Last updated on July 18th, 2025
The derivative of y/x helps us understand how the ratio of y to x changes with a slight variation in x. This concept is crucial in applications like calculating rates, optimizing functions, and understanding real-world relationships between variables. We will now explore the derivative of y/x in detail.
Understanding the derivative of y/x involves applying the quotient rule. It is commonly represented as d/dx (y/x) or (y/x)'. The derivative is calculated using the formula for the quotient of two functions, indicating that y/x is differentiable within its domain.
The key concepts are mentioned below:
Quotient Rule: The primary rule for differentiating y/x, as it involves the division of two functions.
Derivatives: Measures of how a function changes as its input changes.
The derivative of y/x can be denoted as d/dx (y/x) or (y/x)'. The formula for differentiating y/x is: d/dx (y/x) = (x * d/dx(y) - y)/x² The formula applies to all x where x ≠ 0.
We can derive the derivative of y/x using proofs. To show this, we will use the quotient rule of differentiation.
Several methods can be used to prove this, such as:
Let's demonstrate that the differentiation of y/x results in the formula (x * d/dx(y) - y)/x² using these methods:
To differentiate y/x using the quotient rule, we consider y as the numerator and x as the denominator.
The quotient rule states: d/dx (u/v) = (v * u' - u * v')/v²
Applying this to y/x, where u = y and v = x, d/dx (y/x) = (x * d/dx(y) - y * d/dx(x))/x² Since d/dx(x) = 1,
we have: d/dx (y/x) = (x * d/dx(y) - y)/x²
To prove the differentiation of y/x using the product rule, we can rewrite y/x as y * (1/x). Given that u = y and v = 1/x,
Using the product rule formula: d/dx [u * v] = u' * v + u * v' u' = d/dx(y) v = 1/x v' = d/dx(1/x) = -1/x²
Using the product rule: d/dx (y/x) = d/dx(y) * (1/x) + y * (-1/x²)
Simplifying, we get: d/dx (y/x) = (x * d/dx(y) - y)/x²
Thus, the derivative is proved.
When a function is differentiated multiple times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives can be complex. To understand them better, consider a car where the speed changes (first derivative) and the acceleration (second derivative) also changes. Higher-order derivatives help us understand functions like y/x more deeply.
For the first derivative of a function, we write f′(x), indicating how the function changes at a certain point. The second derivative is derived from the first derivative, denoted as f′′(x). Similarly, the third derivative, f′′′(x), is derived from the second derivative, and this pattern continues.
For the nth derivative of y/x, we generally use fⁿ(x) to represent the nth derivative of a function f(x), which shows the change in the rate of change.
When x = 0, the derivative is undefined because y/x has a division by zero. When y = 0, the derivative of y/x simplifies to 0, since the numerator is zero.
Students often make mistakes when differentiating y/x. These errors can be mitigated by following the correct procedures. Here are some common mistakes and how to solve them:
Calculate the derivative of (y/x)²
Here, let f(x) = (y/x)².
Using the chain rule, f'(x) = 2 * (y/x) * d/dx(y/x)
We know d/dx(y/x) = (x * d/dx(y) - y)/x².
Substitute this into the equation: f'(x) = 2 * (y/x) * ((x * d/dx(y) - y)/x²)
Simplify to find the final answer: f'(x) = 2 * (y * (x * d/dx(y) - y))/x³.
We find the derivative by applying the chain rule to (y/x)². We then substitute the derivative of y/x and simplify to reach the final result.
A company's profit is given by the function y/x, where y is the revenue and x is the number of products sold. If revenue y increases at a rate of 5 units per product when 50 products are sold, calculate the rate of change of profit.
Given y/x = profit. Differentiate the function using the quotient rule: d/dx (y/x) = (x * d/dx(y) - y)/x²
We know d/dx(y) = 5 (rate of increase of revenue).
Substitute x = 50.
d/dx (y/x) = (50 * 5 - y)/50²
Assume y = 250 (for example), d/dx (y/x) = (250 - 250)/2500 = 0
Therefore, the rate of change of profit is 0.
We calculate the rate of change of profit by differentiating y/x and substituting the given rates and values to determine the result.
Derive the second derivative of the function y/x.
First, find the first derivative: d/dx (y/x) = (x * d/dx(y) - y)/x²
Now, differentiate this to find the second derivative: d²/dx² (y/x) = d/dx [(x * d/dx(y) - y)/x²]
Use the quotient rule again: d²/dx² (y/x) = ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴)
Therefore, the second derivative is ((x² * d²/dx²(y) - 2x * d/dx(y) + 2y)/x⁴).
We use step-by-step differentiation, first finding the first derivative, then applying the quotient rule again to obtain the second derivative.
Prove: d/dx (y²/x) = (2y * d/dx(y) * x - y²)/x².
Start using the quotient rule: Consider y²/x.
Let u = y² and v = x. d/dx (u/v) = (v * d/dx(u) - u * d/dx(v))/v² d/dx (y²/x) = (x * d/dx(y²) - y² * 1)/x²
Since d/dx(y²) = 2y * d/dx(y), d/dx (y²/x) = (2y * d/dx(y) * x - y²)/x²
Thus proved.
We apply the quotient rule to differentiate y²/x, substituting d/dx(y²) and simplifying to arrive at the proof.
Solve: d/dx (y/x²)
To differentiate y/x², use the quotient rule: d/dx (y/x²) = (x² * d/dx(y) - y * 2x)/x⁴
Simplify: d/dx (y/x²) = (x² * d/dx(y) - 2xy)/x⁴
Therefore, d/dx (y/x²) = (x * d/dx(y) - 2y)/x³.
In this process, we differentiate the given function using the quotient rule, simplifying the equation 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.
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