Problem 20 - The Calculus with Analytic Geometry

Question

In Exercises 13 through 24 , find the indicated partial derivatives by holding all but one of the variables constant and applying theorems for ordinary differentiation. $$ u=\tan ^{-1}(x y z w) ; \frac{\partial u}{\partial w} $$

Step-by-Step Solution

Answer

\( \frac{xyz}{1+(xyzw)^2} \)

Step 1: Understand the Function Given

The function given is \(u = \tan^{-1}(xyzw)\). This is an inverse tangent function.

Step 2: Identify the Variable for Partial Differentiation

We need to find the partial derivative with respect to \(w\), so consider \(x, y,\) and \(z\) as constants during differentiation.

Step 3: Recall the Derivative of Inverse Tangent

The derivative of \( \tan^{-1}(v) \) with respect to \(v \) is \( \frac{1}{1+v^2} \).

Step 4: Apply the Chain Rule

Using the chain rule, differentiate \(u = \tan^{-1}(xyzw) \) with respect to \(w \). Let \ v = xyzw \: \[ \frac{\text{d}}{\text{d}w} \tan^{-1}(v) = \frac{1}{1+v^2} \frac{\text{d}v}{\text{d}w} = \frac{1}{1+(xyzw)^2} \frac{\text{d}(xyzw)}{\text{d}w} \]

Step 5: Differentiate the Inner Function

Differentiate \ v = xyzw \ with respect to \ w \: \[ \frac{\text{d}(xyzw)}{\text{d}w} = xyz \]

Step 6: Combine Results

Substitute the derivative of the inner function into the chain rule expression: \[ \frac{\text{d}}{\text{d}w} \tan^{-1}(xyzw) = \frac{1}{1+(xyzw)^2} \times xyz \]

Step 7: Write the Final Answer

The partial derivative of \ u \ with respect to \ w \ is: \[ \frac{\text{partial} u}{\text{partial} w} = \frac{xyz}{1+(xyzw)^2} \]

Key Concepts

inverse functions

An inverse function essentially undoes the action of the original function. When you have a function, say \( f(x) \), that maps \( x \) to \( y \), the inverse function, typically denoted as \( f^{-1}(y) \), maps \( y \) back to \( x \). In this exercise, we deal with the inverse tangent function, \( \tan^{-1}(v) \). This function returns the angle whose tangent is \( v \). Inverse functions are important in calculus for solving equations and modeling real-world scenarios. To find the inverse function, you swap the roles of \( x \) and \( y \), then solve for the new \( y \) (which is the original \( x \)). For example, the inverse of \( y = 2x + 3 \) is found by solving \( x = 2y + 3 \) for \( y \), resulting in \( y = (x - 3) / 2 \). As routines, inverse functions vastly simplify complex calculations.

chain rule

The chain rule is a fundamental differentiation rule in calculus used when composing functions. It allows us to differentiate a composite function. In simple terms, if you have a function \( f(g(x)) \), the chain rule states that the derivative of this function is \( f'(g(x)) \times g'(x) \). Applied to our problem, we have \( u = \tan^{-1}(xyzw) \). Here, \( \tan^{-1} \) is our outer function and \( xyzw \) is the inner function. To differentiate, we:

• First, find the derivative of the outer function \( \tan^{-1} \) with respect to the inner function \( xyzw \), which is \( \frac{1}{1+(xyzw)^2} \).
• Next, multiply this by the derivative of the inner function \( xyzw \) with respect to \( w \), which simplifies to \( xyz \).

Combining these, the chain rule yields \( \frac{1}{1+(xyzw)^2} \times xyz \). The chain rule is powerful, particularly for breaking down intricate functions into simpler, manageable parts.

differentiation

Differentiation is the process of finding the derivative of a function. The derivative measures how a function changes as its input changes. In the exercise, our target is the partial derivative of the function \( u = \tan^{-1}(xyzw) \) with respect to \( w \). Partial differentiation generalizes differentiation to functions of multiple variables, focusing on how a function changes as one specific variable changes while holding others constant. Here’s how we approach it:

• Step 1: Recognize that \( x \), \( y \), and \( z \) are treated as constants.
• Step 2: Apply the chain rule to find the derivative of the outer function \( \tan^{-1} \) and then of the inner product \( xyzw \).
• Step 3: Combine results to get the final answer, \( \frac{xyz}{1+(xyzw)^2} \).

By breaking down each part, we can see how the derivative captures the rate of change. Differentiation, including partial differentiation, is a critical tool in fields like physics, economics, and engineering, making it easier to understand how things change in response to varying conditions.