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Last updated on September 15, 2025
We use the derivative of f/g, which involves applying the quotient rule, as a tool to understand how the ratio of two functions changes in response to a slight change in x. Derivatives help us calculate rates of change in various real-life situations. We will now talk about the derivative of f/g in detail.
We now understand the derivative of f/g. It is commonly represented as d/dx (f/g) or (f/g)', and its value is given by the quotient rule: (f'g - fg')/g². The function f/g has a clearly defined derivative, indicating it is differentiable where g(x) ≠ 0. The key concepts are mentioned below:
Quotient Rule: A rule for differentiating a ratio of two functions (f/g).
Differentiation: The process of finding the derivative.
Function: A relation that uniquely associates members of one set with members of another set.
The derivative of f/g can be denoted as d/dx (f/g) or (f/g)'. The formula we use to differentiate f/g is: d/dx (f/g) = (f'g - fg')/g² The formula applies to all x where g(x) ≠ 0.
We can derive the derivative of f/g using proofs. To show this, we will use differentiation rules. There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of f/g results in (f'g - fg')/g² using the above-mentioned methods:
The derivative of f/g can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of f/g using the first principle, we will consider h(x) = f(x)/g(x). Its derivative can be expressed as the following limit. h'(x) = limₕ→₀ [h(x + h) - h(x)] / h … (1) Given that h(x) = f(x)/g(x), we write h(x + h) = f(x + h)/g(x + h).
Substituting these into equation (1), h'(x) = limₕ→₀ [f(x + h)/g(x + h) - f(x)/g(x)] / h = limₕ→₀ [f(x + h)g(x) - f(x)g(x + h)] / [h g(x)g(x + h)]
We now use the formula f(x + h)g(x) - f(x)g(x + h) = f'(x)g(x) - f(x)g'(x), h'(x) = limₕ→₀ [f'(x)g(x) - f(x)g'(x)] / [g²(x)]
As h approaches 0, we have, h'(x) = (f'g - fg')/g²
Hence, proved.
To prove the differentiation of f/g using the quotient rule, We use the formula: h(x) = f(x)/g(x)
By quotient rule: d/dx [f(x) / g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]²… (1)
Let’s substitute f(x) and g(x) in equation (1), d/dx (f/g) = [(f'g - fg')]/g²
Thus, the derivative of f/g is (f'g - fg')/g².
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 f/g.
For the first derivative of a function, we write h′(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 h′′(x) Similarly, the third derivative, h′′′(x) is the result of the second derivative and this pattern continues.
For the nth Derivative of f/g, we generally use hⁿ(x) for the nth derivative of a function h(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).
When g(x) = 0, the derivative is undefined because h(x) becomes undefined. When the x is such that g(x) ≠ 0, the derivative of f/g is (f'g - fg')/g², ensuring differentiability.
Students frequently make mistakes when differentiating f/g. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (x²/x³).
Here, we have h(x) = x²/x³.
Using the quotient rule, h'(x) = (f'g - fg')/g² In the given equation, f(x) = x² and g(x) = x³.
Let’s differentiate each term, f′= d/dx (x²) = 2x g′= d/dx (x³) = 3x²
substituting into the given equation, h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²
Thus, the derivative of the specified function is -1/x².
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative using the quotient rule to get the final result.
A company is analyzing the cost efficiency of two products, where the costs are represented by the functions C1(x) = x² and C2(x) = x³. Calculate the rate of change of the cost efficiency ratio C1(x)/C2(x) at x = 1.
We have C(x) = C1(x)/C2(x) = x²/x³...(1)
Now, we will differentiate the equation (1) Using the quotient rule: dC/dx = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²
Given x = 1, substitute this into the derivative: dC/dx = -1/1² = -1
Hence, the rate of change of the cost efficiency ratio at x = 1 is -1.
We find the rate of change of the cost efficiency ratio at x = 1 as -1, which indicates a decrease in efficiency as x increases.
Derive the second derivative of the function y = x²/x³.
The first step is to find the first derivative, dy/dx = -1/x²...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-1/x²] = 2/x³
Therefore, the second derivative of the function y = x²/x³ is 2/x³.
We use the step-by-step process, where we start with the first derivative. Then, we differentiate again to find the second derivative and simplify to get the final answer.
Prove: d/dx (x²/x³) = -1/x².
Let’s start using the quotient rule: Consider h(x) = x²/x³
To differentiate, we use the quotient rule: h'(x) = (2x·x³ - x²·3x²)/x⁶ = (2x⁴ - 3x⁴)/x⁶ = -x⁴/x⁶ = -1/x²
Hence proved.
In this step-by-step process, we used the quotient rule to differentiate the equation. We then simplified the resulting expression to derive the equation.
Solve: d/dx (x²/x).
To differentiate the function, we simplify first: x²/x = x
Now, differentiate x: d/dx(x) = 1 Therefore, d/dx (x²/x) = 1.
In this process, we simplify the given function and then differentiate using basic differentiation rules to obtain the final result.
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.
: He loves to play the quiz with kids through algebra to make kids love it.