3.1 Differentiation
***Chapter 2  defined the slope of a curve at a point as the limit of secant slopes. This
limit was called the DERIVATIVE  and it measures the rate at which the function
changes
** we use notation f(x) instead of just f in the definition to emphasize the independent
variable, x, which we are differentiating with respect to.
 in the domain, if f has a limit at a particular value of x, then it has a derivative
Differentiation = the process of calculating the derivative, which exists if only if limits
exist for all points in the domain of a given function.
Notation: d/dx
or f′(x)
or
y′ or dy/dx (derivative of y with respect to x)
Indication of the value of a derivative at a specified #, x = a use:
f′(a) = dy│
dx│x = a
Differentiable on an Interval: 1 sided derivative
 Function is differentiable on an open interval (finite/infinite) – if it has a
derivative at each point on the interval

Differentiable on closed interval [a,b] if it is differentiable only on interior
Points and if limits exist for right and left hand
Lim f(a+h) – f(a)
x 0+
h
= right hand derivative at a
lim f(b+h) – f(b)
x 0h
= left hand derivative at b
Ex. y = │x│ is not differentiable at the origin
a) to right of origin f′│x│ = 1 lim as approaches 0+ = 1
b) to left of origin f′′│x│ = -1 lim as approaches 0- = -1
one sided derivatives differ at the origin, so there cannot be a derivative at
the origin
When does a derivative exist?
** a function has a derivative at a point if the slopes of the secant lines through
P(x0,f(x0)) and a nearby point Q on the graph approach a limit as Q approaches
P (must have a limit to exist) –
** smoothness is also a condition of differentiability
** Differentiability implies continuity, if has a derivative, then continuous
When Doesn’t a derivative Exist?
1) Whenever the secants fail to take up a limiting position.
2) If the secant becomes vertical before Q approaches P
3) If the following occur, there will not be a derivative at the point:
a. A Corner – where 1 sided derivatives differ
b. A Cusp – where slope of PQ approaches ∞ from 1 side, -∞ from other
c. Vertical Tangent – where slope of PQ approaches ∞ from both sides or
Approaches -∞ from both sides.
d. Discontinuity occurs
3.2 Differentiation Rules:
Rule #1. Derivative of a Constant Function:
Rule # 2. Power Rule for Positive Integers
Rule #3 Constant Multiple Rules
Rule #4 Derivative Sum Rule
Rule #5 Derivative Product Rule (udv + vdu)
Rule #6 Derivative Quotient Rule (vdu – udv)
v²
Rule #7 Power Rule for Negative Integers
Second and Higher Order Derivatives
3.3 Derivatives as Rate of Change
** We now study how derivatives are used to model the rates at which things change in
the world around us:
1) a physician may want to know how change in dosage affects the body’s
response to a drug
Instantaneous Rate of Change:
f´(x0) = lim f(x0 + h) – f(x0)
h 0
h
 instantaneous rates are limits of average rates
Position Function = function s, that gives the position (relative to the origin) of an object
as a function of time s=f(t)
**The Displacement of the object over the time interval for t to t + Δt is:
Δs = f(t + tΔ) – f(t)
Average Velocity = change in distance(displacement) = Δs =f(t+Δt) – f(t)
change in time(travel time)
Δt
Δt
(the 1st derivative of the position function)
** so to find the body’s velocity  take the limit of the average velocity over the interval
from t to t+Δt as Δt shrinks to zero. This limit is the derivative of f with respect to t
Ex. if a billiard ball is dropped from a height of 100 feet, its height s, at time t, is
given by the position function
s = -16t² + 100
where s is measured in feet and t is measured in seconds. Find the average
velocity over each time interval.
a) [1,2]
s = -16(1)² + 100 = 84
S = -16(2)² + 100 = 36
Δs = 36 – 84 = -48 ft/sec
2-1
**velocity (instantaneous velocity) v=ds/dt is the derivative of position with respect to
time.



Besides telling how fast an object is moving  velocity tells the direction
of motion.
When an object is moving forward(positive slope and s increasing) 
velocity positive
When an object is moving backward (negative slope and s
decreasing)velocity negative
Instantaneous Velocity
Velocity Function = derivative of the position function (**can be +/-/0)
Ex. s = -16t² + 100
V(t) = -32t
Speed = absolute value of its velocity **cannot be negative
**measures the rate of progress regardless of direction**
|v(t)| = |ds/dt|
Acceleration = derivative of velocity (second derivative of position)—the rate at
which the body’s velocity changes – measures how quickly the
body picks up or loses speed.
a(t) = dv = d²s
dt
dt²
Jerk = a sudden change in acceleration. The derivative of acceleration with
respect to time.
j(t) = da = d³s
dt dt³
Velocity (v) = derivative of position
Acceleration (a) = derivative of velocity
Jerk (j) = derivative of acceleration
Speed = │ds│ absolute value of velocity
│dt│
Ex. #2 p. 179
a) Find the body’s displacement and average velocity for the given time interval
b) Find the body’s speed and acceleration at the endpoints of the interval
c) When, if ever, during the interval does the body change direction?
s = 6t-t²
0≤t≤6
Ex. At time t=0, a diver jumps from a platform diving board that is 32 feet above the
water. The position of the diver is given by: s(t) = -16t² + 16t +32 where s is measured
in feet and t is measured in seconds.
a) when does the diver hit the water?
b) What is the divers velocity at impact?
a) to find the time, t when the diver hits the water, let s =0 and solve for t
-16t² + 16t + 32 = 0
-16(t² - t – 2) = 0
-16(t – 2)(t+1)= 0
t= -1 or 2
b/c t ≥ 0 chose the positive value and
Conclude the diver hits the water at t=2 sec.
b) velocity at time t is given by v′(t) = -32t + 16
**impact is 2 from above so v = -32(2) + 16 = -48 feet per second
Ex. A rock thrown vertically upward from the surface of the moon at a velocity of 24
m/sec (about 86 km/h) reaches a height of s = 24t - .8t² meters in t second
a. Find the rock’s velocity and acceleration at time t (the acceleration in this case
is the acceleration of gravity on the moon)
b. How long does it take the rock to reach its highest point?
c. How high does the rock go?
d. How long does it take the rock to reach half its maximum height
e. How long is the rock aloft?
**If a particle moves along a line according to the function s=f(t) where s represents the
position of the particle P on the line at time t, then the velocity v, of P at time t, is given
by ds/dt and its acceleration by dv/dt or by d²s/dt². The speed of the particle is |v|, the
magnitude of v. If the line of motion is directed positively to the right, then the motion of
the particle P is subject to the following. At any instant:
a) If v>0 then P is moving to the right and its distance s is increasing:
If v<0 then P is moving to the left and its distance s is decreasing
b) If a>0, then v is increasing
If a<0, then v is decreasing
c) If a and v are both positive or both negative, then (a) and (b) imply that the speed of P
is increasing, or that P is accelerating: if a and v have opposite signs, then the speed of P
is decreasing or P is decelerating
d) If s is a continuous function of t, then P reverses direction whenever v is zero and a is
different from zero, note that zero does not necessarily imply a reversal in direction
ex. a particle moves along a line according to the function: s=2t³ - 9t² + 12t – 4 where
t≥0
a) find all t for which the distance s is increasing
b) find all t for which the velocity is increasing
c) find all t for which the speed of the particle is increasing
d) find the speed when t = 3/2
e) find the total distance traveled between t=0 and t=4
v = 6t² - 18t +12  6(t-2)(t-1)=0
 t=1, 2
**check intervals: t<1 6(0-2)(0-1) = 12  v>0
1<t<2 6(1.5-2)(1.5-1) = -  v<0
t>2
6(3-2)(3-1) = +  v>0
a = 12t – 18  6(2t-3)=0  t = 3/2
**check intervals: t<3/2
t>3/2
6(2(1)-3) = -  a<0
6(2(2) -3) = +  a>0
These signs of v and a yield:
1) s increases when t<1 or t>2
2) v increases when t>3/2
3) the speed |v| is increasing when v and a are both +, that is for t>2 and when v and
a are both -, that is for 1<t<3/2
4) the speed when t = 3/2 equals |v|=|-3/2|=3/2
5) Graphically: P’s moves to the right if t<1, reverses its direction at t=1, moves to
the left when 1<t<2, reverses again at t=2, and continues to the right for all t>2.
The position P at certain times is shown:
(t=2)------------------------------------------(t=3)-----
(t=0)----------------------(t=1)
-4
T
S:
0
-4
1
1
0
1
2
0
4
28
5
**so P travels a total of 34 units between times t= 0 and t=4
Marginal Cost of Production – the rate of change of cost with respect to the level of
production (dc/dx)
**to find increased cost production – take the derivative of the cost function and plug
value in.
**to find increased revenue – take the derivative of the revenue function and plug value
in
-----------------------------------------------------------------------------------------------------------3.4 Derivatives of Trig.Functions
Derivative of Sine Function = d (sin x) = cos x
dx
Derivative of Cosine Function = d (cos x) = - sin x
dx
Derivatives of Tangent = d (tan x) = sec²x
dx
Derivative of Secant = d (sec x) = secxtanx
dx
Derivative of Cotangent = d (cot x) = -csc²x
dx
Derivative of Cosecant = d (csc x ) = -cscxcotx
dx
3.5 The Chain Rule
Chain Rule – says the derivative of the composite of 2 differentiable functions is the
product of their derivatives evaluated at appropriate points.
**used on composite functions**
Formal Def: the derivative of the composite function f(g(x)) at x is the derivative of f at
g(x) · derivative of g at x
(f◦g)′(x) = f′(g(x)) · g′(x)
Leibniz Notation if y = f(u) and u= g(x)
Then dy = dy · du
dx
du dx
IN Words: “differentiate the outside function f and evaluate if at the inside function
g(x) which is left alone, then multiply by the derivative of the inside
function.”
Ex. differentiate sin(x² + x) with respect to x
F(x) = sin u
u=x² + x
Power Rule Combined with Chain Rule: if n is a positive integer and f(u) = u, then the
power rule tells us that f′(u) = nun-1. If u is a differentiable function then
we use the power Chain rule here.
Ex…
Finding Tangent Slopes: Use the chain rule
Ex. Find the slope of the line tangent to the curve y=sin5x at point where x=π/3
--------------------------------------------------------------------------------------------------------Parametric Equations – expressing both coordinates as a function of the 3rd variable t
x = f(t) and y = g(t)
(t usually denotes time)
*are better than a Cartesian formula because they tell us the particles position at any
time*
Parametric Curve –if x and y are given as a function
(f(t), g(t)) defined by these points is a Parametric Curve
t= parameter for the curve, and its domain is I = parameter interval
** if I is a closed interval a≤t≤b, the point (f(a), g(a)) = initial point of the curve
and the terminal point is (f(b), g(b)).
Parametrized The Curve = when we give the parametric equation and a parameter
Interval for a curve.
Parametrization of Curve = the equation and interval given together
Parametrized Curve x = f(t), y=g(t) is differentiable at t if f and g are differentiable at t
Parametric Formula for dy/dx: if all 3 derivative exist and dx/dy≠0
dy = dy/dt
dx dx/dt
Ex. find d²y as a function of t if x = t-t² and y = t-t³
dx²
Steps: first express y′=dy/dx in terms of t
y′ = dy = dy/dt
1-3t²
dx
dx/dt
1-2t
now differentiate y′ with respect to t
dy′ = d 1-3t²
dt dt 1-2t
= (1-2t)(-6t) – (1-3t²)(-2) =
(1-2t)²
6t² - 6t +2
(1-2t)²
divide dy′/dt by dx/dt
d²y = dy′/dt = (6t²-6t+2)/(1-2t)² = 6t² - 6t + 2
dx²
dx/dt
1 – 2t
(1-2t)³
Standard Parametrizations and Derivative Rules
Circle x² + y² = a²
X = accost
Y = asint
0≤t≤2π
Rest in book on page: 201
-----------------------------------------------------------------------------------------------------------3.6 Implicit Differentiation
** Previous derivatives that we have dealt with thus far have been in the form of y=f(x)
which expresses y explicitly in terms of the variable x.
Implicit Differentiation --- used when we can’t put equation in the form y=f(x)
Steps:
#1. Differentiate both sides of equation with respect to x, treating y as a differentiable
function of x.
#2. Collect the terms with dy/dx on one side of the equation (goal: to isolate)
#3. Solve for dy/dx
Folium of Descartes – curves that are in the form of x³+y³ = 6xy
f) implicitly defines y as several functions of x
Normal Line - the line perpendicular to the tangent to the profile curve at the point of
entry.
------------------------------------------------------------------------------------------------------------
3.3 The Derivative as a Rate of Change
** Application problems involving functions that are describe by motion
Position Function = a function that gives the position (relative to the origin) of an object
as a function of time (t)
Average Velocity Vav (position) = displacement Δs
Travel time Δt
Velocity (Instantaneous) – derivative of Position with respect to time
V= ds/dt
Speed = absolute value of velocity
Speed = │v(t)│= │ds/dt│
Acceleration = derivative of velocity with respect to time
a(t) = dv/dt (or view as the 2nd derivative of velocity)
Jerk = derivative of acceleration with respect to time
j(t) = da/dt (or view as the 3rd derivative of velocity)
Hints to solving:
1.** Set Velocity = 0 to solve when trying to find how long it takes to do something
because when something is at highest point the v = 0, nothing is
rising or falling yet
2. To find velocity and speed when an object is certain feet above ground(up and down)
– solve for each quadratic (by factoring, a quadratic formula, etc) – should get 2 answers.
---- that tells time of object on way up and way down – just plug the t’s into velocity
equation to find exact ft/sec.
3. Acceleration – always negative because as something rises, it slows down, and speeds
up as it falls.
4. To find out when an object will hit land, take original position equation and set = 0
5. Velocity can be +, -, 0 but speed can only be +
Ex. If a billard ball is dropped from a height of 100 feet, its height, s, at time t, is given
by: s = -16t² + 100, where s is measured in feet and t is measured in seconds. Find the
average velocity over each time interval.
A) [1,2]
B) [1,1.5]
Ex. s = t²-3t + 2 0≤t≤2 find the following
a) body’s displacement and average velocity for the given time interval
b) the body’s speed and acceleration at the endpoints of the interval
c) When, if ever, during the interval does the body change direction?
Ex. At time t =0 a diver jumps from a platform diving board that is 32 feet above the
water. The position of the diver is given by : s(t) = -16t² + 16t + 32
a) When does the diver hit water
b) What is the diver’s velocity at impact
Ex. It takes 12 hours to drain a storage tank by opening the valve at the bottom. The
depth y, of fluid in the tank t, hours after the valve is opened is given by the formula:
Y = 6(1 – t/12)³ meters
a) find the rate dy/dt at which the tank is draining at time t
b) when is the fluid level in the tank falling fastest? Slowest? What are the values of
dy/dt at these times?
c) graph y and dy/st together and discuss the behavior
Ex. Suppose that the distance an aircraft travels along a runway before takeoff is given
by D=(10/9)t², where D is measured in meters from the starting point, and t is measured
in seconds from the time the brakes are released. The aircraft will become airborne when
its speed reaches 200 km/h. How long will it take to become airborne and what distance
will it travel in that time?
3.8 Linearization
Linearization – when we approximate complicated functions with simpler ones that give
the accuracy we want and are simpler to work with.
**Purpose: its ability to replace a complicated formula by a simpler one over
an entire interval of values.
** As long as the tangent remains close to the graph of f(x) then the linearization
Will give a good approximation to f(x)
If f is differentiable at x=a, then
Linearization of f at a = L(x) = f(a) + f′(a)(x-a)
Center of approximation = the point x=a
Standard linear approximation of f at a f(x) ≈ L(x)
Ex. find linearization of f(x) = √1 + x at x = 0
For Roots and Powers: can use the following formula: (1 + x)k ≈1+ kx
Ex. ³√1 + 5x4
Differentials - Let y = f(x) be a differentiable function then:
Differential dx = an independent variable
Differential dy = f′(x)dx (dependent variable)
(dy is used to estimate error in measurement and sensitivity of a function to change)
** represents the change in the linearization of f – the amount the tangent line
Rises or falls when x changes by an amount dx = Δx
** also written as the differential of f --- df
Ex. df = (3x²-6) = 6x dx
Ex. find the differential of df=(tan 2x)
Estimating with differentials – want to predict how much the value of a differentiable
function will change if we move to a point a+dx
a) if dx is small  then Δy is ≈ equal to differential dy
so f(a+dx)≈ f(a) + dy when dx = Δx
Ex. radius of a circle increases form a=10 m to 10.1m. Use dA to estimate the increase
in the circle’s Area A. Estimate the area of the enlarged circle and compare your
estimate to the true area.
Error in Differential Approximation:
2 ways to describe change in f as x changes from a to a + Δx
1. true change Δf = f(a + Δx) – f(a)
2. differential estimate - df = f′(a)Δx
Є = error
So Δy = f′(a)Δx + ЄΔx- approximation error: ∣Δf-df∣
Sensitivity to change – df = f′(x) dx tells how sensitive the output of f is to a change in
The input at different values of x
** larger values of f′ at x – greater the effect of a given change on dx
3 ways to describe change;
True
Δf=f(a+dx) – f(a)
Estimate
df = f′(a)dx
2. Relative Change
Δf
f(a)
df
f(a)
3. Percentage Change
Δf
f(a) x 100
df
f(a) x 100
1. Absolute Change
3.7 Related Rates:
To compute the rate of change of 1 quantity in terms of the rate of change Of another
quantity: (**rates of change = derivatives)
Procedures:
1. find an equation that relates the 2 quantities
2. Use the chain rule to differentiate both sides with respect to times
Strategies:
1. Draw a picture and name the variables and constants – use t for time, and
then assume all variables are differentiable functions of t.
2. Write down the numerical information
3. Write down what you are asked to find (usually a rate expressed as a
derivative)
4. Write an equation that relates the variables
5. Differentiate with respect to t
6. Evaluate
Ex. Air is being pumped into a spherical balloon so that its volume increases at a rate of
100cm³/sec. How fast is the radius of the balloon increasing when the diameter is 50
cm.?
To solve: Identify given info: the rate of increase of the volume of air is 100 cm³/s
Unknown: The rate of increase of the radius when the diameter is 50 cm r=25
Let v = volume of balloon
r = radius of balloon (in this prob. vol. and radius are both functions of time)
dv/dt = rate of increase of the volume with respect to time
dr/dt = rate of increase of the radius with respect to time
So: dv/dt = 100 cm³/s
dv = 4πr³ dr
dt
3 dt
100 = 4πr² dr
dt
100 1 = dr
4πr²
dt
= 1
4π(25)²
· (100) = 1
25π
Ex. Water pours into a fish tank at a rate of 3 ft³/min. How fast is the water level rising
if the base of the tank is a rectangle of dimensions 2 x 3 ft?
Ex. The radius and height of a right circular cylinder are related to the cylinders volume,
V, by the formula V = πr²h
a) How is the dV/dt related to dh/dt if r is constant?
b) how is dV/dt related to dr/dt if h is constant?
c) How is dV/dt related to dr/dt and dh/dt if neither r nor h is constant?
Ex. 1 train travels west toward Chicago at 120 mph, while a 2nd train travels north away
from Chicago at 90 mph. At time t=0, the 1st train is 10 miles east and the 2nd train is 20
miles north of the Chicago station. Calculate the rate at which the distance between the
trains is changing:
a) at time t=0
b) 10 minutes later