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

\(\left.-\frac{dw}{dx}\right|_{x=1/4} = -\frac{1 + \pi}{2}\)

Step 1: Express w in terms of x

Substitute the given expressions for \(u\) and \(v\) in terms of \(x\): \(u = x\) and \(v = \pi x\), into the equation for \(w\). This gives:\[w = x^2 - x \tan(\pi x)\]

Step 2: Differentiate w with respect to x

To find \(\frac{dw}{dx}\), apply the derivative to each term:- The derivative of \(x^2\) is \(2x\).- For the term \(-x \tan(\pi x)\), use the product rule: \(\frac{d}{dx}[u*v] = u'v + uv'\), where \(u = x\) and \(v = \tan(\pi x)\).Differentiating \(x\) gives \(1\) and differentiating \(\tan(\pi x)\) gives \(\pi \sec^2(\pi x)\) by the chain rule.

Step 3: Continue to apply the product rule

Apply the product rule for \(-x \tan(\pi x)\):\[-x \frac{d}{dx}\tan(\pi x) + \left(-\tan(\pi x)\right) = -(1*\pi \sec^2(\pi x) * x) - \tan(\pi x)\]Combine the derivatives to find:\[\frac{dw}{dx} = 2x - (x \pi \sec^2(\pi x) + \tan(\pi x))\]

Step 4: Evaluate at x = 1/4

Substitute \(x = \frac{1}{4}\) into the derivative expression:\[\frac{dw}{dx}\bigg|_{x=1/4} = 2\left(\frac{1}{4}\right) - \left(\frac{1}{4}\pi \sec^2\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right)\right)\]Since \(\tan\left(\frac{\pi}{4}\right) = 1\) and \(\sec^2\left(\frac{\pi}{4}\right) = 2\), simplify the expression:\[\frac{1}{2} - \left(\frac{1}{4} \pi \cdot 2 + 1\right)\]Which equals:\[\frac{1}{2} - \left(\frac{\pi}{2} + 1\right)\]Final computation gives:\[-\frac{1}{2} - \frac{\pi}{2}\]

Step 5: Simplify the final expression

Combine the fractions to get the final answer:\[\frac{dw}{dx}\bigg|_{x=1/4} = -\frac{1 + \pi}{2}\]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function changes. In simple terms, it calculates how one quantity changes in relation to another. Imagine you are climbing a hill: differentiation helps you understand how steep the hill is at any given point.

When we differentiate a function with respect to a variable, we are finding the derivative, often denoted as \( \rac{df}{dx} \). This process involves applying rules and formulas to calculate these changes, such as the power rule for polynomials or chain and product rules for more complex situations.

• The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). For example, the derivative of \(x^2\) is \(2x\).
• When functions involve more than simple powers of \(x\), additional rules like the chain rule or product rule become essential.

The main takeaway here is that differentiation transforms a function into another that describes its rate of change or slope at each point.

Chain Rule

The chain rule is a vital tool in differentiation used to find the derivative of composite functions. A composite function is essentially a function within another function, like \(f(g(x))\). Think of this as unwrapping layers, like peeling an onion.

The chain rule states that if you have a composite function \(f(g(x))\), the derivative is the product of the derivative of \(f\) with respect to \(g\) and the derivative of \(g\) with respect to \(x\). This can be written as:
\[\frac{d}{dx}f(g(x)) = f'(g(x)) \, g'(x)\]
In our example, we had to apply the chain rule to differentiate \(\tan(\pi x)\). Here, \(\pi x\) is the inside function, and separating them using chain rule simplifies this differentiation process.

• First, differentiate the outer function (tangent), which results in \(\sec^2(\pi x)\).
• Then, multiply it by the derivative of the inside function \(\pi x\), which is \(\pi\). So, \(\frac{d}{dx}\tan(\pi x) = \pi\sec^2(\pi x)\).

Mastering the chain rule helps manage more complex functions, making it a critical part of the differentiation toolkit.

Product Rule

Differentiating products of two functions involves the product rule. This rule allows us to take the derivative of functions multiplied together, such as \(u(x) \, v(x)\).

The product rule states that the derivative of the product of two functions \(u(x) \, v(x)\) is given by:
\[ \frac{d}{dx}[u(x) \, v(x)] = u'(x)v(x) + u(x)v'(x) \]
Think of it like this: when multiplying two things, you need to consider how changes in each of those things impact the product.

In our exercise, the product rule was applied to differentiate the term \(-x\tan(\pi x)\). Here, thinking of \(u = x\) and \(v = \tan(\pi x)\):

• First, find the derivative of \(x\), which is \(1\).
• Then find the derivative of \(\tan(\pi x)\) using the chain rule, which is \(\pi \sec^2(\pi x)\).

Combine these using the product rule to get the derivative. The product rule streamlines solving problems where functions are being multiplied, expanding our ability to tackle more intricate mathematical situations.

Trigonometric Functions

Trigonometric functions show up frequently in calculus and are an essential area of study in understanding wave patterns and oscillations. Functions like sine, cosine, and tangent tell us about angles and triangles' sides, but within calculus, their derivatives reveal even more.

In differentiation, knowing the derivatives of basic trigonometric functions is fundamental. For example:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• The derivative of \(\tan(x)\) is \(\sec^2(x)\).

In the given exercise, understanding that \(\tan(\pi x)\)'s derivative involves \(\sec^2(x)\) was crucial. Furthermore, the derivative resulted in \(\pi\sec^2(\pi x)\) due to an additional step involving the chain rule.

Such trigonometric derivatives are key tools when dealing with phenomena involving cyclic or periodic movements. This makes them invaluable for solving real-world problems and understanding nature's rhythmic patterns.