Lesson 6:
The definition of derivative and shortcuts to the same
Announcements
Homework 6
Quizzes
Popper 04 today
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Now we’ll start with the calculus part of the course. We will learn all about
derivatives – finding all about the tangent line to a point on a curve.
We want to find first the slope of the tangent line to a point at a curve. And we’ll
use limits to do this. Note that the actual slope formula requires 2 points and we’re
talking about ONE point. Clearly we need to bridge this gap.
Let’s look at how:
First let’s take a curve and run a secant line across it. A secant line crosses at two
points.
Let’s pick our two leftmost intersections and label them (x, f (x)) and
(x + h, f (x + h)). We want the slope of the tangent line at the leftmost intersection.
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Now, using those values, what is the slope of the secant line?
DQ 
f ( x  h)  f ( x ) f ( x  h )  f ( x )

( x  h)  x
h
This is also called the Difference Quotient.
(subtract and divide!)
This is also called the Average Rate of Change
from x to x + h
(AROC)
Now we want to reduce h to zero! And some things should immediately pop into
mind…if we reduce h to zero we will be dividing the DQ by zero won’t we?
We’ll get around that by using a LIMIT where we get close to but not necessarily
equal to our target number 0
lim DQ  lim
h 0
h 0
f ( x  h)  f ( x )
h
This limit is known as the derivative and denoted with a tick mark superscripted by
the f in f (x)…the derivative is f ‘(x)…and we say “f prime” lots of times because
it’s fewer syllables!
This is also called the Instantaneous Rate of Change
at x.
(IROC)
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Popper 04 Question 1
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Let’s do one all the way out and then see it on a graph:
f ( x)  x 2  x
f '( x)  ________
Let’s write out steps for our process and do it as we go along:
lim DQ  lim
h 0
h 0
f ( x  h)  f ( x )
h
Step 1
Step 2
Step 3
Step 4
Step 5
Find f (x + h)
Subtract the original function from this
Divide by h
Take the limit
Report the answer
f ( x)  x 2  x
Find f (x + h)
f '( x)  ________
Subtract the original function from this
5
Divide by h
Take the limit
Let’s check our work in GGB!
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Now let’s find and graph the tangent line at x = −1. What is the slope of the
tangent line at this point? NOTE the derivative is the slope of the tangent line!
What is this tangent line telling us?
Popper 03 Question 2
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Popper 04 Question 3
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Now let’s review some more vocabulary:
Velocity – velocity is a “rate of change”
Average velocity - use the DQ!
Instantaneous velocity - take the limit of the DQ!
Example:
Suppose the distance covered by a car can be measured by the function
s(t )  4t 2  32t where s is measured in feet and t is in seconds.
A
Find the average rate of velocity of the car over the interval [0, 4]
What is h for this problem?
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B
Find the instantaneous rate of change at t = 1. Use GGB to get the
derivative.
What is this line f’(x)?
What does f’(1) mean? Where is it on the graph?
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Shortcuts to the derivative:
First some new notation:
“take the derivative of f with respect to x and evaluate it at x = a”
d
[ f (a)]
dx
Shortcut #1
Now, the derivative of a constant c is always zero.
d
[c ]
dx
Let’s look at the graph of y = 3 to see why:
So
d
[ 5] 
dx
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Popper 03 Question 4
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Shortcut #2 The derivative of a power function is always one less than the original
function.
d n
[ x ]  nx n 1
dx
d 9
[ x ]  nx n 1  9 x8
dx
d
[ x ]  nx n 1 
dx
If f ( x) 
1
, find f’(4).
x3
What is f’(4)?
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Popper 04 Question 5
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Shortcut #3 The derivative of a constant times a function is that constant times the
derivative of the function.
d
d
[cf ( x)]  c [ f ( x)]
dx
dx
d
[ 3 x 4 ] 
dx
If f ( x) 
7 3 11
x
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Popper 03 Question 6
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Shortcut #4
The derivative of a sum is the sum of the derivatives.
d
d
d
[ f ( x)  g ( x)]  [ f ( x)]  [ g ( x)]
dx
dx
dx
Find the derivative of
f ( x)  3x5 
2
1

 2051
3
x
x
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Popper 03 Question 7
Popper 03 Question 8
Popper 03 Question 9
Popper 03 Question 10
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