
(ex) = exdy
dx
(ax) = ax (a > 0)
dy
dx(ln |x|) = 1
x
dy
dx(loga |x|) =
1
x ln a(a > 0, a = 1)
Example 7.3 Find the derivative of
(a) f(x) = esin x (b) g(x) = 103x (c) h(x) = log2
x2 + 1.
If we combine the above formulae with the Chain Rule, we get
dy
dx(eu) = eu
du
dx
dy
dx(au) = au
du
dxdy
dx(ln |u|) = 1
u
du
dx
dy
dx(loga |u|) =
1
uln a
du
dx
Dr. Nguyen Ngoc Hai CALCULUS I
2.7 DERIVATIVES OF INVERSE FUNCTIONS2.7.4 LOGARITHMIC DIFFERENTIATION
Th l l i f d i i f li d f i i l i
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The calculation of derivatives of complicated functions involvingproducts, quotients, or powers can often be simplified by taking
logarithms. The method used in the following examples is calledlogarithmic differentiation.
Example 7.4 Find the derivative of
f(x) = (sin x)x, 0 < x < .
In general, to differentiate a function of the form
y = u(x)v(x)
we take natural logarithms of both sides and differentiate implicitly.
Example 7.5 Find the derivative of
y =(x + 1)2(2x2 3)
x2 + 1.
Dr. Nguyen Ngoc Hai CALCULUS I
2.8 LINEAR APPR XIMATI NAND APPLICATIONS
2 8 1 LINEAR APPROXIMATION
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2.8.1 LINEAR APPROXIMATION
Definition 8.1 The approximation
f(x) f(a) + f(a)(x a) (4)
is called the linear approximation or tangent lineapproximation of f at x = a, and the function
L(x) = f(a) + f(a)(x a)
is called the linearization of f at x = a.
Example 8.1 Find the linearization of the function f(x) =
x
at a = 1 and use it to approximate the number
1.001.
Dr. Nguyen Ngoc Hai CALCULUS I
2.8 LINEAR APPR XIMATI NAND APPLICATIONS
2 8 1 LINEAR APPROXIMATION
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2.8.1 LINEAR APPROXIMATION
Let x = x
a be the change in x and y = f(x)
f(a) the
corresponding change in y. Then (4) can be written as
y f(a)x.The right side f(a)x is called the differential of f at a anddenoted by df or dy:
dy = df = f(a)x (5)
In particular, if f(x) = x, then df = dx = x at every point a.Substituting dx = x into Equation (5) we get
dy = df = f(a)dx (6)We call dx the differential of x, and df the correspondingdifferential of y. If we divide both sides of the Equation (6) bydx, we obtain the familiar equation
df = Dr. Nguyen Ngoc Hai CALCULUS I
2.8 LINEAR APPR XIMATI NAND APPLICATIONS
2.8.1 LINEAR APPROXIMATION
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2.8.1 LINEAR APPROXIMATION
Let P(x, f(x)) and Q(x + x, f(x + x)) be points on the graph
of f and let dx = x. The corresponding change in y is
y = f(x + x) f(x).
Therefore,
dy represents the amount that the tangent line rises or fall,whereas y represents the amount that the curve y = f(x)rises or falls when x changes an amount dx .
Sincey = f(x + x) f(x) dy,
we havef(x + x) f(x) + dy.
Dr. Nguyen Ngoc Hai CALCULUS I
2.8 LINEAR APPR XIMATI NAND APPLICATIONS
2.8.2 DIFFERENTIALS
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Example 8.2 The radius of a circle increases from an initialvalue of r0 = 10 by an amount dr = 0.1. Estimate thecorresponding increase in the circles area A = r2 by calculatingdA. Compare dA with the true change A.
The percentage error is defined by
percentage error =
error
actual value
100%
The percentage error is often more important than the error itself.
Dr. Nguyen Ngoc Hai CALCULUS I
2.8 LINEAR APPR XIMATI NAND APPLICATIONS
2.8.3 THE SIZE OF THE ERROR
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In any approximation, the error is defined by
error =true value approximate value
If the linearization of y = f(x) about x = a,
L(x) = f(a) + f(a)(x
a),
is used to approximate f(x) near a, then the error E(x) in thisapproximation is
E(x) = |f(x) L(x)| = |f(x) f(a) f(a)(x a)| = |y dy|.
Theorem 8.1 (Error Bound) If|f(x)| K for all x inthe interval between a and a + h, then
E(x) 12
Kh2.
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Pages Exercises Assignments
148-150 6, 8 11, 13, 16, 22, 24, 25155-156 3, 5, 8, 21, 35 6, 22, 28, 31, 36167-170 3, 21, 31 11, 30, 32, 41, 43197-199 14, 17, 42, 63 18, 20, 34, 41, 45,
47, 54, 55, 60, 61204-205 9, 17, 27, 40, 4, 12, 15, 20, 21,28, 30, 32, 34, 38
215-218 1, 3, 6, 2, 4, 5, 7, 10, 11, 14,15, 16, 19, 22,23
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P E i A i
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Pages Exercises Assignments
223-225 4, 12, 20, 24, 31, 11, 17, 25, 26, 29,
33, 37 32, 34233-236 10, 16, 33, 42, 19, 20, 22, 30, 32,
45, 54 34, 37, 39, 40,46, 47, 48, 49, 5558, 60, 71, 74
243-245 8, 10, 13, 17, 6, 11, 12, 14, 16,20, 26, 29, 46 18, 19, 23, 24, 32,
38, 45, 47, 48, 53250-251 4, 6, 12, 28 11, 16, 19, 20, 23,
29, 33, 35, 38, 39256-257 3, 6, 8 16, 18, 19
Dr. Nguyen Ngoc Hai CALCULUS I
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