Problem 17 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ z=\frac{4 \sqrt{x}}{3 y^{2}+1} $$

Step-by-Step Solution

Answer

The partial derivatives are \( \frac{2}{\sqrt{x}(3y^2 + 1)} \) and \( \frac{-24y\sqrt{x}}{(3y^2 + 1)^2} \).

Step 1: Identify the function

We are given the function \(z\) which is defined as \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \). Our task is to find the first partial derivatives with respect to \( x \) and \( y \).

Step 2: Find the partial derivative with respect to x

To find \( \frac{\partial z}{\partial x} \), we treat \( y \) as a constant and differentiate \( z \) with respect to \( x \). Using the quotient rule for derivatives, where \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2} \), let \( u(x) = 4\sqrt{x} \) and \( v(y) = 3y^2 + 1 \). Thus, \( u'(x) = \frac{2}{\sqrt{x}} \) and \( v(y) \) is a constant when differentiating with respect to \( x \). We get: \[\frac{\partial z}{\partial x} = \frac{\frac{2}{\sqrt{x}}(3y^2 + 1)}{(3y^2 + 1)^2} = \frac{2}{\sqrt{x}(3y^2 + 1)}.\]

Step 3: Find the partial derivative with respect to y

To find \( \frac{\partial z}{\partial y} \), treat \( x \) as a constant and differentiate \( z \) with respect to \( y \) using the quotient rule. In this case, \( u(x) = 4\sqrt{x} \) (which is constant when differentiating with respect to \( y \)) and \( v(y) = 3y^2 + 1 \), so \( v'(y) = 6y \). Applying the quotient rule, we have: \[\frac{\partial z}{\partial y} = \frac{0 \cdot (3y^2 + 1) - 4\sqrt{x} \cdot 6y}{(3y^2 + 1)^2} = \frac{-24y\sqrt{x}}{(3y^2 + 1)^2}.\]

Step 4: Conclusion

The first partial derivatives of the function \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \) are: \[\frac{\partial z}{\partial x} = \frac{2}{\sqrt{x}(3y^2 + 1)}\]and \[\frac{\partial z}{\partial y} = \frac{-24y\sqrt{x}}{(3y^2 + 1)^2}.\]

Key Concepts

Quotient Rule

The quotient rule is essential when differentiating functions represented as a fraction. This rule states that if you have a function expressed as a quotient such as \( \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable functions, the derivative is given by:\[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\]In our exercise, the function \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \) can be separated into these components:

• \( u(x) = 4 \sqrt{x} \)
• \( v(x) = 3y^2 + 1 \)

The first step in applying the quotient rule is differentiating \( u(x) \) and \( v(x) \):- If \( u(x) = 4 \sqrt{x} \), then \( u'(x) = \frac{2}{\sqrt{x}} \).- Since \( v(x) \) is constant in terms of \( x \), there is no need to differentiate \( v(x) \) directly with respect to \( x \) for the \( x \) partial derivative.Applying the quotient rule allows us to neatly derive the derivative of a quotient function, revealing how both the numerator and denominator influence the result.

Function of Two Variables

A function of two variables is a mathematical expression involving two separate variables which can affect the outcome of the function. In our scenario, the function is:\[ z = \frac{4 \sqrt{x}}{3y^2 + 1} \]Here, \( x \) and \( y \) are the two variables. Understanding functions of this kind is crucial because they allow us to calculate rates of change in multiple directions.With two variables:

• \( x \) affects the square root in the numerator.
• \( y \) influences the polynomial in the denominator.

Functions of two variables can be visualized as surfaces in three dimensions where \( z \) is the output, dependent on our two inputs \( x \) and \( y \). Modifying either \( x \) or \( y \) causes changes to \( z \), showing the intricate relationship these variables share. By examining how each variable independently contributes to the change in \( z \), we gain insights into the behavior and properties of the function.

First Order Derivatives

First order derivatives are essential tools used to understand how a function changes as its inputs change. For functions of two variables, this involves partial derivatives:

• \( \frac{\partial z}{\partial x} \): Represents how \( z \) changes as \( x \) changes, keeping \( y \) constant.
• \( \frac{\partial z}{\partial y} \): Shows the change in \( z \) as \( y \) changes, with \( x \) held constant.

In our specific example, the partial derivatives calculated are:- \( \frac{\partial z}{\partial x} = \frac{2}{\sqrt{x}(3y^2 + 1)} \) - \( \frac{\partial z}{\partial y} = \frac{-24y\sqrt{x}}{(3y^2 + 1)^2} \)These derivatives reveal how \( z \), the output of the function, reacts to changes in \( x \) and \( y \) independently. Calculating first order partial derivatives is like peeling back layers of a complex function to see immediate effects of varying one variable while locking the other. This is crucial in fields like economics and physics, where understanding individual influences on a system's behavior aids in predictive modeling.