Question

Find all the second partial derivatives in each of the following cases: (a) \(z=8 \mathrm{e}^{x y}\) (b) \(z=-3 \mathrm{e}^{x} \sin y\) (c) \(z=4 \mathrm{e}^{y} \cos x\)

Short answer

Question: Find the second partial derivatives of the following functions: (a) \(z=8e^{xy}\) (b) \(z=-3e^{x}\sin{y}\) (c) \(z=4e^{y}\cos{x}\) Answer: (a) \(\frac{\partial^2 z}{\partial x^2} = 8y^2e^{xy}\), \(\frac{\partial^2 z}{\partial x \partial y} = 8e^{xy}(x+y)\), \(\frac{\partial^2 z}{\partial y^2} = 8x^2e^{xy}\) (b) \(\frac{\partial^2 z}{\partial x^2} = -3e^x \sin{y}\), \(\frac{\partial^2 z}{\partial x \partial y} = -3e^x \cos{y}\), \(\frac{\partial^2 z}{\partial y^2} = 3e^x \sin{y}\) (c) \(\frac{\partial^2 z}{\partial x^2} = -4e^{y} \cos{x}\), \(\frac{\partial^2 z}{\partial x \partial y} = -4e^{y} \sin{x}\), \(\frac{\partial^2 z}{\partial y^2} = 4e^{y} \cos{x}\)

Step-by-step solution

Case (a): \(z=8e^{xy}\)#

First, find the first partial derivatives with respect to x and y: \(\frac{\partial z}{\partial x} = 8ye^{xy}\) \(\frac{\partial z}{\partial y} = 8xe^{xy}\) Now, find the second partial derivatives: \(\frac{\partial^2 z}{\partial x^2} = 8y^2e^{xy}\) \(\frac{\partial^2 z}{\partial x \partial y} = 8ye^{xy} + 8xe^{xy}\) \(\frac{\partial^2 z}{\partial y^2} = 8x^2e^{xy}\) So, the second partial derivatives for case (a) are: \(\frac{\partial^2 z}{\partial x^2} = 8y^2e^{xy}\) \(\frac{\partial^2 z}{\partial x \partial y} = 8e^{xy}(x+y)\) \(\frac{\partial^2 z}{\partial y^2} = 8x^2e^{xy}\)

Case (b): \(z=-3e^{x}\sin{y}\)#

First, find the first partial derivatives with respect to x and y: \(\frac{\partial z}{\partial x} = -3e^x \sin{y}\) \(\frac{\partial z}{\partial y} = -3e^x \cos{y}\) Now, find the second partial derivatives: \(\frac{\partial^2 z}{\partial x^2} = -3e^x \sin{y}\) \(\frac{\partial^2 z}{\partial x \partial y} = -3e^x \cos{y}\) \(\frac{\partial^2 z}{\partial y^2} = 3e^x \sin{y}\) So, the second partial derivatives for case (b) are: \(\frac{\partial^2 z}{\partial x^2} = -3e^x \sin{y}\) \(\frac{\partial^2 z}{\partial x \partial y} = -3e^x \cos{y}\) \(\frac{\partial^2 z}{\partial y^2} = 3e^x \sin{y}\)

Case (c): \(z=4e^{y}\cos{x}\)#

First, find the first partial derivatives with respect to x and y: \(\frac{\partial z}{\partial x} = -4e^{y} \sin{x}\) \(\frac{\partial z}{\partial y} = 4e^{y} \cos{x}\) Now, find the second partial derivatives: \(\frac{\partial^2 z}{\partial x^2} = -4e^{y} \cos{x}\) \(\frac{\partial^2 z}{\partial x \partial y} = -4e^{y} \sin{x}\) \(\frac{\partial^2 z}{\partial y^2} = 4e^{y} \cos{x}\) So, the second partial derivatives for case (c) are: \(\frac{\partial^2 z}{\partial x^2} = -4e^{y} \cos{x}\) \(\frac{\partial^2 z}{\partial x \partial y} = -4e^{y} \sin{x}\) \(\frac{\partial^2 z}{\partial y^2} = 4e^{y} \cos{x}\)

Key concepts

Partial Differentiation

Partial differentiation is a fundamental technique in calculus, especially useful when dealing with functions of several variables. It involves finding the derivative of a function with respect to one variable while keeping the others constant. Consider a student facing a problem involving a multivariable function like \(z = f(x,y)\), where both \(x\) and \(y\) influence the value of \(z\). The first partial derivatives, denoted as \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), measure the rate of change of \(z\) with respect to \(x\) and \(y\) respectively, while each is varied independently of the other.
When tackling second partial derivatives, the process delves deeper, involving taking the derivative of the first partial derivatives. Essentially, you are examining how the rate of change itself is changing. In case (a) of the given exercise, \(\frac{\partial^2 z}{\partial x^2}\) reveals how the rate at which \(z\) changes with respect to \(x\) is affected by further changes in \(x\). Similarly, \(\frac{{\text{partial}}}^2 z}{\partial x \partial y}\) shows how a change in \(y\) influences the rate of change of \(z\) with respect to \(x\), which is often referred to as the mixed partial derivative. As evidenced from the solution process, understanding each step from first to second partial derivatives enhances comprehension and guarantees correct application in more complex scenarios of multivariable calculus.

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to higher dimensions. Students are often introduced to this branch of calculus when they encounter functions like \(z = f(x, y)\), as shown in the exercise. This broadens their computational toolkit, enabling the analysis of systems where multiple variables interact simultaneously. A key feature of multivariable calculus is that it can describe surfaces and curves in three dimensions and beyond, which is crucial for advanced fields such as physics, engineering, and economics.
In multivariable calculus, the concept of partial differentiation plays a central role, particularly when calculating the curvature or bending of surfaces, which is explored through second partial derivatives. The exercise showcases different mathematical functions and commands the application of the second partial derivative test. This test can determine whether a given point is a local minimum, local maximum, or saddle point by analyzing the sign and value of second partial derivatives, greatly enriching a student's understanding of the topography of multivariable functions.

Mathematical Functions

Mathematical functions are the building blocks of calculus and are essential in describing the relationship between variables. In the context of the given exercise, the functions include exponential and trigonometric expressions combined with multivariable dependency. Each function type, such as \(8e^{xy}\), \( -3e^{x}\text{sin} y\), and \(4e^{y}\text{cos} x\), has its own set of rules for differentiation that students must apply correctly when solving for first and second partial derivatives.
These functions illustrate the diverse ways in which variables can be related and the complexity that can arise in multivariable calculus. By working through exercises like these, students gain a deeper understanding of how to manipulate and compute derivatives of functions involving not just polynomials but also more complex relationships involving exponentials and trigonometric functions. Mastery of these concepts is crucial as they form the foundation for further studies in mathematical analysis, applied mathematics, and other STEM fields.