Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(w=u^{2}-u \tan v, u=x\), and \(v=\pi x\), find \(\left.\frac{d w}{d x}\right|_{x=1 / 4}\)

Step-by-Step Solution

Answer

\(-\frac{1}{2} - \frac{1}{2}\pi\)

Step 1: Introduce the Function

We are given the function for which we need to find the derivative: \[ w = u^2 - u \tan(v) \]with the substitutions \( u = x \) and \( v = \pi x \). We want to find \( \frac{dw}{dx} \).

Step 2: Apply the Chain Rule

We will use the chain rule to find the derivative. The chain rule states that:\[ \frac{dw}{dx} = \frac{\partial w}{\partial u} \cdot \frac{du}{dx} + \frac{\partial w}{\partial v} \cdot \frac{dv}{dx} \]First, we'll calculate the partial derivatives and these derivatives, knowing \( u = x \) and \( v = \pi x \).

Step 3: Compute Partial Derivatives

First, compute \( \frac{\partial w}{\partial u} \) and \( \frac{\partial w}{\partial v} \):- \( \frac{\partial w}{\partial u} = 2u - \tan(v) \)- \( \frac{\partial w}{\partial v} = -u \sec^2(v) \)

Step 4: Compute Derivatives of Substitutions

Next, calculate the derivatives of \( u \) and \( v \) with respect to \( x \):- \( \frac{du}{dx} = \frac{d}{dx}(x) = 1 \)- \( \frac{dv}{dx} = \frac{d}{dx}(\pi x) = \pi \)

Step 5: Substitute and Simplify

Substitute all of the calculated derivatives into the chain rule equation:\[ \frac{dw}{dx} = (2u - \tan(v)) \cdot 1 + (-u \sec^2(v)) \cdot \pi \]Simplify to get:\[ \frac{dw}{dx} = 2u - \tan(v) - u \pi \sec^2(v) \]

Step 6: Evaluate at the Specific Point

We need to evaluate \( \frac{dw}{dx} \) at \( x = \frac{1}{4} \), where \( u = x = \frac{1}{4} \) and \( v = \pi x = \frac{\pi}{4} \).Calculate:- \( \tan(\pi/4) = 1 \)- \( \sec^2(\pi/4) = 2 \)Substitute these into our derivative:\[ \frac{dw}{dx} = 2 \left(\frac{1}{4}\right) - 1 - \frac{1}{4} \pi \cdot 2 \]Simplify:\[ \frac{dw}{dx} = \frac{1}{2} - 1 - \frac{1}{2}\pi = \frac{1}{2} - 1 - \frac{1}{2}\pi \]

Step 7: Final Calculation

Calculate the final expression:\[ \frac{dw}{dx} = \frac{1}{2} - 1 - \frac{1}{2}\pi = -\frac{1}{2} - \frac{1}{2}\pi \].

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, used primarily for differentiating composite functions. A composite function is when you have two or more functions combined into a single function. For instance, if a function \( w \) is dependent on \( u \) and \( v \), and each of these are dependent on \( x \), the chain rule provides a structured way to find \( \frac{dw}{dx} \), which is the rate of change of \( w \) with respect to \( x \).Using the chain rule means you look at how each part of the function contributes to the derivative. This involves calculating the partial derivatives of \( w \) with respect to each of its own variables \( u \) and \( v \), and then factoring in how each of those variables changes with respect to \( x \). The mathematical expression for the chain rule for this calculation is:

• \( \frac{dw}{dx} = \frac{\partial w}{\partial u} \cdot \frac{du}{dx} + \frac{\partial w}{\partial v} \cdot \frac{dv}{dx} \)

This reinforces that to properly apply the chain rule, each part of the function's dependency has to be understood and differentiated separately.

Partial Derivatives

Partial derivatives are used in calculus to understand how a multivariable function changes when one of the variables changes and others are held constant. They are essential when dealing with functions that depend on more than one variable, like \( w(u, v) = u^2 - u \tan(v) \) in our exercise.When calculating the partial derivative of \( w \) with respect to \( u \), denoted \( \frac{\partial w}{\partial u} \), you consider \( v \) as a constant. This gives the expression:

• \( \frac{\partial w}{\partial u} = 2u - \tan(v) \)

Similarly, when taking the partial derivative with respect to \( v \), noted as \( \frac{\partial w}{\partial v} \), \( u \) is treated as a constant:

• \( \frac{\partial w}{\partial v} = -u \sec^2(v) \)

These computations allow us to see how changes in \( u \) and \( v \) independently affect \( w \), providing a crucial step towards finding the full derivative \( \frac{dw}{dx} \). Partial derivatives thus break down complex functions into simpler, one-variable chain reactions, easing the path to understanding how the function behaves as a whole.

Trigonometric Derivatives

Trigonometric derivatives are derivatives of functions that involve trigonometric functions, such as sine, cosine, and tangent. They are indispensable in solving calculus problems that involve periodic or wave functions.In our specific example, we make use of the derivatives of \( \tan(v) \). The derivative of \( \tan(v) \) with respect to \( v \) can be expressed as:

• \( \frac{d}{dv}(\tan(v)) = \sec^2(v) \)

This relationship is vital for our calculation of \( \frac{\partial w}{\partial v} \) because it implies that changes in \( v \) are magnified by \( \sec^2(v) \) when seeing how they affect \( w \). When evaluating the expression at specific points, like \( v = \frac{\pi}{4} \), the trigonometric identity \( \tan(\frac{\pi}{4}) = 1 \) and \( \sec^2(\frac{\pi}{4}) = 2 \) simplify the calculation immensely, allowing us to seek out the pattern in how small alterations in \( v \) influence the function \( w \).By mastering trigonometric derivatives, you gain valuable tools for understanding the behavior and characteristics of a wide range of functions encountered in calculus.