/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Find \(f_{x}\) and \(f_{y}\) whe... [FREE SOLUTION] | 91Ó°ÊÓ

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Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=y^{8}+2 x^{6}+2 x y\).

Short Answer

Expert verified
The partial derivatives of the function are: \(f_x = 12x^5 + 2y\) \(f_y = 8y^7 + 2x\)

Step by step solution

01

Recall how to find partial derivatives

To find the partial derivative of a function, we treat one variable as a constant while differentiating with respect to the other. The partial derivative of the function \(f(x, y)\) with respect to \(x\) is denoted as \(f_x\) and with respect to \(y\) is denoted as \(f_y\).
02

Find the partial derivative \(f_x\)

In this step, we will treat \(y\) as a constant and differentiate the function \(f(x, y) = y^8 + 2x^6 + 2xy\) with respect to \(x\). Using the sum rule and the power rule, we get: \(f_x = \frac{\partial}{\partial x}\left( y^8 \right) + \frac{\partial}{\partial x}\left( 2x^6 \right) + \frac{\partial}{\partial x}\left( 2xy \right)\) Now, apply the power rule and constant rule to each term: - \(\frac{\partial}{\partial x}\left( y^8 \right) = 0\), since \(y^8\) is a constant with respect to \(x\). - \(\frac{\partial}{\partial x}\left( 2x^6 \right) = 12x^5\), using the power rule. - \(\frac{\partial}{\partial x}\left( 2xy \right) = 2y\), since the derivative of \(x\) is 1 and we leave \(y\) as a constant. Now, combining all the terms: \(f_x = 0 + 12x^5 + 2y\)
03

Find the partial derivative \(f_y\)

This time, we will treat \(x\) as a constant and differentiate the function \(f(x, y) = y^8 + 2x^6 + 2xy\) with respect to \(y\). Using the sum rule and the power rule, we get: \(f_y = \frac{\partial}{\partial y}\left( y^8 \right) + \frac{\partial}{\partial y}\left( 2x^6 \right) + \frac{\partial}{\partial y}\left( 2xy \right)\) Now, apply the power rule and constant rule to each term: - \(\frac{\partial}{\partial y}\left( y^8 \right) = 8y^7\), using the power rule. - \(\frac{\partial}{\partial y}\left( 2x^6 \right) = 0\), since \(2x^6\) is a constant with respect to \(y\). - \(\frac{\partial}{\partial y}\left( 2xy \right) = 2x\), since the derivative of \(y\) is 1 and we leave \(x\) as a constant. Now, combining all the terms: \(f_y = 8y^7 + 0 + 2x\)
04

Write the answer

The partial derivatives of the function \(f(x, y) = y^8 + 2x^6 + 2xy\) are: \(f_x = 12x^5 + 2y\) \(f_y = 8y^7 + 2x\)

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Key Concepts

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

Multivariable Calculus
In the realm of calculus, multivariable calculus is an extension of single-variable calculus to higher dimensions. It involves functions that have more than one variable, like the example function, which depends on both variables, 'x' and 'y'. The goal in this branch of mathematics is to analyze changes in a function when there are multiple inputs. This is particularly useful in fields like physics, engineering, economics, and any other area that deals with systems influenced by multiple factors at once.

When dealing with such functions, the concept of partial derivatives comes into play. Unlike the standard derivative of a single-variable function, a partial derivative tells you how the function changes as one of the variables changes, while all other variables are held constant. Understanding how to compute and interpret partial derivatives is a foundational skill in multivariable calculus, as you learn not just how a function moves, but how it moves in multiple dimensions simultaneously. This multidimensional movement can be critical for optimizing functions, understanding physical laws, or even predicting economic trends when multiple variables are involved.
Power Rule
A commonly used technique in calculus is the power rule, which simplifies taking the derivative of a function that is a power of a variable. When you have a term like \(x^n\), where 'n' is any real number, the power rule states that the derivative of that term with respect to 'x' is \(nx^{n-1}\).

Applied to multivariable functions, the power rule must be paired with the understanding that the variable with respect to which you are differentiating must be treated differently than the others. So, in the context of partial derivatives for multivariable functions, if you have a function like \( f(x, y) = x^6 \), the partial derivative with respect to 'x' is found by applying the power rule directly, getting \( 6x^5 \), while the partial derivative with respect to 'y' is zero because the term does not depend on 'y'.

It’s essential to recognize that each term in the function must be differentiated separately, and the power rule simplifies this process greatly when terms of the function are in the form of a variable raised to a power.
Sum Rule
Another fundamental concept in calculus is the sum rule. This rule is particularly straightforward: when finding the derivative of a function that is the sum of several terms, you simply take the derivative of each term individually and then sum those derivatives. The sum rule applies just as well to partial derivatives as it does to regular derivatives.

In our example with the function \(f(x, y) = y^8 + 2x^6 + 2xy\), we used the sum rule to break the problem into manageable parts. We differentiated \(y^8\), \(2x^6\), and \(2xy\) with respect to 'x' and 'y' separately. This made the process easier and more methodical, as opposed to trying to differentiate the entire function at once. By understanding and applying the sum rule, you can simplify complex derivative problems into a series of easier computations, which is a valuable technique not only in mathematics but also in breaking down complex problems in real-world applications.

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Most popular questions from this chapter

Challenge domains Find the domain of the following functions. Specify the domain mathematically, and then describe it in words or with a sketch. $$f(x, y, z)=\ln \left(z-x^{2}-y^{2}+2 x+3\right)$$

The closed unit ball in \(\mathbb{R}^{3}\) centered at the origin is the set \(\left\\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 1\right\\} .\) Describe the following alternative unit balls. a. \(\\{(x, y, z):|x|+|y|+|z| \leq 1\\}\) b. \(\\{(x, y, z): \max \\{|x|,|y|,|z|\\} \leq 1\\},\) where \(\max \\{a, b, c\\}\) is the maximum value of \(a, b,\) and \(c\)

Use Lagrange multipliers in the following problems. When the constraint curve is unbounded, explain why you have found an absolute maximum or minimum value. Maximum perimeter rectangle in an ellipse Find the dimensions of the rectangle of maximum perimeter with sides parallel to the coordinate axes that can be inscribed in the ellipse \(2 x^{2}+4 y^{2}=3\)

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) $$ \text { b. } f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}} $$ This property has the following interpretation. Suppose a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May \(1985,\) and Calculus and Analytical Geometry, 2 nd ed., Philip Gillett, 1984 )

Extreme points on flattened spheres The equation \(x^{2 n}+y^{2 n}+z^{2 n}=1,\) where \(n\) is a positive integer, describes a flattened sphere. Define the extreme points to be the points on the flattened sphere with a maximum distance from the origin. a. Find all the extreme points on the flattened sphere with \(n=2 .\) What is the distance between the extreme points and the origin? b. Find all the extreme points on the flattened sphere for integers \(n>2 .\) What is the distance between the extreme points and the origin? c. Give the location of the extreme points in the limit as \(n \rightarrow \infty\) What is the limiting distance between the extreme points and the origin as \(n \rightarrow \infty ?\)

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