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Chapter 12: Problem 22

Find the first partial derivatives of the following functions. $$g(x, z)=x \ln \left(z^{2}+x^{2}\right)$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function $$g(x, z) = x \ln\left(z^2 + x^2\right)$$ with respect to $$x$$ and $$z$$. Answer: The first partial derivatives of the given function are: $$\frac{\partial g}{\partial x} = \ln\left(z^2 + x^2\right) + x \cdot \frac{2x}{z^2 + x^2}$$ $$\frac{\partial g}{\partial z} = x \cdot \frac{2z}{z^2 + x^2}$$

Step by step solution

01

Compute the partial derivative with respect to x

To compute the partial derivative of $$g(x, z)$$ with respect to $$x$$, we treat $$z$$ as a constant. We'll use the product rule of differentiation, which states that the derivative of a product of two functions is the derivative of the first function multiplied by the second function plus the first function multiplied by the derivative of the second function: $$\frac{\partial g}{\partial x} = \frac{\partial}{\partial x}(x \ln\left(z^2 + x^2\right))$$ Applying the product rule, we get: $$\frac{\partial g}{\partial x} = 1 \cdot \ln\left(z^2 + x^2\right) + x \cdot \frac{\partial}{\partial x}(\ln\left(z^2 + x^2\right))$$ Now, we need to find the derivative of the natural logarithm part: $$\frac{\partial}{\partial x}(\ln\left(z^2 + x^2\right)) = \frac{1}{z^2 + x^2} \cdot \frac{\partial}{\partial x}(z^2 + x^2)$$ The derivative of $$z^2 + x^2$$ is: $$\frac{\partial}{\partial x}(z^2 + x^2) = 0 + 2x$$ Substitute this back into the equation: $$\frac{\partial g}{\partial x} = \ln\left(z^2 + x^2\right) + x \cdot \frac{2x}{z^2 + x^2}$$
02

Compute the partial derivative with respect to z

Next, we'll compute the partial derivative of $$g(x, z)$$ with respect to $$z$$, treating $$x$$ as a constant: $$\frac{\partial g}{\partial z} = \frac{\partial}{\partial z}(x \ln\left(z^2 + x^2\right))$$ Since $$x$$ is constant, we can bring it out of the derivative: $$\frac{\partial g}{\partial z} = x \cdot \frac{\partial}{\partial z}(\ln\left(z^2 + x^2\right))$$ Now, we need to find the derivative of the natural logarithm part: $$\frac{\partial}{\partial z}(\ln\left(z^2 + x^2\right)) = \frac{1}{z^2 + x^2} \cdot \frac{\partial}{\partial z}(z^2 + x^2)$$ The derivative of $$z^2 + x^2$$ is: $$\frac{\partial}{\partial z}(z^2 + x^2) = 2z + 0$$ Substitute this back into the equation: $$\frac{\partial g}{\partial z} = x \cdot \frac{2z}{z^2 + x^2}$$
03

Final answer

The first partial derivatives of the given function are: $$\frac{\partial g}{\partial x} = \ln\left(z^2 + x^2\right) + x \cdot \frac{2x}{z^2 + x^2}$$ $$\frac{\partial g}{\partial z} = x \cdot \frac{2z}{z^2 + x^2}$$

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Key Concepts

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

Product Rule in Calculus
The Product Rule is a fundamental tool in calculus, especially when dealing with derivatives of functions that are the product of two or more functions. In its simplest form, it states that if you have a function that is a product of two functions, say \( u(x) \) and \( v(x) \), then the derivative of their product is:
  • The derivative of the first function \( u'(x) \)
  • times the second function \( v(x) \)
  • plus the first function \( u(x) \)
  • times the derivative of the second function \( v'(x) \)
This can be written as:\[(uv)' = u'v + uv'\]In the context of partial derivatives, like those of the function \( g(x, z) = x \ln(z^2 + x^2) \), we apply the product rule by treating one of the variables as a constant. For instance, when finding the partial derivative with respect to \( x \), we treat \( z \) as a constant, and vice versa. This application helps us successfully find the partial derivatives by breaking down the function into manageable parts that involve products.
Understanding Natural Logarithms
The natural logarithm, represented as \( \ln \), is the logarithm to the base of Euler's number \( e \), which is approximately 2.718. This type of logarithm appears frequently in calculus and its applications because of its natural properties related to growth rates and exponential functions.
  • Firstly, it's important to know that \( \ln(e) = 1 \) because the logarithm represents the power to which the base must be raised to produce a given number.
  • Secondly, the derivative of \( \ln(x) \) is straightforward: it's \( 1/x \).
When dealing with expressions like \( \ln(z^2 + x^2) \) in multivariable calculus, you need to be familiar with these properties. Applying derivatives to logarithms often requires the chain rule, especially if the argument of the logarithm is more complex than a single variable. In our given problem, we see this when we differentiate \( \ln(z^2 + x^2) \) with respect to either \( x \) or \( z \). Understanding these core properties of logarithms simplifies the process of finding derivatives in multivariable functions.
Multivariable Calculus Concepts
Multivariable Calculus is an extension of calculus to more than one variable. It deals with functions that have multiple variables and allows us to explore complex situations that involve varying more than one parameter simultaneously.
  • Partial derivatives are a central concept. They represent the derivative of a function with respect to one variable while keeping other variables constant.
  • For example, given \( g(x, z) = x \ln(z^2 + x^2) \), computing \( \frac{\partial g}{\partial x} \) means finding the change in the function as \( x \) changes, with \( z \) held constant.
  • Similarly, \( \frac{\partial g}{\partial z} \) measures the rate of change as \( z \) changes with \( x \) being fixed.
Understanding these partial derivatives requires leveraging rules from single-variable calculus, like the product rule and chain rule, to accommodate functions of several variables. By understanding the dynamics of each variable independently, we can draw insights into how multivariable functions behave and interact. This knowledge is crucial for applications in fields like economics, physics, and engineering.

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Most popular questions from this chapter

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 that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) 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 that 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, 2nd ed., Philip Gillett, 1984)

Use the definition of the gradient (in two or three dimensions), assume that \(f\) and \(g\) are differentiable functions on \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\), and let \(c\) be a constant. Prove the following gradient rules. a. Constants Rule: \(\nabla(c f)=c \nabla f\) b. Sum Rule: \(\nabla(f+g)=\nabla f+\nabla g\) c. Product Rule: \(\nabla(f g)=(\nabla f) g+f \nabla g\) d. Quotient Rule: \(\nabla\left(\frac{f}{g}\right)=\frac{g \nabla f-f \nabla g}{g^{2}}\) e. Chain Rule: \(\nabla(f \circ g)=f^{\prime}(g) \nabla g,\) where \(f\) is a function of one variable

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).

In the advanced subject of complex variables, a function typically has the form \(f(x, y)=u(x, y)+i v(x, y),\) where \(u\) and \(v\) are real-valued functions and \(i=\sqrt{-1}\) is the imaginary unit. A function \(f=u+i v\) is said to be analytic (analogous to differentiable) if it satisfies the Cauchy-Riemann equations: \(u_{x}=v_{y}\) and \(u_{y}=-v_{x}\). a. Show that \(f(x, y)=\left(x^{2}-y^{2}\right)+i(2 x y)\) is analytic. b. Show that \(f(x, y)=x\left(x^{2}-3 y^{2}\right)+i y\left(3 x^{2}-y^{2}\right)\) is analytic. c. Show that if \(f=u+i v\) is analytic, then \(u_{x x}+u_{y y}=0\) and \(v_{x x}+v_{y y}=0 .\) Assume \(u\) and \(v\) satisfy the conditions in Theorem 12.4.

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). $$u(x, t)=10 e^{-t} \sin x$$
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