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Chapter 29: Problem 17

Find the partial derivative of the dependent variable or function with respect to each of the independent variables. $$f(x, y)=\frac{2 \sin ^{3} 2 x}{1-3 y}$$

Short Answer

Expert verified
The partial derivatives are \( \frac{12 \sin^2(2x) \cos(2x)}{1-3y} \) with respect to \( x \) and \( \frac{6 \sin^3(2x)}{(1-3y)^2} \) with respect to \( y \).

Step by step solution

01

Identify the Variables

The function given is \( f(x, y) = \frac{2 \sin^3(2x)}{1-3y} \). We need to find the partial derivative with respect to each independent variable, \( x \) and \( y \).
02

Partial Derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and differentiate \( f(x, y) \) with respect to \( x \): \[ \frac{\partial}{\partial x} \left( \frac{2 \sin^3(2x)}{1-3y} \right) = \frac{1}{1-3y} \cdot \frac{\partial}{\partial x} (2 \sin^3(2x)) \]Using the chain rule, differentiate \( 2 \sin^3(2x) \):\[ 2 \cdot 3 \sin^2(2x) \cdot \cos(2x) \cdot 2 = 12 \sin^2(2x) \cos(2x) \]Thus, the partial derivative with respect to \( x \) is:\[ \frac{12 \sin^2(2x) \cos(2x)}{1-3y} \]
03

Partial Derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant and differentiate \( f(x, y) \) with respect to \( y \): \[ \frac{\partial}{\partial y} \left( \frac{2 \sin^3(2x)}{1-3y} \right) = 2 \sin^3(2x) \cdot \frac{\partial}{\partial y} \left(\frac{1}{1-3y}\right) \]Differentiating \( \frac{1}{1-3y} \) with respect to \( y \) gives:\[ \frac{d}{dy} \left(\frac{1}{1-3y}\right) = \frac{3}{(1-3y)^2} \]Thus, the partial derivative with respect to \( y \) is:\[ \frac{6 \sin^3(2x)}{(1-3y)^2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It helps us understand how a change in one variable affects another related variable by layering different functions. To visualize, consider you have a function inside another, such as \( g(u) = u^3 \) and \( u = ext{sin}(2x) \). To find the derivative of the exterior function with respect to the innermost variable \( x \,\) you use the chain rule. Here's a simple breakdown of the chain rule steps:
  • Differentiate the outer function: find the derivative of \( g(u) = u^3 \), which gives \(3u^2\).
  • Differentiate the inner function: for \( u = ext{sin}(2x) \), the derivative is \(2 ext{cos}(2x)\).
  • Multiply these results together: you combine the derivatives to get the extrapolated effect i.e. \(3u^2 imes 2 ext{cos}(2x) = 6 ext{sin}^2(2x) ext{cos}(2x)\).
This rule is crucial when working with functions of multiple variables, helping unravel complex interaction between the variables.
Differentiation
Differentiation is the process of finding the derivative of a function, and it measures how a function changes as its input changes. In the realm of calculus, derivatives are used to find local extremes and understand the behavior of functions. Every time you see a slope or a tangent, that鈥檚 differentiation at work.The derivative of a simple function like \( f(x) = x^2 \) with respect to \( x \) is straightforward. However, in functions of multiple variables, we seek partial derivatives. These focus on how a function changes as one variable changes, holding the others constant. For example:
  • To differentiate \( f(x, y) = rac{2 ext{sin}^3(2x)}{1-3y} \) with respect to \( x \,\) treat \( y \) as a constant.
  • For the partial derivative with respect to \( y \,\) consider \( x \) as fixed.
The essence of differentiation allows calculus to deal effectively with multi-tiered relationships in functions with several variables involved.
Functions of Multiple Variables
Functions of multiple variables are functions that have more than one input, creating a multi-dimensional output. Understanding these functions is key in exploring the vast interdependencies present in real-world scenarios.In the example given, \( f(x, y) = rac{2 ext{sin}^3(2x)}{1-3y} \), both \( x \) and \( y \) are independent variables affecting the function's outcome. They allow us to explore complex behaviors not visible in single-variable functions. Key points to note when dealing with functions of multiple variables include:
  • Each input can independently change the output, hence the need for partial derivatives.
  • Visualizing these functions often involves graphs in 3D space to represent each variable's influence.
Grasping this concept means you're better equipped to conduct analyses that are true to multi-variable reality, such as optimizing conditions in engineering and economics.

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Most popular questions from this chapter

Solve the given problems. An isothermal process is one during which the temperature does not change. If the volume \(V\), pressure \(p\), and temperature \(T\) of an ideal gas are related by the equation \(p V=n R T,\) where \(n\) and \(R\) are constants, find the expression for \(\partial p / \partial V,\) which is the rate of change of pressure with respect to volume for an isothermal process.

In Exercises \(37-44,\) sketch the indicated curves and surfaces. Sketch the curve in space defined by the intersection of the surfaces \(x^{2}+(z-1)^{2}=1\) and \(z=4-x^{2}-y^{2}\)

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Solve the given problems. In a simple series electric circuit, with two resistors \(r\) and \(R\) connected across a voltage source \(E,\) the voltage \(v\) across \(r\) is \(v=r E /(r+R) .\) Assuming \(E\) to be constant, find \(\partial v / \partial r\) and \(\partial v / \partial R\).
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