42.1 Partial Derivatives Concept Overview

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

CONCEPT INTRO: When we encounter a function made up of two independent variables, say “𝑥” and “𝑦”, we are able to determine the derivative of this function with respect to one of the variables while assuming the other variable remains constant. If we are given the function: 𝑧 = 𝑓(𝑥, 𝑦) And we choose to consider “y” as a constant, the function becomes one of a single variable “𝑥”, and its derivative (if it exists) can be found. It official terms, this would be stated as “The PARTIAL DERIVATIVE of “𝑧” WITH RESPECT to “𝑥”. On the flip side, if we were to consider “x” to be our constant, then we would be finding “The PARTIAL DERIVATIVE of “𝑧” WITH RESPECT to “𝑦”.

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The partial derivative with respect to “𝑥” is generically expressed as: 𝜕𝑧 𝜕𝑓(𝑥, 𝑦) = 𝜕𝑥 𝜕𝑥 When given a function of multiple variables, the PARTIAL DERIVATIVES of that function can be defined by concentrating exclusively on changing one of the variables at a time, while holding the remaining variables fixed and treating them as constants. So in essence, determining the derivatives of functions of more than one variable is done in pretty much the same way as taking derivatives of a single variable. Illustrating a set of general GUIDELINES, it can be presented like this: • To define the PARTIAL DERIVATIVE of 𝑓(𝑥, 𝑦) with respect to “𝑥”, treat all the “𝑦” terms as constants and then differentiate the “𝑥” terms. • To define the PARTIAL DERIVATIVE of 𝑓(𝑥, 𝑦) with respect to “𝑦”, treat all the “𝑥” terms as constants and then differentiate the “𝑦” terms. These two partial derivatives are often referred to as FIRST ORDER PARTIAL DERIVATIVES. Just as with functions of one variable, derivatives of all orders, as long as they exist, are possible. It’s important to note that the NOTATION for PARTIAL DERIVATIVES is different than that for derivatives of functions of a single variable.

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With functions of a single variable, a single prime is commonly used to denote a derivative. With PARTIAL DERIVATIVES we will always need to remember the context revolving around the variable that we are differentiating with respect to, and for this reason, a variable subscript is often used to note that characteristic. There are a potentially a number of different notations that you will encounter when working with PARTIAL DERIVATIVES, some being:

𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑥 =

𝜕𝑓 𝜕 = 𝑓(𝑥, 𝑦) = 𝐷𝑥 𝑓 𝜕𝑥 𝜕𝑥

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PARTIAL DERIVATIVES | CONCEPT EXAMPLE The following problem introduces the concept reviewed within this module. Use this content as a primer for the subsequent material. Determine the partial derivative with respect to x of the following function:

𝑓(𝑥, 𝑦) =

A.

𝑥−𝑦 𝑥+𝑦

2𝑦 (𝑥+𝑦)2 2𝑦

B. − (2+𝑦)2 2𝑥

C. − (2+𝑥)2 5𝑦

D. (5+𝑦)2

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SOLUTION: The TOPIC of PARTIAL DERIVATIVES can be referenced under the SUBJECT of MATHEMATICS on page 27 of the NCEES Supplied Reference Handbook, Version 9.4 for Computer Based Testing. Recall that to find the partial derivative of 𝑓(𝑥, 𝑦) with respect to “𝑥”, we will need to treat all the “𝑦” terms as constants and then differentiate the “𝑥” terms. The original function is presented as:

𝑓(𝑥, 𝑦) =

𝑥−𝑦 𝑥+𝑦

We will we use the QUOTIENT RULE to determine our PARTIAL DERIVATIVE, which states: 𝑓(𝑥) 𝑔(𝑥)𝑓 ′ (𝑥) − 𝑓(𝑥)𝑔′(𝑥) [ ]= [𝑔(𝑥)]2 𝑔(𝑥) Using this template, we can define that: 𝑓(𝑥) = (𝑥 − 𝑦)

𝑔(𝑥) = (𝑥 + 𝑦)

We don’t suggest that you do this, but for illustration purposes, let’s remember that as we are determining this PARTIAL DERIVATIVE with respect to “x”, that all y terms will be considered to be a constant. Made with

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We could then rewrite these formulas to read: 𝑓(𝑥) = (𝑥 − 1) 𝑔(𝑥) = (𝑥 + 1) We just made the y-terms equal to 1. These aren’t the official formulas we will plug back in to the QUOTIENT RULE, we are just illustrating how we will treat the y-terms. With that, we can determine the DERIVATIVES of each of these functions, such that: 𝑓 ′ (𝑥) = 1 𝑔′ (𝑥) = 1 Taking these iterations and plugging them back in to the standard template of the QUOTIENT RULE, we get: 𝜕𝑓 (𝑥 + 𝑦)1 − (𝑥 − 𝑦)1 𝑥 + 𝑦 − 𝑥 + 𝑦 = = (𝑥 + 𝑦)2 (𝑥 + 𝑦)2 𝜕𝑥 And simplifying, we get: 𝜕𝑓 2𝑦 = 𝜕𝑥 (𝑥 + 𝑦)2 𝟐𝒚

The correct answer choice is A. (𝒙+𝒚)𝟐

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