Question

Calculate \(\frac{\partial z}{\partial y}\) when \(z=\frac{x^{2}-3 y^{2}}{x^{2}+y^{2}}\).

Short answer

Question: Calculate the partial derivative of the function \(z(x, y) = \frac{x^2 - 3y^2}{x^2 + y^2}\) with respect to y. Answer: The partial derivative of z with respect to y, \(\frac{\partial z}{\partial y}\), is \(\frac{-8y^3 - 8x^2y}{(x^2 + y^2)^2}\).

Step-by-step solution

Identify the u and v functions

Here, we have u as the numerator and v as the denominator. \(u(x, y) = x^2 - 3y^2\) \(v(x, y) = x^2 + y^2\)

Calculate partial derivatives

We need to compute the partial derivatives of u with respect to y, and v with respect to y, which are: \(\frac{\partial u}{\partial y} = \frac{\partial (x^2 - 3y^2)}{\partial y} = -6y\) \(\frac{\partial v}{\partial y} = \frac{\partial (x^2 + y^2)}{\partial y} = 2y\)

Apply the Quotient Rule

Now, let's apply the quotient rule for differentiation: \(\frac{\partial z}{\partial y} = \frac{v\frac{\partial u}{\partial y} - u\frac{\partial v}{\partial y}}{v^2}\)

Substitute the values into the formula

Plug in the values of u, v, and their respective partial derivatives, and simplify: \(\frac{\partial z}{\partial y} = \frac{(x^2 + y^2)(-6y) - (x^2 - 3y^2)(2y)}{(x^2 + y^2)^2}\)

Simplify the expression

Finally, simplify the expression to obtain the partial derivative of z with respect to y. \(\frac{\partial z}{\partial y} = \frac{-6y(x^2 + y^2) - 2y(x^2 - 3y^2)}{(x^2 + y^2)^2} = \frac{-8y^3 - 8x^2y}{(x^2 + y^2)^2}\)

Key concepts

Quotient Rule for Differentiation

In calculus, the quotient rule is a method for finding the derivative of a function that is expressed as the quotient of two other functions. This is especially useful when you are dealing with complex functions involving fractions. The rule states that if you have a function \( z = \frac{u(x, y)}{v(x, y)} \) where \( u \) and \( v \) are both differentiable functions of \( x \) and \( y \), the partial derivative of \( z \) with respect to \( y \) is given by:\[ \frac{\partial z}{\partial y} = \frac{v\frac{\partial u}{\partial y} - u\frac{\partial v}{\partial y}}{v^2} \]
In the exercise provided, to find \( \frac{\partial z}{\partial y} \) for the function \( z=\frac{x^{2}-3y^{2}}{x^{2}+y^{2}} \), the first step is identifying the functions \( u(x, y) \) and \( v(x, y) \), which are, respectively, the numerator and denominator of \( z \). Next, we compute the necessary partial derivatives of \( u \) and \( v \) with respect to \( y \) and then apply the quotient rule formula by substituting these values into the formula to find the required partial derivative.

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus to functions of several variables. Instead of dealing with functions that depend on one variable, multivariable calculus tackles functions that have two or more variables, such as \( f(x, y) \). This type of calculus is essential for analyzing systems where multiple factors are at play simultaneously.
The problem at hand requires knowledge of multivariable calculus since the function \( z \) depends on both \( x \) and \( y \). When calculating partial derivatives in multivariable calculus, it's important to consider each variable individually while treating the other variables as constants. This becomes clear by looking at the exercise where partial derivatives are found by differentiating \( u \) and \( v \) with respect to \( y \) one at a time, effectively ignoring \( x \) during the differentiation process.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, used to measure how a function changes as only one variable is varied, keeping others fixed. For example, for a function \( f(x, y) \) the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \) and with respect to \( y \) as \( \frac{\partial f}{\partial y} \). Calculating partial derivatives involves taking the derivative with respect to one variable at a time.
In the context of the provided exercise, we compute the partial derivatives \( \frac{\partial u}{\partial y} = -6y \) and \( \frac{\partial v}{\partial y} = 2y \) by treating \( y \) as the variable and \( x \) as a constant. Understanding how to calculate these is critical in applying the quotient rule for differentiation in the realm of multivariable functions.