34.1 Differential Calculus Concept Overview

DIFFERENTIAL CALCULUS | CONCEPT OVERVIEW The topic of DIFFERENTIAL CALCULUS can be referenced on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

CONCEPT INTRO: DIFFERENTIAL CALCULUS is the mathematics of change, analyzing how a function or application changes over time or other defined independent variable. A DERIVATIVE is the instantaneous RATE OF CHANGE of a given function. A variable 𝑦 is said to be a function of another variable 𝑥 if, when 𝑥 is given, 𝑦 is determined. DIFFERENTIATION is the process of finding the derivative of a function. The derivative of a function is based on the idea of calculating the slope of a function, where we calculate the rate of change of one variable with respect to another variable. We know from our studies in algebra, that the slope of a line is defined as RISE over RUN, or rather, the change in the y-coordinates over the change in the x-coordinates, as shown in the formula:

𝑆𝑙𝑜𝑝𝑒 =



𝑅𝑖𝑠𝑒 𝑦! − 𝑦! = 𝑅𝑢𝑛 𝑥! − 𝑥! Made with

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Using this relationship, we can express the RATE OF CHANGE of a function as the DIFFERENCE QUOTIENT. The DIFFERENCE QUOTIENT of a function represents the limit of a function as the independent and dependent variables change throughout the function. Illustrating this further, let: 𝑦=𝑓 𝑥 If 𝛥𝑥 is any incremental change (increase or decrease) of 𝑥, and 𝛥𝑦 is the corresponding incremental change of 𝑦, then the derivative of 𝑦 with respect to 𝑥 is the limit of the ratio of 𝛥𝑦 to 𝛥𝑥 as 𝛥𝑥 approaches zero. The DIFFERENCE QUOTIENT for any function 𝑦 = 𝑓 𝑥 can be expressed generically as: ∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆!→! ∆𝑥 ∆!→! ∆𝑥

𝑦 ! = lim

The DERIVATIVE of a function may be expressed in many different a ways, all of which are equivalent:

𝐷! 𝑦 =

𝑑𝑦 𝑑 = 𝑦! = 𝑓! 𝑥 = 𝑓 𝑥 𝑑𝑥 𝑑𝑥

When evaluating derivatives, there are general RULES we must follow to correctly carry out the DIFFERENTIATION. Made with by Prepineer | Prepineer.com

Let’s work through some of the most common.

CONSTANT RULE: The GENERAL GUIDELINES for the CONSTANT RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. A function that represents a horizontal line is considered to be a CONSTANT FUNCTION, as there is no change in the rise or the run of the function. A constant function does not have a variable in the expression, and is represented by a single numerical value. Therefore, if 𝑓(𝑥) is considered a CONSTANT FUNCTION, where 𝑓 𝑥 = 𝑐, then: 𝑑𝑐 = 𝑓! 𝑥 = 0 𝑑𝑥 Below are some examples of constant functions that when differentiated, would result in a value zero: • 𝑓 𝑥 =5 !

• 𝑓 𝑥 =! • 𝑓 𝑥 =π

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POWER RULE: The GENERAL GUIDELINES for the POWER RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To define the derivative of a function that has an exponent, we bring the original power of the exponent down to act as the coefficient of the variable, and then reduce the original power of the exponent by 1. To illustrate this further, if 𝑓(x) is a power function, where 𝑓 𝑥 = 𝑥 ! , and 𝑛 is any real number, then: 𝑓 ! 𝑥 = 𝑛 𝑥 !!! Which can also be written as: 𝑑(𝑢! ) 𝑑𝑢 = 𝑛𝑢!!! 𝑑𝑥 𝑑𝑥 The derivative of the function 𝑓 𝑥 = 𝑥 is 1. While the exponent is not explicitly stated in the original form of the function, we should be quick to realize that we can assume that it is 1. When we different a single variable, such as x, it will result in constant value due to the general fact that any number raised to the exponent of 0 will be equal to 1.

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Below are some examples of functions with exponents along with their corresponding derivatives defined using the power rule. • 𝑓 𝑥 = 𝑥 ! ; 𝑓 ! 𝑥 = 2𝑥 !

!

!

• 𝑓 𝑥 = 𝑥!; 𝑓 ! 𝑥 = ! 𝑥! • 𝑓 𝑥 = 𝑥; 𝑓 ! 𝑥 = 1 It’s common practice, and most efficient, to rewrite any functions that have radicals in an equivalent exponential form. Doing so will allow for us to deploy the power rule in evaluating it’s derivative. Some examples of functions in radical form and their corresponding derivatives using the power rule are illustrated as: !

!

!

• 𝑓 𝑥 = 𝑥 = 𝑥 ! ; 𝑓 ! 𝑥 = 𝑥 !! !

• 𝑓 𝑥 =



!

!

!

!

𝑥 = 𝑥 ! ; 𝑓 ! 𝑥 = ! 𝑥 !!

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DIFFERENTIAL CALCULUS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.

The first derivative of the defined function is most close to:

𝑓 𝑥 =−

8 𝑥! !

A. 𝑓 ! 𝑥 = ! !! !"

B. 𝑓 ! 𝑥 = ! !! C. 𝑓 ! 𝑥 =

!" !! !

D. 𝑓 ! 𝑥 = !

SOLUTION: The GENERAL GUIDELINES for the POWER RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Our problem presents the formula:

𝑓 𝑥 =−

8 𝑥! Made with

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The way this function is originally written poses as it is more difficult than it actually is. Once you rewrite the function in a more familiar linear form, we can use the power rule to define the derivative. We can rewrite this function as: 𝑓 𝑥 = −8𝑥 !! From the POWER RULE, we know that if 𝑓(x) is a power function, where 𝑓 𝑥 = 𝑥 ! , and 𝑛 is any real number, then: 𝑓 ! 𝑥 = 𝑛 𝑥 !!! In verbal terms, the POWER RULE states that to define the derivative of a function that has an exponent, we bring the original power of the exponent down to act as the coefficient of the variable, and then reduce the original power of the exponent by 1. Using this as our template, we now differentiate the function using the POWER RULE to find: 𝑓 ! 𝑥 = −2 (−8)𝑥

!!!!

= 16𝑥 !!

Which can be rewritten as:

𝑓′ 𝑥 =

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The correct answer choice is C. 𝒇′ 𝒙 =

𝟏𝟔 𝒙𝟑

CONSTANT MULTIPLE RULE: The GENERAL GUIDELINES for the CONSTANT MULTIPLE RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When a function begins with a coefficient, the coefficient has no effect on the process of differentiation. We simply ignore the coefficient and differentiate according to the appropriate rule. The coefficient stays where it is until the final step when we will simplify the expression by multiplying the derivative with the coefficient in place. If 𝑐𝑓(𝑥) is a DIFFERENTIABLE FUNCTION and “𝑐” is constant, then: 𝑐 𝑓′(𝑥) = 𝑐 𝑓′(𝑥) Otherwise stated as: 𝑑(𝑐𝑢) 𝑑𝑢 =𝑐 𝑑𝑥 𝑑𝑥



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Some example of functions and their corresponding derivatives using the CONSTANT MULTIPLE RULE are: !

!

• 𝑓 𝑥 = ! 𝑥!; 𝑓 ! 𝑥 = ! 𝑥! • 𝑓 𝑥 = 4𝑥 ! ; 𝑓 ! 𝑥 = 12𝑥 ! • 𝑓 𝑥 = 5𝑥; 𝑓 ! 𝑥 = 5

SUM AND DIFFERENCE RULES: The GENERAL GUIDELINES for the SUM AND DIFFERENCE RULES can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When we evaluate functions that have multiple terms, we need to evaluate the derivative of each term individually. The first step in evaluating a function with multiple terms, is to rewrite the function with each term expanded out as much as possible, so that you can identify which rule you will need to use to evaluate the derivative on a term by term basis. If you have a DIFFERENCE or SUM, we use the same process of expanding out all of the terms, but must remember to use caution to make sure each term has the corresponding “ + ” or “ − “ sign applied properly.



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For example, if 𝑢, 𝑣, 𝑎𝑛𝑑 𝑤 are differentiable functions, then: 𝑑(𝑢 + 𝑣 − 𝑤) 𝑑𝑢 𝑑𝑣 𝑑𝑤 = + − 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 Some example of functions and their corresponding derivatives using the SUM AND DIFFERENCE RULES are: • 𝑓 𝑥 = 𝑥 ! + 𝑥 ! − 𝑥 − 10; 𝑓 ! 𝑥 = 6𝑥 ! + 3𝑥 ! − 1 !

!

• 𝑓 𝑥 = 𝑥 + 3𝑥 ! − 𝜋; 𝑓 ! 𝑥 = ! 𝑥 !! + 6𝑥 PRODUCT RULE: The GENERAL GUIDELINES for the PRODUCT RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If “𝑢” and “𝑣” are differentiable functions, then as a product: 𝑑(𝑢𝑣) 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥 A common phrase that we are taught when learning this rule during our time in University, and most definitely one to remember is this: First times the derivative of the second, plus the second times the derivative of the first.



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That is our go-to when we are looking to quickly replicate the form of the PRODUCT RULE.



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DIFFERENTIAL CALCULUS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.

Determine the first derivative of the following function: 𝑓 𝑥 = 𝑥 ! sin 𝑥 A. 3𝑥 ! sin 𝑥 + 𝑥 ! cos 𝑥 B. 6𝑥 ! sin 𝑥 + 𝑥 ! cos 𝑥 C. 3𝑥 ! sin 𝑥 − 𝑥 ! cos 𝑥 D. 𝑥 ! sin 𝑥 + 𝑥 ! cos 𝑥

SOLUTION: The GENERAL GUIDELINES for the PRODUCT RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If “𝑢” and “𝑣” are differentiable functions, then: 𝑑(𝑢𝑣) 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥



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The problem gives us the function: 𝑓 𝑥 = 𝑥 ! sin 𝑥 Where we can define: 𝑢 = 𝑥! 𝑣 = sin 𝑥 Remember our MANTRA for the PRODUCT RULE: First times the derivative of the second, plus the second times the derivative of the first. Although this function comes across difficult at first site, when we break it down according to this MANTRA, everything will fall in to place. Using the PRODUCT RULE, we can rewrite the derivative of the function as: 𝑓 ! 𝑥 = 𝑥 ! ! sin 𝑥 + 𝑥 ! (sin 𝑥)′ Evaluating the derivative of each term individually, we find the derivation of the function is expressed as: 𝑓 ! 𝑥 = 3𝑥 ! sin 𝑥 + 𝑥 ! cos 𝑥 The correct answer choice is A. 𝟑𝒙𝟐 𝒔𝒊𝒏 𝒙 + 𝒙𝟑 𝒄𝒐𝒔 𝒙



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QUOTIENT RULE: The GENERAL GUIDELINES for the QUOTIENT RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The QUOTIENT RULE is used to evaluate the derivative of a function that is presented in the form of a fraction with a defined NUMERATOR and DENOMINATOR. The QUOTIENT RULE tells us that if “𝑢” and “𝑣” are differentiable functions, then:

𝑑

𝑢 𝑑𝑢 𝑑𝑣 𝑣 − 𝑢 𝑣 = 𝑑𝑥 𝑑𝑥 ! 𝑑𝑥 𝑣

CHAIN RULE: The GENERAL GUIDELINES for the CHAIN RULE can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If a composite function is made up of two unique functions, where 𝑔(𝑥) is substituted for “𝑥” into a function 𝑓(𝑥) such as: 𝑓 ∙ 𝑔 = 𝑓(𝑔 𝑥 )



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The CHAIN RULE states that if both 𝑓(𝑥) and 𝑔(𝑥) are differentiable functions and 𝐹 𝑥 = 𝑓(𝑔 𝑥 ), then “𝐹” is differentiable with the derivative defined as: 𝐹 ! 𝑥 = 𝑓 ! 𝑔 𝑥 𝑔′(𝑥) It is sometimes easier to think of the functions 𝑓(𝑥) and 𝑔(𝑥) as “layers” of the function with 𝑓 𝑥 being the “outer layer” and 𝑔(𝑥) being the “inner layer”. The chain rule tells us to first differentiate the “outer layer”, leaving the “inner layer” unchanged, then to add on the derivative of the inner layer. If the derivative of a function (𝑓 ! 𝑥) is differentiable, the derivative of this differentiated function can be taken as well. The new function, 𝑓"(𝑥) is referred to as the SECOND DERIVATIVE of 𝑓 𝑥 . This process of differentiation can continue as long as the function can be differentiated. In general: 𝑓 ! (𝑥) is called the nth derivative of f



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TANGENT LINES: The GENERAL GUIDELINES for the SLOPE of a TANGENT LINE can be referenced under the topic of DIFFERENTIAL CALCULUS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. When we want to find the slope of a curve at a point, we are solving for the INSTANTANEOUS RATE OF CHANGE in a function for a specific value of the independent variable. The derivative of a function 𝑓(𝑥) at some number 𝑥 = 𝑐, written as 𝑓′(𝑐), is the slope of the tangent line to 𝑓 drawn at 𝑐. The instantaneous rate of change is the slope of a TANGENT LINE at that point on the curve. Illustrating this, we have:



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In the standard Cartesian Coordinate System, a tangent line intersects a curve at a point, and has the same slope as the curve at that point. The slope of a curve at point (𝑥! , 𝑦! ) can be approximated by constructing a line through that point and a nearby point (𝑥, 𝑦), and then finding the slope of the line, such that: y ! = the slope of the curve f(x) When defining the equation of a TANGENT LINE, the slope of the curve 𝑓(𝑥) at 𝑎, 𝑓 𝑎

is equal to the derivative of “𝑓” at “𝑎”.

This slope can generally be stated as: ∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆!→! ∆𝑥 ∆!→! ∆𝑥

𝑚 = lim



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DIFFERENTIAL CALCULUS | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material.

Define the equation of the tangent line to the function given below at point 1,2 : 𝑓 𝑥 = 3𝑥 ! − 𝑥 ! A. 3𝑥 − 1 B. 6𝑥 ! − 3𝑥 + 1 C. 6𝑥 ! + 3𝑥 ! − 1 D. 6𝑥 ! + 3𝑥 + 1

SOLUTION: The GENERAL GUIDELINES for the SLOPE of a TANGENT LINE can be referenced under the topic of DIFFERENTIAL CALCULUS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The slope of a curve at point (𝑥! , 𝑦! ) can be approximated by constructing a line through that point and a nearby point (𝑥, 𝑦), and then finding the slope of the line, such that: y ! = the slope of the curve f(x)

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When defining the equation of a TANGENT LINE, the slope of the curve 𝑓(𝑥) at 𝑎, 𝑓 𝑎

is equal to the derivative of “𝑓” at “𝑎”.

This slope can generally be stated as: ∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆!→! ∆𝑥 ∆!→! ∆𝑥

𝑚 = lim

We are given the function: 𝑓 𝑥 = 3𝑥 ! − 𝑥 ! Our first move is to differentiate leaning back on RULES that we had learned previously, specifically in this case, the POWER RULE and the SUM AND DIFFERENCE RULES. The GENERAL GUIDELINES for the POWER RULE and the SUM AND DIFFERENCE RULES can be referenced under the topic of DERIVATIVES AND INDEFINITE INTEGRALS on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The POWER RULE generally states that to define the derivative of a function that has an exponent, we bring the original power of the exponent down to act as the coefficient of the variable, and then reduce the original power of the exponent by 1.



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To illustrate this further, if 𝑓(x) is a power function, where 𝑓 𝑥 = 𝑥 ! , and 𝑛 is any real number, then: 𝑓 ! 𝑥 = 𝑛 𝑥 !!! When we evaluate functions that have multiple terms, we need to evaluate the derivative of each term individually. The first step in evaluating a function with multiple terms, is to rewrite the function with each term expanded out as much as possible, so that you can identify which rule you will need to use to evaluate the derivative on a term by term basis. If you have a DIFFERENCE or SUM, we use the same process of expanding out all of the terms, but must remember to use caution to make sure each term has the corresponding “ + ” or “ − “ sign applied properly. For example, if 𝑢, 𝑣, 𝑎𝑛𝑑 𝑤 are differentiable functions, then: 𝑑(𝑢 + 𝑣 − 𝑤) 𝑑𝑢 𝑑𝑣 𝑑𝑤 = + − 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 Again, in our problem, we are given the function: 𝑓 𝑥 = 3𝑥 ! − 𝑥 !



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Which can be broken down as two functions: 𝑢 = 3𝑥 ! 𝑣 = −𝑥 ! Using the POWER RULE, we can determine the derivative of each of these functions, such that: 𝑢′ = 6𝑥 𝑣′ = −3𝑥 ! The slope of the TANGENT LINE at any given point can be determined using the differentiated function: 𝑚!"#$%#! = 𝑦 ! = 𝑢! + 𝑣 ! Or: 𝑚!"#$%#! = 𝑦 ! = 6𝑥 − 3𝑥 ! We are given the point: 1,2



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Plugging in the value of 𝑥 = 1, we are able to solve for the slope of the TANGENT LINE, such that: 𝑚!"#$%#! = 𝑦 ! = 6 1 − 3 1

!

=3

We can now substitute the calculated slope and given coordinates of our point into the standard equation of a line, which is given as: 𝑦 = 3𝑥 + 𝑏 Plugging in the coordinates of our point 1,2 , we get: 2 = 3(1) + 𝑏 Rearranging and solving for “𝑏” we have: 𝑏 = −1 Therefore, the equation of the TANGENT LINE at point (1,2) is: 𝑦 = 3𝑥 − 1 The correct answer choice is A. 𝟑𝒙 − 𝟏



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