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Chapter 12: Problem 34

Find the four second partial derivatives of the following functions. $$f(x, y)=\cos x y$$

Short Answer

Expert verified
Question: Find the four second partial derivatives of the function $$f(x, y) = \cos(xy)$$. Answer: The four second partial derivatives of the given function are: 1. $$\frac{\partial^2 f}{\partial x^2} = -(y^2\cos(xy))$$ 2. $$\frac{\partial^2 f}{\partial x \partial y} = -\cos(xy) + xy\sin(xy)$$ 3. $$\frac{\partial^2 f}{\partial y \partial x} = -\cos(xy) + xy\sin(xy)$$ 4. $$\frac{\partial^2 f}{\partial y^2} = -(x^2\cos(xy))$$

Step by step solution

01

(Step 1: Find the first partial derivatives)

(Calculate the first partial derivatives of the function with respect to x and y: $$\frac{\partial f}{\partial x} = -y\sin(xy)$$ $$\frac{\partial f}{\partial y} = -x\sin(xy)$$ )
02

(Step 2: Find the second partial derivatives)

(Now, calculate the second partial derivatives using the first partial derivatives. There are four possible combinations: 1. Differentiate the first partial derivative with respect to x: $$\frac{\partial^2 f}{\partial x^2} = -(y^2\cos(xy))$$ 2. Differentiate the first partial derivative with respect to y: $$\frac{\partial^2 f}{\partial x \partial y} = -\cos(xy) + xy\sin(xy)$$ 3. Differentiate the second partial derivative with respect to x: $$\frac{\partial^2 f}{\partial y \partial x} = -\cos(xy) + xy\sin(xy)$$ 4. Differentiate the second partial derivative with respect to y: $$\frac{\partial^2 f}{\partial y^2} = -(x^2\cos(xy))$$ ) These are the four second partial derivatives of the given function: $$\frac{\partial^2 f}{\partial x^2} = -(y^2\cos(xy))$$ $$\frac{\partial^2 f}{\partial x \partial y} = -\cos(xy) + xy\sin(xy)$$ $$\frac{\partial^2 f}{\partial y \partial x} = -\cos(xy) + xy\sin(xy)$$ $$\frac{\partial^2 f}{\partial y^2} = -(x^2\cos(xy))$$

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Key Concepts

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

Partial Differentiation
Partial differentiation is a fundamental technique in calculus used when dealing with multivariable functions, that is, functions with more than one input. In contrast to ordinary differentiation, which considers the rate of change of a function with respect to one variable, partial differentiation focuses on the rate of change with respect to one variable while keeping all other variables constant.

Think of partial derivatives as a way to explore a multivariable landscape: if you want to understand how steep the hill is going in the north direction, disregarding any changes in the east-west direction, you would use partial differentiation. In the exercise, we see the partial derivatives of the function \( f(x,y) = \text{cos}(xy) \) with respect to both \( x \) and \( y \) are computed, giving us insight into how the function's value changes as either \( x \) or \( y \) changes independently.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It includes partial differentiation, as well as multiple integration and topics in vector analysis. In this larger context, partial derivatives play a critical role in understanding the geometry and behavior of multivariable functions.

Just like you can analyze curves in single-variable calculus, in multivariable calculus, you analyze surfaces and higher-dimensional analogues. The second partial derivatives that we calculated in our exercise are similar to the second derivative in single-variable calculus, which can reveal information about the concavity and points of inflection of the function.
Mixed Partial Derivatives
Mixed partial derivatives refer to the second partial derivatives of a function with respect to two different variables. When computing these, the order of differentiation matters. In theory, if a function is nice enough (technically speaking, if it鈥檚 continuously differentiable), the mixed partial derivatives should be equal regardless of the order of differentiation, which is known as Clairaut's theorem.

In our exercise example, we derived \( \frac{\partial^{2} f}{\partial x \partial y} \) and \( \frac{\partial^{2} f}{\partial y \partial x} \), both of which resulted in the same expression, \( -\text{cos}(xy) + xy\text{sin}(xy) \), thus satisfying Clairaut's theorem. This tells us about how the cross-relationships between variables behave when both are changing.
Cosine Function Differentiation
The cosine function, denoted by \( \text{cos} \), is one of the trigonometric functions and appears frequently in calculus, particularly when dealing with oscillations and waves. Differentiating a cosine function follows the basic rules of differentiation, but with trigonometric nuances.

As shown in the exercise, the derivative of \( \text{cos}(xy) \) with respect to \( x \) involves bringing down the \( y \) as per the chain rule and results in \( -y\text{sin}(xy) \). Similarly, when differentiated with respect to \( y \), it produces \( -x\text{sin}(xy) \). These derivatives encapsulate the rate at which the cos function's value is changing as each of the variables changes, which is essential in many physical applications, from engineering to physics.

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Most popular questions from this chapter

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$-\frac{x^{2}}{6}-24 y^{2}+\frac{z^{2}}{24}-6=0$$

Use the definition of differentiability to prove that the following functions are differentiable at \((0,0) .\) You must produce functions \(\varepsilon_{1}\) and \(\varepsilon_{2}\) with the required properties. $$f(x, y)=x+y$$

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 \(k=1\). $$u(x, t)=e^{-t}(2 \sin x+3 \cos x)$$

Let \(E\) be the ellipsoid \(x^{2} / 9+y^{2} / 4+z^{2}=1, P\) be the plane \(z=A x+B y,\) and \(C\) be the intersection of \(E\) and \(P\). a. Is \(C\) an ellipse for all values of \(A\) and \(B\) ? Explain. b. Sketch and interpret the situation in which \(A=0\) and \(B \neq 0\) c. Find an equation of the projection of \(C\) on the \(x y\) -plane. d. Assume \(A=\frac{1}{6}\) and \(B=\frac{1}{2} .\) Find a parametric description of \(C\) as a curve in \(\mathbb{R}^{3}\). (Hint: Assume \(C\) is described by \(\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle\) and find \(a, b, c, d, e, \text { and } f .)\)

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=\tan ^{-1}\left(\frac{y}{x-1}\right)-\tan ^{-1}\left(\frac{y}{x+1}\right)$$
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