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118 LearnersLast updated on September 10, 2025
Derivative of f(x)/g(x)
We use the derivative of a quotient of two functions, f(x) and g(x), as a tool to understand how the ratio of these functions changes in response to a change in x. Derivatives are crucial in calculating rates of change and optimizing solutions in real-life scenarios. We will now discuss the derivative of f(x)/g(x) in detail.
What is the Derivative of f(x)/g(x)?
The derivative of the quotient of two functions, f(x)/g(x), is expressed using the quotient rule. It is represented as d/dx [f(x)/g(x)], and the formula is [f'(x)g(x) - f(x)g'(x)] / [g(x)]².
This indicates that the function f(x)/g(x) is differentiable where g(x) ≠ 0. The key concepts are mentioned below:
Quotient Rule: The rule for differentiating the quotient of two functions.
Product and Chain Rules: These are essential for deriving the quotient rule.
Derivative of f(x)/g(x) Formula
The derivative of f(x)/g(x) is denoted as d/dx [f(x)/g(x)].
The formula is: d/dx [f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]² The formula applies to all x where g(x) ≠ 0.
Proofs of the Derivative of f(x)/g(x)
To derive the derivative of f(x)/g(x), we can use several methods, such as:
By Using the First Principles
- Using the Chain Rule
- Using the Product Rule
We will demonstrate the differentiation of f(x)/g(x) using these methods: By First Principles The derivative can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. Consider f(x)/g(x). Its derivative can be expressed as: f'(x) = limₕ→₀ [(f(x + h)/g(x + h)) - (f(x)/g(x))] / h By simplifying and using the limit laws, we arrive at the quotient rule.
Using Chain Rule
When f(x) = u and g(x) = v, we can express the quotient as u/v and use the chain rule: d/dx [u/v] = (v * du/dx - u * dv/dx) / v²
Using Product Rule
Express the quotient as a product, f(x) * (1/g(x)), and apply the product rule.
Higher-Order Derivatives of f(x)/g(x)
Higher-order derivatives are obtained by repeatedly differentiating a given function. These derivatives can provide deeper insights into the behavior of functions like f(x)/g(x).
For a first derivative, we write f′(x), indicating the rate of change. The second derivative, f′′(x), is obtained from the first derivative and indicates how the rate of change itself changes. This process continues for third derivatives, f′′′(x), and so on.
Special Cases
When g(x) = 0, the derivative is undefined because the function has a vertical asymptote there.
When f(x) = 0, the derivative simplifies because the numerator of the derivative formula becomes zero.
Common Mistakes and How to Avoid Them in Derivatives of f(x)/g(x)
Students often make mistakes when differentiating f(x)/g(x). These mistakes can be resolved by understanding the correct solutions. Here are a few common mistakes and how to solve them:
Mistake 1
Not Simplifying the Expression
Students may forget to simplify the expression, leading to incomplete or incorrect results. They often skip steps, especially when using the product or chain rule.
Ensure each step is written in sequence to avoid errors.
Mistake 2
Forgetting Undefined Points
Students might not remember that the function is undefined at points where g(x) = 0.
Always consider the domain of the function being differentiated.
Mistake 3
Incorrect Use of the Quotient Rule
Students might misapply the quotient rule. For example: Incorrect differentiation: d/dx [f(x)/g(x)] = f'(x)/g'(x).
Correct: d/dx [f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]². To avoid this mistake, carefully apply the quotient rule formula.
Mistake 4
Not Considering Constants and Coefficients
Students sometimes forget to account for constants preceding f(x) or g(x). For example, incorrectly writing d/dx [5f(x)/g(x)] = 5[f'(x)g(x) - f(x)g'(x)]/g(x)².
Ensure constants are correctly multiplied.
Mistake 5
Ignoring the Chain Rule
Students often neglect the chain rule, especially when differentiating functions with inner functions.
For example, d/dx [f(2x)/g(x)] should be handled by considering the derivative of the inner function.
Examples Using the Derivative of f(x)/g(x)
Problem 1
Calculate the derivative of [(2x³ + 3)/(x² + 1)].
Here, f(x) = 2x³ + 3 and g(x) = x² + 1. Using the quotient rule, f'(x) = 6x² and g'(x) = 2x. d/dx [(2x³ + 3)/(x² + 1)] = [(6x²)(x² + 1) - (2x³ + 3)(2x)] / (x² + 1)² = [6x⁴ + 6x² - 4x⁴ - 6x] / (x² + 1)² = [2x⁴ + 6x² - 6x] / (x² + 1)²
Explanation
The derivative of the quotient is found by applying the quotient rule, simplifying each term, and combining them to get the final result.
Problem 2
A company uses the function y = (4x + 5)/(x - 2) to model its cost structure, where y is the cost and x is the production level. Find the rate of change of cost when x = 3.
Given y = (4x + 5)/(x - 2). Using the quotient rule, f(x) = 4x + 5, g(x) = x - 2. f'(x) = 4, g'(x) = 1. d/dx [(4x + 5)/(x - 2)] = [(4)(x - 2) - (4x + 5)(1)] / (x - 2)² = [4x - 8 - 4x - 5] / (x - 2)² = [-13] / (x - 2)² At x = 3, = [-13] / (3 - 2)² = -13. The rate of change of cost at x = 3 is -13.
Explanation
By applying the quotient rule and substituting x = 3, we find the rate of change of cost, indicating a decrease at this level of production.
Problem 3
Derive the second derivative of the function y = (x²)/(2x + 1).
First derivative: f(x) = x², g(x) = 2x + 1. f'(x) = 2x, g'(x) = 2. d/dx [x²/(2x + 1)] = [(2x)(2x + 1) - x²(2)] / (2x + 1)² = [4x² + 2x - 2x²] / (2x + 1)² = [2x² + 2x] / (2x + 1)² Second derivative: Differentiate d²y/dx² = d/dx [(2x² + 2x)/(2x + 1)²] using the quotient rule again. Apply the quotient rule to find the second derivative, simplifying each step for accuracy.
Explanation
We obtain the second derivative by differentiating the first derivative using the quotient rule again, ensuring all terms are simplified correctly.
Problem 4
Prove: d/dx [(x² + 1)/(x - 1)] = [(x² - 2x - 1)/(x - 1)²].
Using the quotient rule, f(x) = x² + 1, g(x) = x - 1. f'(x) = 2x, g'(x) = 1. d/dx [(x² + 1)/(x - 1)] = [(2x)(x - 1) - (x² + 1)(1)] / (x - 1)² = [2x² - 2x - x² - 1] / (x - 1)² = [x² - 2x - 1] / (x - 1)² Hence proved.
Explanation
By applying the quotient rule to differentiate, we simplify the expression to match the given proof, ensuring each step follows logically.
Problem 5
Solve: d/dx [(x² + 3)/(2x)].
Using the quotient rule: f(x) = x² + 3, g(x) = 2x. f'(x) = 2x, g'(x) = 2. d/dx [(x² + 3)/(2x)] = [(2x)(2x) - (x² + 3)(2)] / (2x)² = [4x² - 2x² - 6] / (4x²) = [2x² - 6] / (4x²) = [x² - 3] / (2x²)
Explanation
We differentiate the function using the quotient rule and simplify to find the derivative, ensuring all steps are logical and accurate.
FAQs on the Derivative of f(x)/g(x)
1.What is the derivative of f(x)/g(x)?
2.How is the derivative of f(x)/g(x) useful in real life?
3.Can we find the derivative of f(x)/g(x) if g(x) = 0?
4.Which rule is used to differentiate f(x)/g(x)?
5.Are the derivatives of f(x)/g(x) and g(x)/f(x) the same?
6.Can we derive the formula for the derivative of f(x)/g(x)?
Important Glossaries for the Derivative of f(x)/g(x)
- Derivative: Measures how a function changes as its input changes.
- Quotient Rule: A formula for finding the derivative of a quotient of two functions.
- Product Rule: Used to find the derivative of the product of two functions.
- Chain Rule: Used to differentiate compositions of functions.
- Undefined: Refers to points where a function or its derivative does not exist.
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Jaskaran Singh Saluja
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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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