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In Exercises \(23-34,\) find \(f_{x}, f_{y},\) and \(f_{z}\) $$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}$$

Short Answer

Expert verified
\(f_x = -\frac{x}{(x^2 + y^2 + z^2)^{3/2}}, f_y = -\frac{y}{(x^2 + y^2 + z^2)^{3/2}}, f_z = -\frac{z}{(x^2 + y^2 + z^2)^{3/2}}\)

Step by step solution

01

Identifying Partial Derivative with Respect to x

The function given is \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \). To find \( f_x \), we need the partial derivative of \( f \) with respect to \( x \). We'll apply the chain rule differentiating \( (x^2 + y^2 + z^2)^{-1/2} \) with respect to \( x \).
02

Applying the Chain Rule for f_x

Using the chain rule, differentiate the outer function which is \( u^{-1/2} \) where \( u = x^2 + y^2 + z^2 \), yielding \( -\frac{1}{2}u^{-3/2} \). Differentiate the inner function \( u = x^2 \) with respect to \( x \) resulting in \( 2x \). Thus, \( f_x = -\frac{1}{2}(x^2 + y^2 + z^2)^{-3/2} \cdot 2x = -\frac{x}{(x^2 + y^2 + z^2)^{3/2}} \).
03

Identifying Partial Derivative with Respect to y

To find \( f_y \), take the partial derivative of \( f(x, y, z) \) with respect to \( y \). As before, we will apply the chain rule on \( (x^2 + y^2 + z^2)^{-1/2} \).
04

Applying the Chain Rule for f_y

Again, applying the chain rule: take derivative of outer function \( u^{-1/2} \), which is \( -\frac{1}{2}u^{-3/2} \). The derivative of \( y^2 \) with respect to \( y \) is \( 2y \). Thus, \( f_y = -\frac{1}{2}(x^2 + y^2 + z^2)^{-3/2} \cdot 2y = -\frac{y}{(x^2 + y^2 + z^2)^{3/2}} \).
05

Identifying Partial Derivative with Respect to z

For \( f_z \), we differentiate \( f(x, y, z) \) with respect to \( z \) using the chain rule again on \( (x^2 + y^2 + z^2)^{-1/2} \).
06

Applying the Chain Rule for f_z

Using the chain rule: differentiate \( u^{-1/2} \), resulting in \( -\frac{1}{2}u^{-3/2} \). The derivative of \( z^2 \) with respect to \( z \) is \( 2z \). Therefore, \( f_z = -\frac{1}{2}(x^2 + y^2 + z^2)^{-3/2} \cdot 2z = -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \).

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is an extension of calculus to functions of multiple variables. Such functions depend on more than one input, like our function, which depends on variables \(x\), \(y\), and \(z\). This branch of mathematics focuses on techniques for differentiation and integration involving these multivariable functions.

In single-variable calculus, you have one input changing over time. In contrast, multivariable calculus examines how multiple inputs interact and influence the outcome. This field is fundamental in finding rates of change and areas related to several dynamic quantities. For example, finding the rate of heat spread in a metal plate requires understanding these multivariable functions.

Real-world applications include physics, engineering, and economics, where predicting outcomes based on many factors is crucial. In cases where a function represents a 3D surface or space, it helps determine gradients and optimize processes in spatial contexts.
Chain Rule
The chain rule is a key differentiation technique, especially for composite functions. It allows us to differentiate complex functions by breaking down the parts. In the exercise, we're using it to differentiate the function \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \) with respect to each variable.

The chain rule formula can be succinctly written as \( (f(g(x)))' = f'(g(x)) \, g'(x) \). This means that to find the derivative, you differentiate the outer function and multiply by the derivative of the inner function. In our case, the outer function is \( u^{-1/2} \) and the inner function is \( u=x^2+y^2+z^2 \).

This approach simplifies differentiation by handling each component carefully and is widely applicable and essential in tackling derivatives in multivariable calculus problems.
Differentiation
Differentiation is a process used to find the rate at which a function changes with respect to its variables. In multivariable calculus, differentiation extends to finding how functions change with respect to several variables.

In the exercise, we differentiate the given function with respect to each variable \(x\), \(y\), \(z\). This is known as finding partial derivatives. Partial derivatives signify how a function changes as we vary one variable while keeping others constant.

To find a partial derivative, treat all other variables as constants and apply regular differentiation techniques. Partial derivatives are crucial, forming the basis of gradient vectors, tangent planes, and optimization in multiple dimensions.
Calculus of Functions of Several Variables
The calculus of functions of several variables deals with multiple inputs and outputs. This domain includes finding derivatives and integrals within a multidimensional context.

In our exercise, the function is a scalar field, which is a 3-variable function producing a single value calculated using \( x \), \( y \), and \( z \). You encountered working with scalar fields while finding partial derivatives, indicating how this function changes in each direction.

These methods are pivotal in fields like astrophysics and meteorology, where conditions aren't uniform and vary with multiple parameters. For example, understanding weather patterns involves analyzing numerous factors like pressure and temperature, each dependent on different variables.

The calculus of functions of several variables provides the tools to dissect, decipher, and work with multi-dimensional models efficiently.

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Most popular questions from this chapter

Find the point on the plane \(3 x+2 y+z=6\) that is nearest the origin.

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