Problem 2
Question
\([a(x) b(x)]^{\prime}=a^{\prime}(x) b(x)+a(x) b^{\prime}(x)\)
Step-by-Step Solution
Verified Answer
The formula is the Product Rule for differentiation, used to find the derivative of a product of two functions.
1Step 1: Identify the Product Rule
The given formula \([a(x) b(x)]^{\prime}=a^{\prime}(x) b(x)+a(x) b^{\prime}(x)\) represents the Product Rule for differentiation, which states that the derivative of the product of two functions is equal to the first function's derivative times the second function, plus the first function times the derivative of the second function.
2Step 2: Apply the Product Rule Components
Identify two functions, \(a(x)\) and \(b(x)\). To differentiate the product \(a(x) b(x)\) with respect to \(x\), apply the product rule by multiplying \(a^{\prime}(x)\) by \(b(x)\), and adding it to the product of \(a(x)\) and \(b^{\prime}(x)\).
3Step 3: Substitute and Differentiate
If specific functions for \(a(x)\) and \(b(x)\) are given, substitute them into the formula and compute \(a^{\prime}(x)\) and \(b^{\prime}(x)\) respectively. Then, plug these derivatives back into the product rule formula to find the derivative of \(a(x)b(x)\).
Key Concepts
DifferentiationFunctionsDerivatives
Differentiation
Differentiation is a core concept in calculus and is all about finding the rate at which things change. It helps us determine how a function's output changes as its input changes. At a basic level, differentiation provides a means to calculate the derivative of a function.
- The process involves applying specific rules and formulas to find the derivative of a function.
- One such rule, which is particularly crucial when dealing with products of functions, is the Product Rule.
- The Product Rule allows us to find the derivative of a product of two functions, such as in the original exercise.
Functions
In mathematics, a function is a relation between a set of inputs (known as the domain) and a set of possible outputs (known as the range) where each input is related to exactly one output. Functions are tremendously useful in expressing mathematical models and real-life scenarios.
- Each function can be represented as an equation, graph, table, or verbal description.
- In the context of differentiation, the function is often expressed as an equation, involving variables like \(x\) and outputs \(f(x)\).
- For the Product Rule, recognizing each part of the overall function, such as \(a(x)\) and \(b(x)\), helps us treat them properly when differentiating.
Derivatives
A derivative represents the rate of change of a function's output with respect to changes in the input. Think of it as a mathematical tool to assess how a small change in input affects the function's output.
- The derivative of a function is often denoted by \(f'(x)\) or \(\frac{d}{dx}[f(x)]\).
- In the case of the Product Rule, derivatives such as \(a'(x)\) and \(b'(x)\) represent the rates at which these individual functions change.
- By computing these derivatives and substituting them into the Product Rule, we can find the derivative of the entire product \(a(x)b(x)\).
Other exercises in this chapter
Problem 1
If \(_{c}\) is algebraic over \(F\) so are \(c+1\) and \(k c\) (where \(\left.k \in F\right)\)
View solution Problem 1
Prove that each of the following numbers is algebraic over \(\mathbb{Q}\) : (a) \(i\) (b) \(\sqrt{2}\) (c) \(2+3 i\) (d) \(\sqrt{1+\sqrt[3]{2}}\) (e) \(\sqrt{i-
View solution Problem 3
If \(c\) is transcendental over \(F\), so are \(c+1, k c\) (where \(k \in F)\), and \(c^{2}\).
View solution Problem 3
Find the minimum polynomial of the following numbers over the indicated fields: $$ \begin{array}{ll} \sqrt{3}+i & \text { over } \mathbb{R} ; \text { over } \ma
View solution