Derivatives, Properties, and Formulas
Site:  Saylor Academy 
Course:  MA005: Calculus I 
Book:  Derivatives, Properties, and Formulas 
Printed by:  Guest user 
Date:  Friday, August 9, 2024, 4:14 AM 
Description
Read this section to understand the properties of derivatives. Work through practice problems 111.
This section begins with a look at which functions have derivatives. Then we'll examine how to calculate derivatives of elementary combinations of basic functions. By knowing the derivatives of some basic functions and just a few differentiation patterns, you will be able to calculate the derivatives of a tremendous variety of functions.
This section contains most, but not quite all, of the general differentiation patterns you will ever need.
Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorgresources/wwwresources/site/wpcontent/uploads/2012/12/MA0053.3DerivativesPropertiesandFormulas.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.
Theorem 
If 
a function is differentiable at a point, 

then 
it is continuous at that point. 
The contrapositive form of this theorem tells about some functions which do not have derivatives:
Contrapositive Form of the Theorem: 


If  is not continuous at a point,  

then 
Proof of the Theorem: We assume that the hypothesis, is differentiable at the point , is true so
exists and equals . We want to show that must necessarily be continuous at
Since can be written as , we have
It is important to clearly understand what is meant by this theorem and what is not meant: If the function is differentiable at a point, then the function is automatically continuous at that point. If the function is continuous at a point, then the function may or may not have a derivative at that point.
If the function is not continuous at a point, then the function is not differentiable at that point.
Example 1: Show that is not continuous and not differentiable at 2 (Fig. 1).
Solution: The onesided limits, and , have different values so does not exist, and is not continuous at 2. Since is not continuous at 2, it is not differentiable there.
Lack of continuity is enough to imply lack of differentiability, but the next two examples show that continuity is not enough to guarantee differentiability.
Example 2: Show that is continuous but not differentiable at (Fig. 2)
Solution: so is continuous at , but we showed in Section 2.1 that the absolute value function was not differentiable at .
A function is not differentiable at a cusp or a "corner".
Example 3: Show that is continuous but not differentiable at (Fig. 3)
Solution: so , and is continuous at . which is undefined at so is not differentiable at .
A function is not differentiable where its tangent line is vertical.
Practice 1: At which integer values of is the graph of f in Fig. 4 continuous? differentiable?
Graphically, a function is continuous if and only if its graph is connected and does not have any holes or breaks.
Graphically, a function is differentiable if and only if it is continuous and its graph is smooth with no corners or vertical tangent lines.
Example 4: The derivative of is , and the derivative of is
What are the derivatives of their elementary combinations: and ?
Solution:
Unfortunately, the derivatives of and don't follow the same easy patterns:
These two very simple functions show that, in general, and .
The Main Differentiation Theorem below states the correct patterns for differentiating products and quotients.
Practice 2: For and , what are the derivatives of and ?
The following theorem says that the simple patterns in the example for constant multiples of functions and sums and differences of functions are true for all differentiable functions. It also includes the correct patterns for derivatives of products and quotients of differentiable functions.
Main Differentiation Theorem: If and are differentiable at , then
(a) Constant Multiple Rule: or
The proofs of parts (a), (b), and (c) of this theorem are straightforward, but parts (d) and (e) require some clever algebraic manipulations. Lets look at an example first.
Example 5: Recall that and . Find and .
Solution: is an application of part (a) of the theorem with and so
uses part (c) of the theorem with and so
Practice 4: Fill in the values in the table for , and
0 
3 
–2 
–4 
3 



1 
2 
–1 
1 
0 



2 
4 
2 
3 
1 



Proof of the Main Derivative Theorem (a) and (c): The only general fact we have about derivatives is the definition as a limit, so our proofs here will have to recast derivatives as limits and then use some results about limits. The proofs are applications of the definition of the derivative and results about limits.
The proof of part (b) is very similar to these two proofs, and is left for you as the next Practice Problem.
The proof for the Product Rule and Quotient Rules will be given later.
Practice 5: Prove part (b) of the theorem, the Sum Rule: .
Practice 6: Use the Main Differentiation Theorem and the values in the table to fill in the rest of the table.
0 
3 
–2 
–4 
3 



1 
2 
–1 
1 
0 



2 
4 
2 
3 
1 



Solution: (a) We can use the Product Rule with and :
(b) We can use the Quotient Rule with and :
Proof of the Product Rule: The proofs of parts (d) and (e) of the theorem are complicated but only involve elementary techniques, used in just the right way. Sometimes we will omit such computational proofs, but the Product and Quotient Rules are fundamental techniques you will need hundreds of times.
By the hypothesis, and are differentiable so and .
Also, both and are continuous (why?) so and .
(d) Product Rule: Let . Then .
finding common factors  
taking limit as  
(e) The steps for a proof of the Quotient Rule are shown in Problem 55.
You definitely need to memorize the differentiation rules, but it is vitally important that you also know how to use them. Sometimes it is clear that the function we want to differentiate is a sum or product of two obvious functions, but we commonly need to differentiate functions which involve several operations and functions. Memorizing the differentiation rules is only the first step in learning to use them.
Solution: This function is more difficult because it involves both an addition and a multiplication. Which rule(s) should we use, or, more importantly, which rule should we use first?
applying the Sum Rule and trading one derivative for two easier ones
this expression has no more derivatives so we are done.
If you were evaluating the function for some particular value of , you would (1) raise to the 5th power, (2) calculate , (3) multiply by , and (4) your FINAL evaluation step, SUM the values of and .
The FINAL step of your evaluation of indicates the FIRST rule to use to calculate the derivative of .
Practice 8: Which differentiation rule should you apply FIRST for each of the following:
Example 8: A weight attached to a spring is oscillating up and down.
Over a period of time, the motion becomes "damped" because of friction and air resistance (Fig. 5), and its height at time seconds is feet
What are the height and velocity of the weight after 2 seconds?
Solution: The height is
Practice 10: What are the height and velocity of the weight in the previous example after 5 seconds? What are the height and velocity of the weight be after a "long time"?
Solution: Clearly we need to use the Product Rule since the only operation in this function is multiplication, but the Product Rule deals with a product of two functions and we have the product of three; and and . However, if we think of our two functions as and , then we do have the product of two functions and
We are not done, but we have traded one hard derivative for two easier ones. We know that , and we can use the Product Rule (again) to calculate . Then the last line of our calculation becomes
The derivative of a function is a new function which gives the slope of the line tangent to the graph of at each point . To find the slope of the tangent line at a particular point on the graph of , we should first calculate the derivative and then evaluate the function at the point to get the number . If you mistakenly evaluate first, you get a number , and the derivative of a constant is always equal to 0.
Example 10: Determine the slope of the line tangent to at and :
Solution: When , the graph of goes through the point with slope . When , the graph goes through the point with slope .
Differentiability and Continuity: If a function is differentiable then it must be continuous.
If a function is not continuous then it cannot be differentiable.
A function may be continuous at a point and not differentiable there.
Graphically: CONTINUOUS means connected.
DIFFERENTIABLE means continuous, smooth and not vertical.
Differentiation Patterns:
The FINAL STEP used to evaluate f indicates the FIRST RULE to use to differentiate .
To evaluate a derivative at a point, first differentiate and then evaluate.
Practice 1: is continuous at , and is differentiable at , and
Practice 4:
x 
f(x) 
f '(x) 
g(x) 
g '(x) 
D( 3f(x) ) 
D( 2f(x) + g(x) ) 
D( 3g(x) – f(x) ) 

0 
3 
–2 
–4 
3 
–6 
–1 
11 
1 
2 
–1 
1 
0 
–3 
–2 
1 
2 
4 
2 
3 
1 
6 
5 
1 
Practice 5:
Practice 6:
Practice 7:
Practice 8: (a) difference rule (b) product rule (c) difference rule (d) quotient rule
Practice 9:
Practice 10:
Practice 11: