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Calculus - Santowski
04/20/23Calculus - Santowski1
B.1.1b – Derivatives As Functions
Lesson Objectives
1. Calculate the derivative functions of simple polynomial, root and rational functions from first principles
2. Calculate the derivative of simple polynomial and rational functions using the TI-89
3. Explore the relationship between differentiability and continuity
4. Calculate derivatives and apply to real world scenarios
Fast Five – Table Quiz
04/20/23Calculus - Santowski3
Work through the following questions and make relevant conclusions based upon the results of your investigation:
Determine the value of the derivative for the following:
€
′ f 1( ) if f x( ) =x 2 if x ≤1
1
2x +
1
2if x >1
⎧ ⎨ ⎪
⎩ ⎪
€
dy
dx at x = 0 if y = x3
€
dy
dx x= 0
if y = x
(A) The Derivative as a Function
04/20/23Calculus - Santowski4
Our change now will be to alter our point of view and let the value of a or x (the point at which we are finding the derivative value) vary (in other words, it will be a variable)
Consequently, we develop a new function - which we will now call the derived function (AKA the derivative)
We will do this as an investigation using two different methods: a graphic/numeric approach and a more algebraic approach
(A) The Derivative as a Function
04/20/23Calculus - Santowski5
We will work with first principles - the tangent concept – and draw tangents to given functions at various points, tabulate results, create scatter-plots and do a regression analysis to determine the equation of the curve of best fit.
Ex: f(x) = x² - 4x - 8 for the interval [-3,8]
(A) The Derivative as a Function
04/20/23Calculus - Santowski6
Example: y = x² - 4x - 8. for the interval [-3,8]
1. Draw graph.2. Find the tangent slope at x = -3 using the TI-
893. Repeat for x = -2,-1,….,7,8 and tabulate
X -3 -2 -1 0 1 2 3 4 5 6 7 8Slope -10 -8 -6 -4 -2 0 2 4 6 8 10 12
4. Tabulate data and create scatter-plot5. Find best regression equation
(A) The Derivative as a Function
04/20/23Calculus - Santowski7
(A) The Derivative as a Function
04/20/23Calculus - Santowski8
(A) The Derivative as a Function
04/20/23Calculus - Santowski9
(A) The Derivative as a Function
04/20/23Calculus - Santowski10
Our function equation was f(x) = x² - 4x - 8
Equation generated is f `(x) = 2x - 4
The interpretation of the derived equation is that this "formula" (or equation) will give you the slope of the tangent (or instantaneous rate of change) at every single point x.
The equation f `(x) = 2x - 4 is called the derived function, or the derivative of f(x) = x² - 4x - 8
(B) The Derivative as a Function - Algebraic
04/20/23Calculus - Santowski11
Given f(x) = x2 – 4x – 8, we will find the derivative at x = a using our “derivative formula” of
Our one change will be to keep the variable x in the “derivative formula”, since we do not wish to substitute in a specific value like a€
′ f (a) = limh →0
f (a + h) − f (a)
h
(B) The Derivative as a Function - Algebraic
04/20/23Calculus - Santowski12
€
′ f (a) = limh →0
f (a + h) − f (a)
h
′ f (a) = limh →0
a + h( )2
− 4 a + h( ) − 8( ) − a2 − 4a − 8( )
h
′ f (a) = limh →0
a2 + 2ah + h2 − 4a − 4h − 8( ) − a2 − 4a − 8( )
h
′ f (a) = limh →0
2ah + h2 − 4h
h′ f (a) = lim
h →02a + h − 4( )
′ f (a) = 2a − 4
(B) The Derivative as a Function – TI-89
(C) In-Class Examples
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Use the algebraic method to determine the equations of the derivative functions of the following and then state the domain of both the function and the derivative function
(i) f(x) = (x+5)(ii) g(x) = (x + 1)/(3x – 2)
Confirm your derivative equations using the TI-89
(D) Discontinuity and Differentiability
Question? If a function has a discontinuity at a given point on its domain, is this function therefore differentiable at this point on its domain? Justify your answer.
Use the following fcns to investigate:
(D) Discontinuity and Differentiability
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Work through the following fcns and make relevant conclusions based upon the results of your investigation:
Determine the value of the derivative for the following:
€
′ f 1( ) if f x( ) =x 2 if x ≤1
x + 2 if x >1
⎧ ⎨ ⎩
€
dy
dx at x = 0 if y =
1
x
€
dy
dx x= 2
if y =x 2 − 4
x + 2
(D) Discontinuity and Differentiability
04/20/23Calculus - Santowski17
Recall the fundamental idea that a derivative at a point is really the idea of a limiting sequence of secant slopes (or tangent line) drawn to a curve at a given point
Now , if a function is discontinuous at a given point, try drawing secant lines from the left and secant lines from the right and then try drawing a specific tangent slope at the point of discontinuity in the following diagrams
Conclusion you can only take a derivative at any given point (differentiate a function) if it is continuous at that/every point
(E) Continuity and Differentiability
Question? If a function is continuous on its domain, is this function therefore differentiable at every point on its domain? Justify your answer.
(E) Continuity and Differentiability
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Work through the following questions and make relevant conclusions based upon the results of your investigation:
Determine the value of the derivative for the following:
€
′ f 1( ) if f x( ) =x + 3 if x <1
1− x 2 if x ≥1
⎧ ⎨ ⎩
€
dy
dx at x = 0 if y =
1
x
€
dy
dx x= 2
if y =x 2 − x − 2
x − 2
€
′ f 0( ) if f x( ) =1
x 2if x ≠ 0
1 if x = 0
⎧ ⎨ ⎪
⎩ ⎪
(E) Continuity and Differentiability
04/20/23 Calculus - Santowski 20
(E) Continuity and Differentiability
04/20/23Calculus - Santowski21
One other point to add that comes from our study of the last two examples even if a function is continuous, this does not always guarantee differentiability!!!!
If a continuous function as a cusp or a corner in it, then the function is not differentiable at that point see graphs on the preceding slide and decide how you would draw tangent lines (and secant lines for that matter) to the functions at the point of interest (consider drawing tangents/secants from the left side and from the right side)
As well, included on the graphs are the graphs of the derivatives (so you can make sense of the tangent/secant lines you visualized)
(E) Continuity and Differentiability
Continuous functions are non-differentiable under the following conditions: The fcn has a “corner” (ex 1) The fcn has a “cusp” (ex 2) The fcn has a vertical tangent (ex 3)
This non-differentiability can be seen in that the graph of the derivative has a discontinuity in it!
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(F) Continuity and Differentiability
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Follow this link to One-sided derivatives from IES Software
And then follow this link to Investigating Differentiability of Piecewise Functions from D. Hill (Temple U.) and L. Roberts (Georgia College and State University
(G) Internet Links
04/20/23Calculus - Santowski24
From Paul Dawkins - Calculus I (Math 2413) - Derivatives - The Definition of the Derivative
From UTK - Visual Calculus - Definition of Derivative
Tutorial for Derivatives From Stefan Waner @ Hofstra U
(H) Homework
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Page 191-194
(1) C LEVEL: Q2,3, 11-16 (algebraic work - verify using GDC)
(2) C LEVEL: Q31-34, (graphs)(3) B LEVEL: Q38-44(4) A LEVEL: WORKSHEET (p77): Q15,16(5) A LEVEL: WORKSHEET (p52):
Q1,2,4,5,6