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Find the first partial derivatives of the function. $$ f(x, y)=\frac{2}{3} x^{3 / 2} $$

Short Answer

Expert verified
The partial derivatives are \(x^{1/2}\) with respect to \(x\) and 0 with respect to \(y\).

Step by step solution

01

Identify the Function and Variables

The given function is \( f(x, y) = \frac{2}{3} x^{3/2} \). This is a function of two variables, \(x\) and \(y\). However, it's important to note that the function explicitly depends only on \(x\), and \(y\) does not appear in the expression.
02

Find the Partial Derivative with Respect to x

To find the partial derivative of \( f(x, y) \) with respect to \(x\), we need to differentiate \( f(x, y) = \frac{2}{3} x^{3/2} \) concerning \(x\). Using the power rule: \[ \frac{d}{dx} x^{n} = n x^{n-1} \]we get:\[ \frac{\partial f}{\partial x} = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{3}{2} - 1} = x^{\frac{1}{2}} \].
03

Find the Partial Derivative with Respect to y

Since \( f(x, y) = \frac{2}{3} x^{3/2} \) does not contain the variable \(y\), differentiating with respect to \(y\) will yield zero. Thus, the partial derivative of \( f \) with respect to \(y\) is:\[ \frac{\partial f}{\partial y} = 0 \].

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Key Concepts

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

Power Rule
The power rule is a fundamental aspect of calculus, especially in dealing with derivatives. It provides a straightforward method for finding the derivative of a power function, which has the form \( x^n \). This rule states that if you have a function \( x^n \), the derivative is \( n \cdot x^{n-1} \). This makes differentiation much easier!

Imagine you need to differentiate \( f(x) = x^3 \). Using the power rule, we simply bring down the exponent 3 in front and reduce the exponent by 1. So the derivative is \( 3 \cdot x^{3-1} = 3x^2 \).

This rule is versatile and applies to any real number exponent, whether it's an integer, fraction, or even negative. In our original function \( f(x, y) = \frac{2}{3} x^{3/2} \), we utilize the power rule to find the partial derivative with respect to \( x \). By recognizing \( 3/2 \) as the exponent, we apply the power rule to get \( x^{1/2} \) after multiplying and simplifying the terms.
Differentiation
Differentiation is the process of finding the derivative—essentially, a measure of how a function changes as its inputs change. This is a key tool in calculus, allowing us to understand the rate of change or the slope of the function at any given point.

When you're working with a single variable, let's say \( x \), differentiation involves calculating \( \frac{d}{dx} \) of the function. But when dealing with functions of multiple variables, like \( f(x, y) \), we use partial derivatives. These are similar to regular derivatives but focus on the effect of changing one variable while keeping the others constant.

For example, to determine how \( f(x, y) = \frac{2}{3} x^{3/2} \) changes with \( x \), we compute \( \frac{\partial f}{\partial x} \), treating \( y \) as a constant. Since \( y \) doesn't appear in the function, its partial derivative concerning \( y \) is zero, as the function doesn't change with respect to \( y \). This tells us exactly how each variable affects the function's outcome when varied independently.
Functions of Multiple Variables
Functions of multiple variables are prevalent in math and the real world, where outcomes rely on several inputs. Such functions can generally be expressed as \( f(x, y, z,...) \), indicating dependence on two or more variables.

These functions are useful when modeling realistic scenarios, where several factors affect the result simultaneously. For instance, a function dependent on temperature and humidity might model weather conditions.

When analyzing such functions, partial derivatives are vital. They allow us to examine how the function changes with respect to one specific variable, keeping others constant. This forms the basis for understanding many real-world applications, from economics to physics.

In our example problem, \( f(x, y) = \frac{2}{3} x^{3/2} \), the function is presented with two variables, \( x \) and \( y \). However, it only actually depends on \( x \). This simplifies our differentiation task since altering \( y \) has no impact, reflected by a zero partial derivative with respect to \( y \). Understanding this concept helps focus our efforts on the parts of the function that matter, leading to clearer insights and solutions.

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Most popular questions from this chapter

The ground state energy \(E(x, y, z)\) of a particle of mass \(m\) in a rectangular box with dimensions \(x, y\), and \(z\) is given by $$ E(x, y, z)=\frac{h^{2}}{8 m}\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}\right) $$ where \(h\) is a constant. Assuming that the volume \(V\) of the box is fixed, find the values of \(x, y\), and \(z\) that minimize the value of \(E\).

The mass of a rocket lifting off from earth is decreasing (due to fuel consumption) at the rate of 40 kilograms per second. How fast is the magnitude \(F\) of the force of gravity decreasing when the rocket is 6400 kilometers from the center of the earth and is rising with a velocity of 100 kilometers per second? (Hint: By Newton's Law of Gravitation, \(F=G M m / r^{2}\), where \(G\) is the universal gravitational constant, \(M\) is the mass of the earth, \(m\) is the mass of the rocket, and \(r\) is the distance between the rocket and the center of the earth.)

Find the extreme values of \(f\) in the region described by the given inequalities. In each case assume that the extreme values exist. $$ f(x, y)=x y ; 2 x^{2}+y^{2} \leq 4 $$

Let \(f\) be a function of two variables with partials of all orders. Then \(f\) has four second partials, \(f_{x x}, f_{x y}, f_{y x}\), and \(f_{y y} .\) If the second partials are continuous, only three of them can be distinct. a. How many third-order partials does \(f\) have? If these are continuous, what is the maximum number that can be distinct? b. Generalize (a) to \(n\) th-order partials. (Hint: Observe that if all \(n\) th- order partials are continuous, then any \(n\) th-order partial is equal to another \(n\) th-order partial in which all the differentiations with respect to \(x\) are performed first and the differentiations with respect to \(y\) are done last.)

Use implicit differentiation to find \(\partial z / \partial x\) and \(\partial z / \partial y\) at the given point. Then find an equation of the plane tangent to the level surface at that point. $$ \ln x+\ln y+\ln z=1 ;(1,1, e) $$

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