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103 LearnersLast updated on October 8, 2025
Derivative of uv
We explore the derivative of a product of two functions, uv, using differentiation rules such as the product rule. Derivatives are essential in many real-life applications, like calculating profit or loss. We will now discuss in detail how to find the derivative of uv.
What is the Derivative of uv?
To find the derivative of the product of two functions, uv, we use the product rule. If u and v are differentiable functions of x, then the derivative of uv is represented as d/dx(uv) or (uv)'.
The product rule states that: d/dx(uv) = u'v + uv' This indicates that the derivative of the product uv is the sum of the derivative of u times v and u times the derivative of v.
Derivative of uv Formula
The derivative of the product of two functions uv is given by the product rule formula: d/dx(uv) = u'v + uv'
This formula applies to all x where u and v are differentiable.
Proofs of the Derivative of uv
We can derive the derivative of uv using several methods.
To demonstrate this, we apply differentiation rules such as:
- Using the First Principle
- Using the Product Rule
- Using the Chain Rule
By First Principle
The derivative of uv can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. Let f(x) = uv. Its derivative can be expressed as: f'(x) = limₕ→₀ [(u(x + h)v(x + h) - u(x)v(x))] / h Expanding and simplifying using the definition of limits, we arrive at: f'(x) = u'v + uv'
Using the Product Rule
The product rule directly provides the derivative of uv. If u and v are differentiable, then: d/dx(uv) = u'v + uv' This rule is derived by applying the First Principle and simplifying.
Using the Chain Rule
While the chain rule is typically used for composite functions, it can verify the product rule in more complex derivatives involving uv. However, the primary approach remains the product rule as shown above.
Higher-Order Derivatives of uv
When a function is differentiated multiple times, the results are higher-order derivatives. Higher-order derivatives can be complex but help in understanding changes in a function's rate of change. For instance, if f(x) = uv, then: The first derivative is f′(x) = u'v + uv'.
The second derivative is derived from the first derivative. The nth derivative involves repeatedly applying the derivative rules to find higher-order changes.
Special Cases:
If either u or v is constant, the derivative simplifies significantly. If u or v is zero at a specific point, the derivative at that point is reduced to the derivative of the other function multiplied by zero.
Common Mistakes and How to Avoid Them in Derivatives of uv
Students often make mistakes when differentiating products of functions. Understanding the correct application of differentiation rules can prevent these errors. Here are some common mistakes and how to resolve them:
Mistake 1
Forgetting the Product Rule
Students might forget to apply the product rule when differentiating uv and instead use incorrect methods like simply multiplying the derivatives.
Always remember: d/dx(uv) = u'v + uv'
Mistake 2
Neglecting Constant Multipliers
When one of the functions is a constant, students might neglect to differentiate correctly.
Remember that the derivative of a constant is zero, and apply the product rule accordingly.
Mistake 3
Misapplying the Chain Rule
Some students incorrectly apply the chain rule instead of the product rule.
The chain rule is used for composite functions, while the product rule is for products of functions. Ensure the correct rule is applied.
Mistake 4
Incorrect Order of Differentiation
Students may differentiate in the wrong sequence, leading to errors.
Stick to the correct order: d/dx(uv) = u'v + uv'
Mistake 5
Calculation Errors
Errors in basic arithmetic can lead to incorrect results.
Carefully perform all calculations and double-check your work for accuracy.
Examples Using the Derivative of uv
Problem 1
Calculate the derivative of (x²·sin x)
Here, we have f(x) = x²·sin x. Using the product rule, f'(x) = u'v + uv' In the given equation, u = x² and v = sin x. Let's differentiate each term, u' = d/dx (x²) = 2x v' = d/dx (sin x) = cos x Substituting into the given equation, f'(x) = (2x)·(sin x) + (x²)·(cos x) Let's simplify terms to get the final answer, f'(x) = 2x sin x + x² cos x Thus, the derivative of the specified function is 2x sin x + x² cos x.
Explanation
We find the derivative of the given function by dividing it into two parts.
The first step is finding their derivatives and then combining them using the product rule to get the final result.
Problem 2
A company's revenue is modeled by the function R = p·q, where p is the price per unit and q is the number of units sold. If p = 10 and q = 100, find the rate of change of revenue when p increases by $0.5.
R = p·q (revenue of the company)...(1) Now, we will differentiate the equation (1): dR/dp = q + p(dq/dp) Given p = 10 and q = 100, and assuming dq/dp = 0 (for simplicity), dR/dp = 100 + 10(0) = 100 If p increases by $0.5, the rate of change of revenue is 100(0.5) = $50.
Explanation
We find the rate of change of revenue by differentiating the function with respect to p and evaluating it at the given values.
We assume dq/dp is zero for simplicity, indicating no change in quantity sold with a minor price change.
Problem 3
Derive the second derivative of the function y = x²·e^x.
The first step is to find the first derivative, dy/dx = (d/dx (x²))·e^x + x²·(d/dx (e^x)) = 2x·e^x + x²·e^x Now we will differentiate to get the second derivative: d²y/dx² = (d/dx (2x·e^x)) + (d/dx (x²·e^x)) = [2·e^x + 2x·e^x] + [2x·e^x + x²·e^x] = 2e^x + 4x·e^x + x²·e^x Therefore, the second derivative of the function y = x²·e^x is 2e^x + 4x·e^x + x²·e^x.
Explanation
We use a step-by-step process, starting with the first derivative.
We then apply the product rule to each term and simplify to find the second derivative.
Problem 4
Prove: d/dx ((x³)·(ln x)) = x²(3ln x + 1).
Let's start using the product rule: Consider y = x³·ln x To differentiate, we apply the product rule: dy/dx = (d/dx (x³))·ln x + x³·(d/dx (ln x)) = 3x²·ln x + x³·1/x = 3x²·ln x + x² Hence, d/dx ((x³)·(ln x)) = x²(3ln x + 1).
Explanation
In this step-by-step process, we use the product rule to differentiate the equation.
The derivative of ln x is 1/x, which is substituted to derive the final expression.
Problem 5
Solve: d/dx ((x·cos x)/x²).
To differentiate the function, we use the quotient rule: d/dx ((x·cos x)/x²) = (d/dx (x·cos x)·x² - x·cos x·d/dx (x²))/(x²)² We will substitute: d/dx (x·cos x) = cos x - x sin x d/dx (x²) = 2x = [(cos x - x sin x)·x² - x·cos x·2x]/x⁴ = [x² cos x - x³ sin x - 2x² cos x]/x⁴ = [-x² cos x - x³ sin x]/x⁴ = -(cos x + x sin x)/x² Therefore, d/dx ((x·cos x)/x²) = -(cos x + x sin x)/x².
Explanation
In this process, we differentiate using the product and quotient rules.
We simplify the equation at each step to obtain the final result.
FAQs on the Derivative of uv
1.Find the derivative of uv.
2.Can the derivative of uv be applied in real life?
3.Is it possible to take the derivative of uv if one function is a constant?
4.What rule is used to differentiate the product of two functions?
5.Are the derivatives of uv and u/v the same?
Important Glossaries for the Derivative of uv
- Derivative: The derivative of a function measures how a function changes as its input changes.
- Product Rule: A rule used to differentiate the product of two functions.
- Constant: A fixed value that does not change.
- Quotient Rule: A rule used to differentiate the division of two functions.
- Second Derivative: The derivative of the derivative, showing how the rate of change itself changes.
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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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: He loves to play the quiz with kids through algebra to make kids love it.
