17. 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 changes, and 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. The derivative of a function is based on the idea of calculating slope of a function, where calculate the rate of change of one variable with respect to another variable. 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. We know that from our studies of algebra, that the slope of a line is defined as the change in rise over the run, as shown in the formula:

𝑆𝑙𝑜𝑝𝑒 =



𝑅𝑖𝑠𝑒 𝑦. − 𝑦0 = 𝑅𝑢𝑛 𝑥. − 𝑥0

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Using this relationship, we can express the rate of change of the function as the DIFFERENCE QUOTIENT. The difference quotient of a function represents the limit of the function as the independent and dependent variables change throughout the function. Let 𝑦 = 𝑓 𝑥 . If 𝛥𝑥 is any increment (increase or decrease) given to 𝑥, and 𝛥𝑦 is the corresponding increment in 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 is represented by the expression: ∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆:→< ∆𝑥 ∆:→< ∆𝑥

𝑦 5 = lim

The derivative of a function is expressed as:

𝐷: 𝑦 =

𝑑𝑦 = 𝑦′ 𝑑𝑥

There are a few ways to express the derivative of a function, all of which are equivalent:

𝑓5 𝑥 = 𝑦5 =

𝑑𝑦 𝑑 = 𝑓 𝑥 𝑑𝑥 𝑑𝑥

When evaluating derivatives, there are general formulas we must follow to correctly differentiate a function:



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CONSTANT RULE: The FORMULA FOR THE CONSTANT RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on 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 run of the function. A constant function does not have a variable in the expression, and is represented by a single numerical value. If “𝑓(𝑥)” is a CONSTANT FUNCTION, where 𝑓 𝑥 = 𝑐, then: 𝑑𝑐 = 𝑓5 𝑥 = 0 𝑑𝑥 Below are some examples of constant functions that would have a derivative equal to zero: • 𝑓 𝑥 =5 • 𝑓 𝑥 =

0 F

• 𝑓 𝑥 =π



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POWER RULE: The FORMULA FOR THE POWER RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. To find the derivative of a function that has an exponent, we bring the power of the exponent down from the exponent to act as the coefficient of the variable, and then reduce the power of the exponent by 1. If “𝑓” is a power function, where 𝑓 𝑥 = 𝑥 H and 𝑛 is any real number, then: 𝑓 5 𝑥 = 𝑛 𝑥 HJ0 𝑑(𝑢H ) 𝑑𝑢 = 𝑛𝑢HJ0 𝑑𝑥 𝑑𝑥 The derivative of the function 𝑓 𝑥 = 𝑥 is 1. While the exponent is not explicitly stated, we should realize that the exponent is assumed to be 1, and any number raise to the exponent of 0 will be equal to 1. Below are some examples of functions with exponent and their corresponding derivatives calculated using the power rule: • 𝑓 𝑥 = 𝑥 . ; 𝑓 5 𝑥 = 2𝑥 M N

F

5

• 𝑓 𝑥 = 𝑥 ; 𝑓 𝑥 = 𝑥 O

P N

• 𝑓 𝑥 = 𝑥; 𝑓 5 𝑥 = 1



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For any functions that radicals, we can re-write the radicals in exponential form, so we can easily use the power rule to evaluate the derivative of the function. Some example of functions in radical form and their corresponding derivatives using the power rule: R

0

R

• 𝑓 𝑥 = 𝑥 = 𝑥 P ; 𝑓 5 𝑥 = 𝑥 JP .

• 𝑓 𝑥 =

N

R N

5

0

𝑥 = 𝑥 ; 𝑓 𝑥 = 𝑥 O

J

P N

CONCEPT EXAMPLE: Determine the first derivative of the following function:

𝑓 𝑥 =−

A. 𝑓 5 𝑥 = B. 𝑓 5 𝑥 = C. 𝑓 5 𝑥 = D. 𝑓 5 𝑥 =



8 𝑥. . : TN 0 : TU 0 : TN 0 : TM

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SOLUTION: The FORMULA FOR THE POWER RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If “𝑓” is a power function, where 𝑓 𝑥 = 𝑥 H and 𝑛 is any real number, then: 𝑓 5 𝑥 = 𝑛 𝑥 HJ0 𝑑(𝑢H ) 𝑑𝑢 = 𝑛𝑢HJ0 𝑑𝑥 𝑑𝑥 This function appears more difficult than it actually is. Once you rewrite the function in a more familiar linear form, we can use the power rule to figure out the derivative. We are given the function:

𝑓 𝑥 =−

8 𝑥.

We can then rewrite this function as: 𝑓 𝑥 = −8𝑥 J.



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Recall that the power rule states that if 𝑓(𝑥) is a power function, where 𝑓 𝑥 = 𝑥 H and 𝑛 is real number, then: 𝑓 5 𝑥 = 𝑛 𝑥 HJ0 We now differentiate the function using the power rule to find: 𝑓 5 𝑥 = −2 − 8𝑥

J.J0

= 16𝑥 JO

This derivative can then be rewritten to match the form of the answer choices such that:

𝑓′ 𝑥 =

1 𝑥 JO

Therefore, the correct answer choice is C. 𝒇′ 𝒙 =



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CONSTANT MULTIPLE RULE: The FORMULA FOR THE CONSTANT MULTIPLE RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on 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. If “𝑓” is a DIFFERENTIABLE FUNCTION and “𝑐” is constant, then: 𝑐 𝑓′(𝑥) = 𝑐 𝑓′(𝑥) 𝑑(𝑐𝑢) 𝑑𝑢 =𝑐 𝑑𝑥 𝑑𝑥 Some example of functions and their corresponding derivatives using the constant multiple rule are: 0

F

O

O

• 𝑓 𝑥 = 𝑥F; 𝑓 5 𝑥 = 𝑥[ • 𝑓 𝑥 = 4𝑥 O ; 𝑓 5 𝑥 = 4𝑥 . • 𝑓 𝑥 = 5𝑥; 𝑓 5 𝑥 = 5 SUM AND DIFFERENCE RULES: The FORMULA FOR THE SUM AND DIFFERENCE RULES can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing.

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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, so that you can identify which rule you will need to use to evaluate the derivative of that particular term. 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. 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 a are: • 𝑓 𝑥 = 𝑥 b + 𝑥 O − 𝑥 − 10; 𝑓 5 𝑥 = 6𝑥 F + 3𝑥 . − 1 0

R

• 𝑓 𝑥 = 𝑥 + 3𝑥 . − 𝜋; 𝑓 5 𝑥 = 𝑥 JP + 6𝑥 .



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PRODUCT RULE: The FORMULA FOR THE PRODUCT RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on 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 A phrase to remember the product rule is: 𝐹𝑖𝑟𝑠𝑡 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑, 𝑝𝑙𝑢𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡

CONCEPT EXAMPLE: Determine the first derivative of the following function: 𝑓 𝑥 = 𝑥 O sin 𝑥 A. 𝑓 5 𝑥 = 3𝑥 . sin 𝑥 + 𝑥 O cos 𝑥 B. 𝑓 5 𝑥 = 6𝑥 . sin 𝑥 + 𝑥 O cos 𝑥 C. 𝑓 5 𝑥 = 3𝑥 . sin 𝑥 − 𝑥 O cos 𝑥 D. 𝑓 5 𝑥 = 𝑥 . sin 𝑥 + 𝑥 O cos 𝑥



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SOLUTION: The FORMULA FOR THE PRODUCT RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on 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

This function appears more difficult than it actually is. Once you rewrite the function in a more familiar linear form, we can use the product rule to figure out the derivative. We are given the function: 𝑓 𝑥 = 𝑥 O sin 𝑥 We see that there are two functions that we need to evaluate, requiring us to use the product rule where 𝑢 = 𝑥 O and 𝑣 = sin 𝑥. If “𝑢” and “𝑣” are differentiable functions, then: 𝑑(𝑢𝑣) 𝑑𝑣 𝑑𝑢 =𝑢 +𝑣 𝑑𝑥 𝑑𝑥 𝑑𝑥



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Using the product rule, we can rewrite the derivative of the function as: 𝑓 5 𝑥 = 𝑥 O 5 sin 𝑥 + 𝑥 O (sin 𝑥)′ Evaluating the derivative of each term, we find the derivation of the function is expressed as: 𝑓 5 𝑥 = 3𝑥 . sin 𝑥 + 𝑥 O cos 𝑥

Therefore, the correct answer choice is A. 𝒇5 𝒙 = 𝟑𝒙𝟐 𝒔𝒊𝒏 𝒙 + 𝒙𝟑 𝒄𝒐𝒔 𝒙



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QUOTIENT RULE: The FORMULA FOR THE QUOTIENT RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on 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. If “𝑢” and “𝑣” are differentiable functions, then:

𝑑

𝑢 𝑑𝑢 𝑑𝑣 𝑣 − 𝑢 𝑣 = 𝑑𝑥 𝑑𝑥 . 𝑑𝑥 𝑣

CHAIN RULE: The FORMULA FOR THE CHAIN RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If a composite function made up of two other functions, where 𝑔(𝑥) is substituted for “𝑥” into a function 𝑓(𝑥) such as: 𝑓 ∙ 𝑔 = 𝑓(𝑔 𝑥 ) The chain rules states if “𝑓” and ”𝑔” are differentiable functions and 𝐹 𝑥 = 𝑓(𝑔 𝑥 ), then “𝐹” is differentiable and the derivative of “𝐹” is given by: 𝐹 5 𝑥 = 𝑓 5 𝑔 𝑥 𝑔′(𝑥)

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It it is sometimes easier to think of the functions 𝑓(𝑥) and 𝑔(𝑥) as “layers” of a problem. Function 𝑓 𝑥 is the “outer layer” and function 𝑔(𝑥) is the “inner layer.” Thus, the chain rule tells us to first differentiate the outer layer, leaving the inner layer unchanged (the term 𝑓′(𝑔 𝑥 ), then differentiate the inner layer (the term 𝑔′(𝑥) ) . If the derivative (𝑓 5 𝑥) is differentiable, the derivative of it can be taken as well. The new function, 𝑓"(𝑥)is called the SECOND DERIVATIVE of 𝑓 𝑥 . This process can continue as long as the function is differentiable. In general, 𝑓 H (𝑥) is called the 𝑛𝑡ℎ 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓. TANGENT LINES: The FORMULA FOR THE SLOPE OF A TANGENT LINE can be referenced under the reference topic of DIFFERENTIAL CALCULUS on 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.



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In the 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 (𝑥w , 𝑦w ) can be approximated by constructing a line through that point and a nearby point (𝑥, 𝑦), and then finding the slope of the line. 𝑦 5 = 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒 𝑓(𝑥) When finding equations of tangent lines, the slope of the curve 𝑓(𝑥) at 𝑎, 𝑓 𝑎

is equal

to the derivative of “𝑓” at “𝑎” When finding rates of change, the instantaneous rate of change of a function 𝑓(𝑥) when 𝑥 = 𝑎 is equal to the derivative of “𝑓” at “𝑎” given by:

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∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆:→< ∆𝑥 ∆:→< ∆𝑥

𝑚 = lim

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

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

SOLUTION: The FORMULA FOR THE SLOPE OF A TANGENT LINE can be referenced under the reference topic of DIFFERENTIAL CALCULUS on on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. The slope of a curve at point (𝑥w , 𝑦w ) can be approximated by constructing a line through that point and a nearby point (𝑥, 𝑦), and then finding the slope of the line. 𝑦 5 = 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒 𝑓(𝑥)

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When finding equations of tangent lines, the slope of the curve 𝑓(𝑥) at 𝑎, 𝑓 𝑎

is equal

to the derivative of “𝑓” at “𝑎” When finding rates of change, the instantaneous rate of change of a function 𝑓(𝑥) when 𝑥 = 𝑎 is equal to the derivative of “𝑓” at “𝑎” given by: ∆𝑦 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥) = lim ∆:→< ∆𝑥 ∆:→< ∆𝑥

𝑚 = lim

We differentiate the provided function: 𝑓 𝑥 = 3𝑥 . − 𝑥 O The FORMULA FOR THE POWER RULE can be referenced under the reference topic of DERIVATIVES AND INDEFINITE INTEGRALS on on page 29 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. If 𝑓 is a power function, where 𝑓 𝑥 = 𝑥 H and “𝑛” is a real number, then: 𝑓 5 𝑥 = 𝑛 𝑥 HJ0 𝑑(𝑢H ) 𝑑𝑢 = 𝑛𝑢HJ0 𝑑𝑥 𝑑𝑥 Using the POWER RULE, we find the slope of the tangent line’s slope at any given point is given as: 𝑚xyHz{Hx = 𝑦 5 = 6𝑥 − 3𝑥 .



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Plugging in the value of 𝑥 = 1, we are able to solve for the slope of the tangent line as: 𝑚xyHz{Hx = 𝑦 5 = 6(1) − 3(1). =3 We can now substitute the calculate slope and given values for the point into the standard equation of a line to calculate the y-intercept of the line: 𝑦 = 𝑚𝑥 + 𝑏 Plugging in the values for point 1,2 , we find 2=3+𝑏 Solving for “𝑏” we find: 𝑏 = −1 Therefore, the equation of the tangent line at point (1,2) is: 𝑦 = 3𝑥 − 1

Therefore, the correct answer choice is A. 𝒇 𝒙 = 𝟑𝒙 − 𝟏



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