Differentiation
CHAPTER 4
The Derivative and the Tangent
Line Problem
SECTION 4.1
The Tangent Line Problem
To find the tangent line of a curve at a point P, you
need to find the slope of the tangent line at point
P. You can approximate this slope using the
secant line through the point of tangency and a
second point on the curve where (c, f(c)) is the
point of tangency and (c + ∆x, f (c + ∆x)) is a
second point on the graph of f.
Slope formula: m = (y2 –y1)/(x2-x1) = ∆y/ ∆x
Slope formula: msec =( f (c + ∆x) – f(c))/(c + ∆x – c)
msec = f (c + ∆x) – f(c))/ ∆x
Definition of Tangent Line with Slope m
If f is defined on an open interval containing c,
and if the limit
lim ∆y/ ∆x = lim (f(c + ∆x) – f(c))/ ∆x = m
∆x0
∆x0
Exists, then the line passing through (c,f(c)) with
slope m is the tangent line to the graph of f at
the point (c,f(c))
The slope of the tangent line is also called the
slope of the graph of f at x =c
Example
Find the slope of the graph of f(x) = 2x -3 at the
point (2,1)
Use the definition of Tangent Line with Slope m
Then c = 2
lim ∆y/ ∆x = lim (f(c + ∆x) – f(c))/ ∆x = m
∆x0
∆x0
lim (f(2 + ∆x) – f(2))/ ∆x = {[2(2 + ∆x) -3]-[2(2) -3]}/ ∆x
∆x0
=
lim (4 + 2 ∆x - 3 – 4 + 3)/ ∆x = 2 ∆x / ∆x = 2
∆x0
m=2
Example
Find the slope of the tangent line to the graph of
1. g(x) = 3x – 6 at the point (2,0)
2. g(x) = -4x + 1 at the point (-1,5)
3. f(x) = x2 + 1 at the point (-1,2)
4. f(x) = x2 + 1 at the point (0,1)
Definition of the Derivative of a Function
The derivative of f at x is given by
f′(x)= lim (f(x + ∆x) – f(x))/ ∆x
∆x0
Provided the limit exists. For all x for which this
limit exists, f′ is a function of x.
f′ is a new function of x which gives the slope of
the tangent line to the graph of f at the point
(x,f(x))
Differentiation
Differentiation the process of finding the
derivative of a function.
A function is differentiable at x if its derivative
exists at x and is differentiable on an open
interval (a,b) if it is differentiable at every point
in the interval
Notations for Derivatives
f′(x)
dy/dx
y′
d/dx[f(x)]
Dx[y]
Derivatives
dy/dx = lim ∆y/ ∆x
∆x 0
= lim (f(x + ∆x) – f(x))/ ∆x
∆x0
= f′(x)
Example
Find the derivative by the limit process
1. g(x) = 2x2 - 9
2. f(x) = x3 + 2x
3. f(x) = √x
4. y = 2/t
Differentiability and Continuity
The following alternative limit form of the
derivative is useful in investigating the
relationship between differentiability and
continuity. The derivative of f at c is
f′(c) = lim (f(x) –f(c))/(x – c)
x c
provided the limit exists.
The existence of the limit in this alternative form
requires that the one-sided limits exist and are
equal
Differentiability and Continuity – con’d
lim (f(x) –f(c))/(x – c)
x c-
and
lim (f(x) –f(c))/(x – c)
x c+
These one-sided limits are called the derivatives
from the left and from the right. It follows that f
is differentiable on the closed interval [a,b] if it
is differentiable on (a,b) and if the derivative from
the right at a and the derivative from the left at b
both exist.
Differentiability and Continuity – con’d
Differentiability implies continuity.
• If a function is differentiable at a point, then the
function is continuous at that point
Continuity does not imply differentiability.
 A function may be continuous at a point, but
not differentiable at that point
 A function that is not continuous at a point is
not differentiable at that point
Example
Determine differentiability and/or continuity
1. g(x) = |x – 2| at x = 2
2. f(x) = x1/3 at x = 0
3. f(x) = [| x |] at x = 0 the greatest integer
Basic Differentiation Rules and
Rates of Change
SECTION 4.2
The Constant Rule
The derivative of a constant function is 0. That
is, if c is a real number, then
d/dx[c] = 0
Example
Find the derivative of these functions
1. y = 7
2. f(x) = 0
3. s(t) = -3
4. y = kπ2, where k is constant
The Power Rule
If n is a rational number, then the function given
by f(x) = xn is differentiable and
d/dx[xn] = nxn-1
For f to be differentiable at x=0, n must be a
number such that xn-1 is defined on an interval
containing 0.
The Power Rule when n = 1
If n =1, then the function given by f(x) = x is
differentiable and
d/dx[x] = 1
Example
Find the derivatives of these functions
1. f(x) = x3
2. f(x) =3√x
3. s(t) = 1/x2
4. y = x4
Finding the Equation of a Tangent Line
 First, evaluate the function at x
 Second, find the slope using the derivative and
evaluate at x
 Third, use point slope form to find the tangent
line
Example
Find the equation of a tangent line for each of
these functions
1.
2.
3.
4.
f(x) = x2 when x = -2
f(x) =3√x when x = 8
s(t) = 1/x2 when x = -3
y = x4
when x = ½
The Constant Multiple Rule
If f is a differentiable function and c is a real
number, then cf is also differentiable and
d/dx[cf(x)] = cf′(x)
Example
Find the derivatives of these functions
1. f(x) =2/x
2. f(x) =4t2/5
3. s(t) =2√x
4. y = -3x/2
5. 1/2( 3√x2 )
6. y = 7/(3x)-2
The Sum and Difference Rules
The sum (or difference) of two differentiable
functions f and g is itself differentiable.
Moreover, the derivative of f + g (or f - g) is the
sum (or difference) of the derivatives of f and g.
d/dx [f(x) + g(x)] = f′ (x)+ g′ (x) Sum Rule
d/dx [f(x) - g(x)] = f′ (x) - g′ (x) Difference Rule
Example
Find the derivatives of these functions
1. f(x) =x3 – 4x + 5
2. f(x) =3x2 - 4
3. s(t) =-2x3 + 9x2 - 2
4. y = -x4/2 + 3x3 – 2x
Rates of Change
The derivative can be used to determine the rate
of change of one variable with respect to
another. A common use of rate of change is to
describe the motion of an object moving in a
straight line. The function s that gives the
position (relative to the origin) of an object as a
function of time t is called a position function.
Then
rate = distance/time or
Rates of Change – cont’d
Average velocity is
Change in distance/change in time
= ∆s/ ∆t
which is the derivative
Example – Finding Average Velocity
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
Find the average velocity over each time interval
1. [1,2]
2. [1,1.5]
3. [1,1.1]
What do the negative answers indicate?
Example – Using Derivative to Find Velocity
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?
The Product and Quotient Rules
and Higher-Order Derivatives
SECTION 4.3
The Product Rule
The product of two differentiable
functions f and g is itself differentiable.
Moreover, the derivative of fg is the
first function times the derivative of
the second, plus the second function
times the derivative of the first.
d/dx [f(x)g(x)] = f(x)g′(x) + g(x)f′(x)
Example
Find the derivative of each function
1.
2.
3.
4.
5.
h(x) =(3x-2x2)(5 + 4x)
y = (3x -1)(2x + 5)
g(x) = (x3 +2x2 -5)(x4 – 4x2)
f(x) = x2(x+1)(2x-3)
y= (2x3 + 5x)(7)
The Quotient Rule
The product of two differentiable functions f
and g is itself differentiable at all values of x
for which g(x) ≠ 0. Moreover, the derivative
of f/g is given by 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.
d/dx [f(x)/g(x)] = g(x)f′(x) - f(x)g′(x)] ÷ [g(x)]2
Example
Find the derivative of each function
1.
2.
3.
4.
5.
h(x) = (5x -2) ÷ (x2 + 1)
y = (x2 + 3x) ÷ 6
g(x) = (12x3 - 3x2) ÷ (6x2)
f(x) = (x+1) ÷ (x-1)
y = (x-3) ÷ (x2-4)
Example – Find an Equation of the Tangent Line
f(x) = (3 – 1/x) ÷ (x+5) at (-1,1)
1. Find f′(x)
2. Evaluate f′(x)
3. Use point-slope form to find equation
Higher-Order Derivatives
s(t)
Position function
v(t) = s′(t)
Velocity Function
a(t) = v′(t)= s″(t) Acceleration Function
Example – Finding Acceleration
A car’s velocity starting from rest is
v(t) = 100t ÷ (2t+15) where v is measured in feet
per second. Find the acceleration at each of
the following times.
1. 5 seconds
2. 10 seconds
3. 20 seconds
The Chain Rule
SECTION 4.4
The Chain Rule
The chain rule is a very powerful differentiation
rule that deals with composite functions
The Chain Rule
Without the Chain Rule
With the Chain Rule
 f(x) =x2 + 1
 y = (x2 +1)½
 g(x)=3x + 2
 f(x) = 3x +2)5
 y =x + 2
 g(x) = (x + 2)1/3
The Chain Rule
If y =f(u) is a differentiable function of u and
u =g(x) is a differentiable function of x, then
y =f(g(x)) is a differentiable function of x and
dy/dx = dy/du · du/dx
or
d/dx[f(g(x))] = f′(g(x))g′(x)
Example – Applying the Chain Rule
 Write the function as a composite of functions
 Apply Chain Rule
y = f(g(x))
y= 1/(x+1)
y = (3x2 – x +1)½
u = g(x)
u=x+1
u = 3x2 – x +1
y =f(u)
1/u
u½
Example – Applying the Chain Rule
 Find dy/dx for y = (x2 +1)3
y = f(g(x))
y = (x2 +1)3
u = g(x)
u = x2 +1
dy/dx = dy/du · du/dx
3(x2 + 1)2(2x) = 6x(x2 +1)2
dy/du du/dx
y =f(u)
u3
The General Power Rule
If y = [u(x)]n where u is a differentiable function
of x and n is a rational number, then
dy/dx = n[u(x)]n-1du/dx
d/dx[un] = nun-1u′
or
Example
Differentiate the functions
1. f(x) = (3x -2x2)3
2. f(x) = [(x2 -1)2 ]1/3
3. g(t) = -7/(2t-3)2
4. y = [(3x -1)÷ (x2 +3)]2
Implication Differentiation
SECTION 4.5
Implicit and Explicit Functions
Up to this point, most in explicit form.
For example in the equation y = 3x2 - 5
the variable y is explicitly written as a
function of x
Implicit Form Explicit Form Derivative
xy = 1
y = 1/x
-1/x2
Sometimes, this procedure doesn’t work,
so you would use implicit differentiation
Implicit Differentiation
To understand how to find dy/dx
implicitly, you must realize that the
differentiation is taking place with
respect to x. This means when you
differentiate terms involving x alone,
you can differentiate as usual, but
when you differentiate terms involving
y, you must apply the Chain Rule,
because you are assuming that y is
defined implicitly as a differentiable
function of x.
Differentiating with Respect to x
1. d/dx[x3] = 3x2
2. d/dx[y3] = 3y2dy/dx where y =u
and n = 3
nun-1 u′
3. d/dx[x +3y] = 1 +3(dy/dx)
4. d/dx[xy2]= xd/dx +y2(d/dx)[x]
=2xy(dy/dx) + y2
Guidelines for Implicit Differentiation
1. Differentiate both sides of the equation
with respect to x.
2. Collect all terms involving dy/dx on the
left side of the equation and move all
other terms to the right side of the
equation.
3. Factor dy/dx out of the left side of the
equation.
4. Solve for dy/dx
Example
Find dy/dx given that y3 +y2 -5y –x2 = - 4
Example – Finding the Slope of a Graph Implicitly
Find the slope of the tangent line to the
graph of 3(x2 + y2)2 = 100xy at the
point (3,1)
Find dy/dx and then use (3,1) to find the
value of the slope
Other Applications of Implicit Differentiation
1. Finding a Differentiable Function
2. Finding the Second Derivative Implicitly
3. Finding a Tangent Line to a Graph
Related Rates
SECTION 4.6
Guidelines for Solving Related-Rate Problems
1. Identify all given quantities and quantities to be
determined. Make a sketch and label the
quantities.
2. Write an equation involving the variables whose
rates of change either are given or are to be
determined.
3. Using the Chain Rule, implicitly differentiate both
sides of the equation with respect to time t.
4. Substitute into the resulting equation all known
values for the variables and their rates of
change. Then solve for the required rate of
change.
Example
An airplane flying at an altitude of 6 miles is on a
flight path that will take it directly over a radar
tracking station. If distance s is decreasing at
a rate of 400 miles per hour when s = 10
miles, what is the speed of the plane?
Draw and label a right triangle.
s is the hypotenuse of the right triangle and x is
the horizontal distance from the station.
Example – Cont’d
s = 10
x = (102 – 62)½ = 8
ds/dt = -400 when s =10
Find dx/dt when x=10 and x = 8
The function is x2 + 62 =s2
1. Differentiate with respect to x
2. Solve for dx/dt
3. Substitute for s,x and ds/dt
END
THE END`