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112 LearnersLast updated on September 27, 2025
Derivative of Multiplication
We use the derivative of multiplication to measure how the product of two functions changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of multiplication in detail.
What is the Derivative of Multiplication?
To understand the derivative of multiplication, we use the product rule. It is commonly represented as d/dx (u·v), where u and v are functions of x. The derivative is given by u'v + uv'. This rule indicates that the product of two differentiable functions is itself differentiable.
The key concepts are mentioned below: Product Rule: A rule for differentiating the product of two functions. Derivative: The rate of change of a function with respect to a variable. Function: A mathematical relation between a set of inputs and outputs, where each input is related to exactly one output.
Derivative of Multiplication Formula
The derivative of multiplication can be denoted as d/dx (u·v). The formula we use to differentiate the product of two functions is: d/dx (u·v) = u'v + uv'
This formula applies to all x where both functions u and v are differentiable.
Proofs of the Derivative of Multiplication
We can derive the derivative of multiplication using proofs. To show this, we will use the rules of differentiation. Here are several methods we use to prove this:
Using the Product Rule
According to the product rule for differentiation, if u(x) and v(x) are differentiable functions, then the derivative of their product is given by: d/dx (u·v) = u'v + uv' This formula comes from considering the limit of the difference quotient of the product of two functions.
Using the First Principle
The derivative of the product u·v can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. Consider f(x) = u(x)·v(x). Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [(u(x + h)v(x + h) - u(x)v(x))] / h We can expand u(x + h) and v(x + h) using their respective derivatives, and upon simplification, the limit resolves to: f'(x) = u'(x)v(x) + u(x)v'(x) Thus, the derivative of the product is the sum of each function's derivative times the other function.
Higher-Order Derivatives of Multiplication
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like multiplication of multiple functions.
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of a product of functions, we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change.
Special Cases:
When one of the functions is a constant, the derivative of the product simplifies to the derivative of the non-constant function multiplied by the constant.
When one of the functions is zero, the derivative is zero, as the product itself is zero.
Common Mistakes and How to Avoid Them in Derivatives of Multiplication
Students frequently make mistakes when differentiating products of functions. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Not applying the product rule correctly
Students may forget to apply the product rule, leading to incorrect results. The rule states that the derivative of a product is not simply the product of the derivatives.
Ensure that you apply the rule correctly: d/dx (u·v) = u'v + uv'.
Mistake 2
Ignoring constants
Students might overlook constants in multiplication. Remember that constants must be included in calculations, as they affect the result.
For example, in d/dx (3x²·x), the constant 3 should be multiplied throughout.
Mistake 3
Misidentifying functions
While differentiating, students may confuse which function is u and which is v, leading to errors in applying the product rule.
Clearly identify each function before applying the rule.
Mistake 4
Overlooking higher-order derivatives
Students may not consider higher-order derivatives when required.
The product rule extends to higher derivatives, and neglecting this can lead to incomplete solutions.
Mistake 5
Forgetting to apply the chain rule when necessary
Students often overlook the need to apply the chain rule when differentiating composite functions within a product.
Ensure that each function is correctly differentiated, considering any nested functions.
Examples Using the Derivative of Multiplication
Problem 1
Calculate the derivative of (x²·e^x)
Here, we have f(x) = x²·e^x. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = x² and v = e^x. Let’s differentiate each term, u′ = d/dx (x²) = 2x v′ = d/dx (e^x) = e^x substituting into the given equation, f'(x) = (2x)·(e^x) + (x²)·(e^x) Let’s simplify terms to get the final answer, f'(x) = 2xe^x + x²e^x Thus, the derivative of the specified function is 2xe^x + x²e^x.
Explanation
We find the derivative of the given function by dividing the function into two parts.
The first step is finding its derivative and then combining them using the product rule to get the final result.
Problem 2
A company produces widgets, and the production cost is represented by the function C(x) = x·ln(x), where x is the number of widgets produced. Calculate the rate of change of the cost when x = 10.
We have C(x) = x·ln(x). Now, we will differentiate the equation using the product rule, dC/dx = u′v + uv′ Here, u = x and v = ln(x). Let’s differentiate each term, u′ = d/dx (x) = 1 v′ = d/dx (ln(x)) = 1/x substituting into the equation, dC/dx = (1)·(ln(x)) + (x)·(1/x) dC/dx = ln(x) + 1 Given x = 10, substitute this into our derivative, dC/dx = ln(10) + 1 Thus, the rate of change of the cost when producing 10 widgets is ln(10) + 1.
Explanation
We find the rate of change of the production cost at x = 10 by differentiating the cost function using the product rule.
We then substitute x = 10 into our derivative to calculate the rate of change.
Problem 3
Derive the second derivative of the function y = x·sin(x).
The first step is to find the first derivative, dy/dx = cos(x) + x·cos(x)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(x) + x·cos(x)] Here we use the product rule, d²y/dx² = -sin(x) + (cos(x) - x·sin(x)) = -sin(x) + cos(x) - x·sin(x) Therefore, the second derivative of the function y = x·sin(x) is -sin(x) + cos(x) - x·sin(x).
Explanation
We use the step-by-step process, where we start with the first derivative.
Using the product rule, we differentiate the terms and simplify to find the second derivative.
Problem 4
Prove: d/dx (x^3·ln(x)) = 3x²ln(x) + x².
Let’s start using the product rule: Consider y = x^3·ln(x) To differentiate, we use the product rule: dy/dx = u′v + uv′ Where u = x^3 and v = ln(x). Differentiating each, u′ = d/dx (x^3) = 3x² v′ = d/dx (ln(x)) = 1/x Thus, dy/dx = (3x²)·(ln(x)) + (x^3)·(1/x) = 3x²ln(x) + x² Hence proved.
Explanation
In this step-by-step process, we used the product rule to differentiate the equation.
We then simplify to derive the final equation.
Problem 5
Solve: d/dx ((x² + 1)·tan(x))
To differentiate the function, we use the product rule: d/dx ((x² + 1)·tan(x)) = (d/dx (x² + 1))·tan(x) + (x² + 1)·d/dx (tan(x)) Differentiating, = (2x)·tan(x) + (x² + 1)·sec²(x) Therefore, d/dx ((x² + 1)·tan(x)) = 2x·tan(x) + (x² + 1)·sec²(x).
Explanation
In this process, we differentiate the given function using the product rule and simplify to obtain the final result.
FAQs on the Derivative of Multiplication
1.What is the derivative of a product of two functions?
2.Can the derivative of multiplication be used in real life?
3.Is it possible to take the derivative of a product at a point where one function is undefined?
4.What rule is used to differentiate a product of more than two functions?
5.Are the derivatives of a product and a quotient the same?
Important Glossaries for the Derivative of Multiplication
- Product Rule: A rule for differentiating the product of two functions.
- Derivative: The rate of change of a function with respect to a variable.
- Function: A relation where each input is related to exactly one output.
- Higher-Order Derivative: Derivatives that are obtained by differentiating a function multiple times.
- Undefined: A term describing a function or point where a mathematical expression has no meaning.
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Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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