Problem 37 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$u=\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}$$

Step-by-Step Solution

Answer

The partial derivative is \( \frac{x_i}{\sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}} \).

Step 1: Express the Function

The function given is \( u = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \). This is essentially the Euclidean norm of the vector \( \mathbf{x} = (x_1, x_2, \ldots, x_n) \).

Step 2: Use the Chain Rule

To find the first partial derivative of \( u \) with respect to each \( x_i \), recognize \( u \) can be rewritten as \( u = (x_1^2 + x_2^2 + \cdots + x_n^2)^{1/2} \). The derivative of \( u \) with respect to a specific variable \( x_i \) requires the chain rule.

Step 3: Differentiate the Outer Function

The outer function is \( f(z) = z^{1/2} \) where \( z = x_1^2 + x_2^2 + \cdots + x_n^2 \). The derivative with respect to \( z \) is \( f'(z) = \frac{1}{2}z^{-1/2} \). Substitute back \( z = x_1^2 + x_2^2 + \cdots + x_n^2 \).

Step 4: Differentiate the Inner Function

The inner function is \( z = x_1^2 + x_2^2 + \cdots + x_n^2 \). The partial derivative of \( z \) with respect to \( x_i \) is \( \frac{\partial z}{\partial x_i} = 2x_i \) because the derivative of \( x_i^2 \) is \( 2x_i \) and the other terms are constants.

Step 5: Combine the Derivatives

Using the chain rule result, \( \frac{\partial u}{\partial x_i} = f'(z) \cdot \frac{\partial z}{\partial x_i} = \left( \frac{1}{2}(x_1^2 + x_2^2 + \cdots + x_n^2)^{-1/2} \right)(2x_i) \).

Step 6: Simplify the Expression

Simplify the partial derivative: \( \frac{\partial u}{\partial x_i} = \frac{x_i}{\sqrt{x_1^2 + x_2^2 + \cdots + x_n^2}} \), which gives the first partial derivative of \( u \) with respect to \( x_i \).

Key Concepts

Euclidean norm

The Euclidean norm, often referred to as the magnitude or length of a vector, is a crucial concept in multivariable calculus. It represents the distance from the origin in a space defined by several dimensions. This can be visualized as the length of a line segment from the origin to a point in 2D, 3D, or n-dimensional space.
In mathematical terms, for a vector \( \mathbf{x} = (x_1, x_2, \ldots, x_n) \), the Euclidean norm is calculated as:

• \( \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \)

This operation involves squaring each component, summing those squares, and then taking the square root of the result. The Euclidean norm is fundamental when seeking the partial derivatives of a function since it often appears in expressions for norm-based functions like the one given. Understanding it well can simplify derivative computations as well as offer geometric insights.

chain rule

The chain rule is a pivotal technique in calculus, especially in functions involving multiple variables and layers of functions. It allows us to differentiate composite functions by breaking them down into simpler parts.
In general terms, if you have a function \( u(g(x)) \), the chain rule stipulates that:

• \( \frac{du}{dx} = \frac{du}{dg} \cdot \frac{dg}{dx} \)

For multivariable functions like the one in our exercise, we consider the outer function \( u = (x_1^2 + x_2^2 + \cdots + x_n^2)^{1/2} \), and inner function \( z = x_1^2 + x_2^2 + \cdots + x_n^2 \). This layered structure is typical for situations encountered in multivariable calculus. Applying the chain rule lets us find the partial derivative by first dealing with the outer function and then the inner function, multiplying the results as shown in the solution.

multivariable calculus

Multivariable calculus extends ordinary calculus to functions of several variables. It plays a significant role in fields like physics, engineering, and economics.
We often deal with functions that are dependent on multiple variables, for example, \( u(x_1, x_2, \dots, x_n) \). This requires understanding how changes in each variable independently affect the function. Hence, we use partial derivatives to assess how the function changes as each individual variable is varied, while the others are held constant.

• Conceptually, a partial derivative can be seen as an input-specific slope for multivariable functions.
• Functions in multivariable calculus are often visualized as surfaces or higher-dimensional analogs.

This understanding is crucial when finding the partial derivatives of complex functions like the Euclidean norm.

derivative computation

Computing derivatives, especially partial derivatives, involves a step-by-step methodology to ensure accuracy and precision. The exercise illustrates this by first expressing the function and then applying differentiation rules strategically.
For the function \( u = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} \), the computation involves:

• Recognizing the structure of the function, akin to an outer and inner function relationship, suited for the chain rule.
• Differentiating the outer function: \( f(z) = z^{1/2} \) giving \( f'(z) = \frac{1}{2}z^{-1/2} \).
• Differentiating the inner function \( z = x_1^2 + x_2^2 + \cdots + x_n^2 \), leading to \( \frac{\partial z}{\partial x_i} = 2x_i \).
• Combining these derivatives by the chain rule to find the partial derivatives.

By understanding each step, from differentiating individual components to combining results, students can accurately compute derivatives for multivariable functions.