Question

Differentiate the following functions: \(\frac{2 a y}{a^{2}-y^{2}}\)

Short answer

Answer: The derivative of the given function with respect to \(y\) is \(\frac{2a(a^2+y^2)}{(a^2-y^2)^2}\).

Step-by-step solution

Identify the numerator and denominator functions

We are given the function: \(\frac{2 a y}{a^{2}-y^{2}}\). Let the numerator be the function \(u\) and the denominator be the function \(v\): \(u(y)=2ay\) and \(v(y)=a^2-y^2\)

Differentiate the functions u(y) and v(y) with respect to y

Find the derivatives of the functions u(y) and v(y) with respect to y: \(\frac{du}{dy} = \frac{d(2ay)}{dy} = 2a\) and \(\frac{dv}{dy} = \frac{d(a^2-y^2)}{dy} = -2y\)

Apply the quotient rule

Now apply the quotient rule to find the derivative of the given function: \(\frac{d(\frac{2 a y}{a^{2}-y^{2}})}{dy} = \frac{v\frac{du}{dy} - u\frac{dv}{dy}}{v^2}\) Substitute \(u\), \(v\), \(\frac{du}{dy}\), and \(\frac{dv}{dy}\) with the values we found in Steps 1 and 2: \(\frac{d(\frac{2 a y}{a^{2}-y^{2}})}{dy} = \frac{(a^2-y^2)(2a) - (2ay)(-2y)}{(a^2-y^2)^2}\)

Simplify the expression

Now, simplify the expression: \(\frac{d(\frac{2 a y}{a^{2}-y^{2}})}{dy} = \frac{2a^3-2ay^3 + 4a y^3}{(a^2-y^2)^2}\) Combine the terms in the numerator: \(\frac{d(\frac{2 a y}{a^{2}-y^{2}})}{dy} = \frac{2a^3 + 2ay^3}{(a^2-y^2)^2}\) Finally, the derivative of the given function with respect to \(y\) is: \(\frac{d(\frac{2 a y}{a^{2}-y^{2}})}{dy} = \frac{2a(a^2+y^2)}{(a^2-y^2)^2}\)

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus, which involves finding the derivative of a function. The derivative measures how a function's output value changes as its input changes, essentially capturing the rate of change or the slope of the function at any given point. This is incredibly useful because it allows us to understand how quantities vary in relation to each other, a key aspect of analyzing real-world phenomena.
In our exercise, differentiation is applied to the function \( \frac{2 a y}{a^{2}-y^{2}} \). To find its derivative, we first need to break the expression into manageable components by identifying the numerator and the denominator as separate functions. Differentiating complex fraction functions typically involves using the quotient rule, a specific method for taking the derivative of the division of two functions.
For simpler functions, differentiation can involve intuitive rules such as the power rule, product rule, or chain rule. Each rule offers a technique for quickly determining derivatives without doing the laborious algebra of first principles. Understanding these rules can significantly simplify the differentiation process and provide a clear path to finding derivatives efficiently.

Calculus

Calculus is a branch of mathematics that explores the concepts of change and motion through derivatives and integrals. It forms the basis for many areas of mathematical analysis concerning rates of change (differentiation) and accumulation (integration). The origins of calculus date back centuries and have enabled major advancements in science, engineering, and economics.
In the context of the given exercise, calculus provides the framework required for applying the quotient rule to differentiate the function \( \frac{2 a y}{a^{2}-y^{2}} \). The quotient rule itself is one of the many tools within calculus designed to handle situations where functions are divided, as in our example.
By systematically using calculus concepts like differentiation, one can not only solve specific mathematical problems but also model dynamic systems such as tracking population growth, predicting financial trends, or even controlling the flow of traffic. The versatility and power of calculus make it a cornerstone of modern mathematical education.

Derivative

The derivative of a function represents how the function value changes with respect to changes in its input, often thought of as the slope of the function at a given point. In mathematical terms, the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
To find the derivative of the function \( \frac{2 a y}{a^{2}-y^{2}} \), we applied the quotient rule. This rule is necessary when the function is a division of two other functions. The quotient rule states: for two functions \( u(y) \) and \( v(y) \), the derivative of their quotient \( \frac{u(y)}{v(y)} \) is given by:

• \( \frac{d}{dy} \left( \frac{u}{v} \right) = \frac{v(y) \cdot \frac{du}{dy} - u(y) \cdot \frac{dv}{dy}}{[v(y)]^2} \).

In the original exercise, we calculated \( u(y) = 2ay \) and \( v(y) = a^2-y^2 \) and found their derivatives \( \frac{du}{dy} = 2a \) and \( \frac{dv}{dy} = -2y \). Following the quotient rule, we solved for the derivative and simplified the expression to reach the final result.
Understanding how to compute derivatives and what they signify helps in many practical applications, such as optimizing functions for the best output or understanding the dynamics of change in physical systems.