Chapter 2 The Derivative:
Tangent Line Problem – basically finding the slope of the tangent line at a Point.
You can approximate this slope by using the secant line (gives 2 points to
calculate m)  the changes in y divided by the changes in x
** you can obtain more and more accurate approximations of the slope of
the tangent line by choosing points closer and closer to the point of tangency
=LIMITING PROCESS
lim
𝑓 (𝑐+ ∆𝑥)− 𝑓(𝑐)
∆𝑥
𝛥𝑥→0
=
m
** example finding derivative using the limiting process.
** example using the derivative to find the slope at a point
Differentiability and Continuity:
1. Differentiability implies continuity, Continuity does not imply
differentiability
a) Graph with a sharp turn: f(x) = Ιx – 2 Ι
 Use alternate form of derivative to check: lim
𝑓(𝑥)− 𝑓(𝑐)
𝑥−𝑐
𝑥→𝑐

lim
|𝑥−2|− 0
𝑥−2
𝑥→2−
|𝑥−2|− 0
 lim
𝑥→2+
𝑥−2
=1
= -1
** these are not equal so it is not differentiable
at (2,0)
b) A graph with a vertical tangent line
 Function: 𝑥 1/3
 Is continuous at x = 0 because the limit lim
𝑥→0
𝑥 1/3 − 0
𝑥
= lim
𝑥→0
1
𝑥 2/3
=∞
Rates of Change:
Average Velocity=
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 ∆𝒔
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆 ∆𝒕
Example: A billiard ball is dropped from a height of 100 ft, 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 of the following time
intervals: a) [1,2] b [1, 1.1
Instantaneous Velocity – can be calculated again by approximating the slope of
tangent line by calculating the slope of the secant line 
finding the first derivative of the position function
Speed – absolute value of the velocity – cannot be negative
Acceleration – derivative of velocity
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 diver’s velocity at impact?
Parametric Equations: Instead of describing a curve by expressing the ycoordinate of a point P(x,y) on the curve as a function of x, it is more convenient
to describe a curve by expressing both coordinates as functions of a third variable,
t.
parametric curve: if x and y are given as functions
x = f(t)
y = g(t)
over an interval of t-values, then the set of points (x,y) = (f(t),g(t)) defined by
these equations is a parametric curve. The equations are parametric equation for
the curve.
** When we give parametric equations and a parameter interval for a curve, we
say that we have parametrized the curve. The equations and the interval
together constitute the parametrization of the curve.
Ex. Find a Cartesian equation of the curve defined by the parametric equations
x=2t – 3 and y=4t – 1 and sketch the curve:
Step 1: eliminate t from the 2 equations by solving the first equation for t to
obtain:
t = 1/2x + 3/2 and then substitute into the y equation
To obtain: y = 2x+5 -- if graph it is just a line, but notice as t
increases so do x and y. Thus a particle moving on the line goes upward and to
the right as indicated by the arrowhead.
****In general, the graph of any pair of parametric equations of the form
X= at + b
and y = ct + d
where either a≠0 or c≠0 ***
Ex. Find the Cartesian equation of the graph of the parametric equations:
X= 2 cos t
and y = 2 sin t
0≤ t≤2π
**to eliminate the parameter – square both sides of the equation and add:
x² + y² = 4cos²t + 4 sin²t
x² + y² = 4(cos²t + sin²t)  x²+ y²=4
A plane curve defined by the parametric equations x=f(t) and y= g(t) is said to be
smooth on the closed interval [a,b] and f´ and g´ are continuous on [a,b] and f´(t)
and g´(t) are not both zero at every number in the open interval (a,b)
A parametrized curve x = f(t) and y = g(t) is differentiable at t if f and g are
differentiable at t. At a point on a differentiable parametrized curve where y is
also a differentiable function of x, the derivatives dy/dt, dx/dt, and dy/dx are
related by the chain rule: dy/dt = dy/dx · dx/dt
If dx/dt≠0 then we may divide both sides of this equation by dx/dt to solve for
dy/dx.
If all 3 derivatives exist and dx/dt ≠0 then:
dy/dx =
𝒅𝒚/𝒅𝒕
𝒅𝒙/𝒅𝒕
Ex. If x= 2t + 3 and y = t² - 1 find dy/dx at t = 6
Dy/dt = 2t
dx/dt= 2
dy/dx =
2𝑡
2
=t
If the equations x = f(t), y=g(t) define y as a twice
differentiable function of x, then at any point
where dx/dt ≠0:
𝒅𝟐 𝒚
𝒅𝒙𝟐
=
𝒅𝒚′
𝒅𝒕
𝒅𝒙
𝒅𝒕
Ex. Find d²y/dx² as a function of t if x = t - t²,
y =t - t³
Step 1: express y´ = dx/dt in terms of t
y’ =
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
=
1−3𝑡²
1−2𝑡
Step 2: differentiate y’ with respect to t
𝑑𝑦′
𝑑𝑡
=
𝑑 1−3𝑡 2
𝑑𝑡 1−2𝑡
=
2−6𝑡+6𝑡 2
(1−2𝑡)2
Step 3: divide dy’/dt by dx/dt
𝑑2 𝑦
𝑑𝑥 2
=
𝑑𝑦′
𝑑𝑡
𝑑𝑥
𝑑𝑡
=
(2−6𝑡+6𝑡 2 )/(1−2𝑡)2
1−2𝑡
=
2−6𝑡+6𝑡 2
(1−2𝑡)3
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
= 1
dt
· (100) = 1
4π(25)²
25π