Derivative of uv
2026-02-28 13:25 Diff
https://brightchamps.com/en-us/math/calculus/derivative-of-uv

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Last updated on October 8, 2025

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.

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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:

Problem 1

Calculate the derivative of (x²·sin x)

Okay, lets begin

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.

Well explained 👍

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.

Okay, lets begin

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.

Well explained 👍

Problem 3

Derive the second derivative of the function y = x²·e^x.

Okay, lets begin

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.

Well explained 👍

Problem 4

Prove: d/dx ((x³)·(ln x)) = x²(3ln x + 1).

Okay, lets begin

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.

Well explained 👍

Problem 5

Solve: d/dx ((x·cos x)/x²).

Okay, lets begin

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.

Well explained 👍

FAQs on the Derivative of uv

1.Find the derivative of uv.

Using the product rule for uv gives: d/dx(uv) = u'v + uv', where u and v are differentiable functions.

2.Can the derivative of uv be applied in real life?

Yes, derivatives of uv are used in real-life scenarios, such as calculating changes in revenue, physics problems involving forces, and much more.

3.Is it possible to take the derivative of uv if one function is a constant?

Yes, if one function is constant, the derivative simplifies to the derivative of the other function multiplied by the constant.

4.What rule is used to differentiate the product of two functions?

The product rule is used to differentiate the product of two functions: d/dx(uv) = u'v + uv'.

5.Are the derivatives of uv and u/v the same?

No, they are different. The derivative of uv is given by the product rule, while the derivative of u/v is given by the quotient rule.

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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