Question

Find all the second partial derivatives in each of the following cases: (a) \(z=\frac{1}{x}\) (b) \(z=\frac{y}{x}\) (c) \(z=\frac{x}{y}\) (d) \(z=\frac{1}{x}+\frac{1}{y}\)

Short answer

Question: Find the second partial derivatives for the following functions: (a) \(z = \frac{1}{x}\), (b) \(z = \frac{y}{x}\), (c) \(z = \frac{x}{y}\), and (d) \(z = \frac{1}{x} + \frac{1}{y}\). Answer: For (a) we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\). For (b) we have: \(\frac{\partial^2 z}{\partial x^2} = \frac{2y}{x^3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{x^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{x^2}\). For (c) we have: \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial y^2}= \frac{2x}{y^3}\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{y^2}\). For (d) we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}= 2y^{-3}\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\).

Step-by-step solution

(a) Function and First Partial Derivatives

Given function: \(z = \frac{1}{x} = x^{-1}\). First, we will find the first partial derivatives with respect to x: \(\frac{\partial z}{\partial x} = -x^{-2}\) Now, let's find the second partial derivatives:

(a) Second Partial Derivative with respect to x

Differentiating with respect to x again: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\)

(a) Second Partial Derivative with respect to y

As our given function contains no y term, \(\frac{\partial z}{\partial y} = 0\). So, all mixed partial derivatives are also 0. Hence, for (a), we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\).

(b) Function and First Partial Derivatives

Given function: \(z = \frac{y}{x}\). First, we will find the first partial derivatives: \(\frac{\partial z}{\partial x} = -\frac{y}{x^2}\) \(\frac{\partial z}{\partial y} = \frac{1}{x}\) Now, let's find the second partial derivatives:

(b) Second Partial Derivative with respect to x

Differentiating with respect to x again: \(\frac{\partial^2 z}{\partial x^2} = \frac{2y}{x^3}\)

(b) Second Partial Derivative with respect to y

Differentiating with respect to y again: \(\frac{\partial^2 z}{\partial y^2} = 0\)

(b) Mixed Partial Derivatives

Differentiating first with respect to x and then with respect to y: \(\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(-\frac{y}{x^2}\right) = -\frac{1}{x^2}\) Differentiating first with respect to y and then with respect to x: \(\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial x}\left(\frac{1}{x}\right) = -\frac{1}{x^2}\) Hence, for (b), we have: \(\frac{\partial^2 z}{\partial x^2} = \frac{2y}{x^3}\), \(\frac{\partial^2 z}{\partial y^2}=0\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{x^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{x^2}\).

(c) Function and First Partial Derivatives

Given function: \(z = \frac{x}{y}\). First, we will find the first partial derivatives: \(\frac{\partial z}{\partial x} = \frac{1}{y}\) \(\frac{\partial z}{\partial y} = -\frac{x}{y^2}\) Now, let's find the second partial derivatives:

(c) Second Partial Derivative with respect to x

Differentiating with respect to x again: \(\frac{\partial^2 z}{\partial x^2} = 0\)

(c) Second Partial Derivative with respect to y

Differentiating with respect to y again: \(\frac{\partial^2 z}{\partial y^2} = \frac{2x}{y^3}\)

(c) Mixed Partial Derivatives

Differentiating first with respect to x and then with respect to y: \(\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{1}{y}\right) = -\frac{1}{y^2}\) Differentiating first with respect to y and then with respect to x: \(\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial x}\left(-\frac{x}{y^2}\right) = -\frac{1}{y^2}\) Hence, for (c), we have: \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial y^2}= \frac{2x}{y^3}\), \(\frac{\partial^2 z}{\partial x \partial y}=-\frac{1}{y^2}\), \(\frac{\partial^2 z}{\partial y \partial x}=-\frac{1}{y^2}\).

(d) Function and First Partial Derivatives

Given function: \(z = \frac{1}{x} + \frac{1}{y} = x^{-1} + y^{-1}\). First, we will find the first partial derivatives: \(\frac{\partial z}{\partial x} = -x^{-2}\) \(\frac{\partial z}{\partial y} = -y^{-2}\) Now, let's find the second partial derivatives:

(d) Second Partial Derivative with respect to x

Differentiating with respect to x again: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\)

(d) Second Partial Derivative with respect to y

Differentiating with respect to y again: \(\frac{\partial^2 z}{\partial y^2} = 2y^{-3}\)

(d) Mixed Partial Derivatives

Differentiating first with respect to x and then with respect to y: \(\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial y}\left(-x^{-2}\right) = 0\) Differentiating first with respect to y and then with respect to x: \(\frac{\partial^2 z}{\partial y \partial x} = \frac{\partial}{\partial x}\left(-y^{-2}\right) = 0\) Hence, for (d), we have: \(\frac{\partial^2 z}{\partial x^2} = 2x^{-3}\), \(\frac{\partial^2 z}{\partial y^2}= 2y^{-3}\), \(\frac{\partial^2 z}{\partial x \partial y}=0\), \(\frac{\partial^2 z}{\partial y \partial x}=0\).

Key concepts

Second Derivatives

When dealing with functions in calculus, second derivatives provide insights into the concavity and curvature of a function's graph. In the realm of partial derivatives, which involve functions of multiple variables, second derivatives are particularly crucial for understanding how the function behaves across different directions or dimensions.

For instance, consider a function of two variables, say, \( z = f(x, y) \). Here, second partial derivatives are calculated twice with respect to the same or different variables:

• \( \frac{\partial^2 z}{\partial x^2} \): Measures the concavity in the x-direction.
• \( \frac{\partial^2 z}{\partial y^2} \): Measures the concavity in the y-direction.
• \( \frac{\partial^2 z}{\partial x \partial y} \) or \( \frac{\partial^2 z}{\partial y \partial x} \): Mixed derivatives, indicating the curvature due to the interaction between x and y.

These second derivatives have a substantial impact in fields like engineering, where understanding the behavior of a system in multiple dimensions is crucial for design and analysis.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value changes as its input changes. In partial differentiation, this principle extends to functions of multiple variables. By holding one variable constant, we can find the derivative of the function concerning another variable. This is essential in multi-variable calculus.

Consider a function \( z = f(x, y) \). To find the partial derivative of \( z \) with respect to \( x \), denoted as \( \frac{\partial z}{\partial x} \), the variable \( y \) is treated as constant. Similarly, \( \frac{\partial z}{\partial y} \) is obtained by holding \( x \) constant. This method helps in understanding how \( z \) changes in isolation with respect to each variable, which is a foundation for solving many engineering and physics problems.

• Partial differentiation allows us to study the effect of a single variable on a multi-variable function.
• This is widely applied in optimization problems and in understanding the behavior of complex systems.

The step-by-step breakdown of partial derivatives is critical for mastering these concepts.

Calculus

Calculus, the mathematical study of change, underpins many advanced mathematics courses and is pivotal in various applied fields. It primarily involves differentiation and integration, offering tools to model and analyze dynamic systems.

The concept of calculus extends to partial derivatives when dealing with functions of several variables. These derivatives help analyze functions that change based on more than one input, enabling detailed insights into how different factors influence a system or model.

• It allows for the understanding of real-world phenomena like heat transfer, fluid dynamics, and economic models.
• Calculus provides a way to express complex geometries and surfaces in a numerical format readily used in computing.

Grasping the concept of partial derivatives within calculus is crucial for students tackling multi-variable problems, as it helps them understand rates of change and system behaviors more profoundly.

Engineering Mathematics

Engineering mathematics is an essential aspect of creating and understanding the explanations within sciences and engineering. It applies mathematical theories and practices to solve complex problems found in engineering domains, from designing structures to developing new technologies.

Within this field, partial derivatives and second derivatives are critical as they provide means to model and simulate the physical world in two or three dimensions. These derivatives help engineers predict behaviors under various conditions, such as stress and strain in materials or forces in dynamic systems.

• They are employed in mechanical, civil, electrical, and chemical engineering to refine processes and innovations.
• Using partial derivatives, engineers can isolate the impact of a single variable, enhancing precision in simulations and analysis.

Thus, mastering partial differentiation techniques forms a significant component of engineering mathematics, enabling improved design and functionality in real-world applications.