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For Activities 17 through \(22,\) write the first and second partial derivatives. $$ j(x, y)=y^{2} \ln x $$

Short Answer

Expert verified
First derivatives: \( \frac{y^2}{x}, 2y \ln x \); second derivatives: \( \frac{-y^2}{x^2}, \frac{2y}{x}, 2 \ln x \).

Step by step solution

01

Identify the Function and Variables

We are given the function \( j(x, y) = y^2 \ln x \). The function depends on two variables, \( x \) and \( y \). We need to find the partial derivatives with respect to both variables.
02

Find the First Partial Derivative with respect to x

To find \( \frac{\partial j}{\partial x} \), differentiate \( j(x, y) = y^2 \ln x \) with respect to \( x \), treating \( y \) as a constant. This results in: \[ \frac{\partial j}{\partial x} = y^2 \cdot \frac{1}{x} = \frac{y^2}{x}. \]
03

Find the First Partial Derivative with respect to y

To find \( \frac{\partial j}{\partial y} \), differentiate \( j(x, y) = y^2 \ln x \) with respect to \( y \), treating \( x \) as a constant. This results in: \[ \frac{\partial j}{\partial y} = 2y \ln x. \]
04

Find the Second Partial Derivative with respect to x

Differentiate the first partial derivative \( \frac{\partial j}{\partial x} = \frac{y^2}{x} \) with respect to \( x \) again: \[ \frac{\partial^2 j}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{y^2}{x} \right) = \frac{-y^2}{x^2}. \]
05

Find the Mixed Partial Derivative with respect to y then x

Differentiate \( \frac{\partial j}{\partial y} = 2y \ln x \) with respect to \( x \): \[ \frac{\partial^2 j}{\partial y \partial x} = \frac{2y}{x}. \] This is same as the derivative of \( \frac{\partial j}{\partial x} = \frac{y^2}{x} \) with respect to \( y \): \[ \frac{\partial^2 j}{\partial x \partial y} = \frac{2y}{x}. \]
06

Verify Second Mixed Partial Derivative

The second mixed partial derivatives, \( \frac{\partial^2 j}{\partial x \partial y} = \frac{2y}{x} \) and \( \frac{\partial^2 j}{\partial y \partial x} = \frac{2y}{x} \), are indeed equal, satisfying Clairaut's theorem, which confirms correctness.
07

Find the Second Partial Derivative with respect to y

Differentiate the first partial derivative \( \frac{\partial j}{\partial y} = 2y \ln x \) with respect to \( y \) again: \[ \frac{\partial^2 j}{\partial y^2} = 2 \ln x. \]

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that focuses on studying change. It provides the tools necessary to understand how quantities vary with one another.
In our example, we are dealing with the function \( j(x, y) = y^2 \ln x \). Here, calculus helps us find how this function changes as the variables \( x \) and \( y \) vary.

In calculus, derivatives are fundamental. They represent the rate of change of a function with respect to one of its variables. For a multivariable function like \( j(x, y) \), this means finding partial derivatives to understand how the function changes with respect to each variable separately.
Understanding these changes helps us grasp various concepts such as slopes of curves and surfaces, which are central to calculus.
Multivariable Functions
Multivariable functions involve more than one input variable. In our example, \( j(x, y) = y^2 \ln x \) is a function of two variables, \( x \) and \( y \).
These functions are essential when dealing with systems where multiple factors vary simultaneously.

Analyzing multivariable functions helps in various fields including physics, engineering, and economics, where multiple factors influence outcomes.
The critical concept here is the ability to explore how a change in one variable impacts the entire function while keeping other variables constant. This understanding comes through partial derivatives, which provide insight into how each variable uniquely contributes to the overall behavior of the function.
Clairaut's Theorem
Clairaut's Theorem is a remarkable result in the calculus of multivariable functions. It states that for a function \( f \) with continuous second partial derivatives, the mixed partial derivatives are equal.
In mathematical terms, \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \).

This theorem simplifies the process of differentiation, saving time and effort. For our function, we calculate mixed partial derivatives \( \frac{\partial^2 j}{\partial x \partial y} \) and \( \frac{\partial^2 j}{\partial y \partial x} \) and find they are the same, thus confirming Clairaut's theorem.
Understanding this theorem fosters deeper insights into the symmetry of differentiable functions, providing a reliable method to cross-verify calculations and ensure their correctness.
Differentiation Techniques
Differentiation techniques refer to strategies used to find derivatives. In our example, we apply these techniques to find partial derivatives.
Each step involves carefully identifying the variable of interest and treating other variables as constants.

For the first partial derivatives, we examined how the function changes with respect to \( x \) and \( y \). This involves using known formulas and rules such as the power rule and properties of logarithms.
For second partial derivatives, the process repeats using the results of the first derivatives, which may involve more complex calculations and simplifications.
Mastery of these techniques is crucial for anyone working with calculus, allowing for effective analysis and interpretation of complex, multivariable systems.

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Most popular questions from this chapter

For Activities 9 through 12 a. Calculate the output associated with the given input values. b. Approximate the change needed in one input variable to compensate for the given change in the other input variable. $$ \begin{array}{l} f(m, n)=3 m^{2}+2 m n+5 n^{2} \text { when } m=2, n=1, \text { and } \\ \Delta m=0.2 \end{array} $$

For Activities 9 through \(16,\) write formulas for the indicated partial derivatives for each of the multivariable functions. \(g(k, m)=k^{3} m^{5}-2 k m\) a. \(g_{k}\) b. \(g_{m}\) c. \(\left.g_{m}\right|_{k=2}\)

Future Value The value \(F(t, r)\) of an investment of \(\$ 1000\) after \(t\) years in an account for which the interest rate \(100 r \%\) is compounded continuously is given by the function \(F(t, r)=1000 e^{r t}\) dollars. a. Write the partial derivatives \(\frac{\partial F}{\partial t}\) and \(\frac{\partial F}{\partial r}\). b. Write each of the second partial derivative formulas and interpret them for \(t=30\) and \(r=0.047 .\)

For Activities 1 through 6 a. Write the mathematical notation for the partial rate-ofchange function needed to answer the question posed. b. Write the units of measure for that rate-of-change function. \(t(x, y)\) degrees Fahrenheit is the mean daily temperature at longitude \(x\) degrees and latitude \(y\) degrees. If longitude is \(23^{\circ}\) and latitude is changing, how rapidly is temperature changing?

Future Value The value \(F(P, r)\) of an investment of \(P\) dollars after 2 years in an account with annual percentage yield \(100 r \%\) is given by the function \(F(P, r)=P(1+r)^{2}\) dollars. a. Write the first partial derivatives of \(F\). b. Write each of the second partial derivative formulas and interpret them for \(P=10,000\) and \(r=0.09 .\)

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