Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\left(\frac{x-2}{x-\pi}\right)^{-3} $$

Step-by-Step Solution

Answer

\(D_{x} y = 3(\pi-2) \cdot \frac{(x-\pi)^2}{(x-2)^4}\)

Step 1: Simplify the Expression

Rewrite the given function in a form that is easier to differentiate. Let\[y = \left(\frac{x-2}{x-\pi}\right)^{-3}\]This can be rewritten using the property of exponents:\[y = \left(\frac{x-\pi}{x-2}\right)^{3}\]

Step 2: Differentiate Using the Chain Rule

Apply the chain rule to differentiate the expression. The chain rule states that if you have \(y = f(u)\) where \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).In this case, set \(u = \frac{x-\pi}{x-2}\) so that \(y = u^{3}\). The derivative \(\frac{dy}{du} = 3u^{2}\).First, we will need the derivative of \(u\) with respect to \(x\).

Step 3: Differentiate the Inner Function Using the Quotient Rule

The function \(u = \frac{x-\pi}{x-2}\) is a quotient, so we need the derivative using the quotient rule: \[\frac{d}{dx}\left(\frac{a}{b}\right) = \frac{a'b - ab'}{b^2}\]For \(u = \frac{x-\pi}{x-2}\), let \(a = x-\pi\) and \(b = x-2\). Therefore, \(a' = 1\) and \(b' = 1\).Substitute these into the quotient rule:\[\frac{du}{dx} = \frac{(1)(x-2) - (x-\pi)(1)}{(x-2)^2} = \frac{x-2 - x + \pi}{(x-2)^2} = \frac{\pi-2}{(x-2)^2}\].

Step 4: Apply the Chain Rule and Simplify

Now substitute back into the chain rule:\[\frac{dy}{dx} = 3u^{2} \cdot \frac{du}{dx}\]\[= 3\left(\frac{x-\pi}{x-2}\right)^{2} \cdot \frac{\pi-2}{(x-2)^2}\]Combine and simplify the expression:\[\frac{dy}{dx} = 3(\pi-2) \cdot \frac{(x-\pi)^2}{(x-2)^4}\]

Key Concepts

Chain Rule in Differentiation

One of the essential tools in calculus for differentiation is the Chain Rule. It is particularly useful when dealing with composite functions, where one function is nested inside another.
The Chain Rule states that if you have a composite function like \( y = f(g(x)) \), the derivative of \( y \) with respect to \( x \) is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). In more mathematical terms, this is expressed as:

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

For instance, in our exercise, the function \( y = u^3 \) is written in terms of \( u \), where \( u = \frac{x-\pi}{x-2} \).
Thus, applying the Chain Rule involves:

• Finding \( \frac{dy}{du} = 3u^2 \)
• Finding \( \frac{du}{dx} \)
• Multiplying these results to get \( \frac{dy}{dx} \)

This approach simplifies finding the derivative of more complex functions by breaking them into simpler parts.

Quotient Rule for Derivatives

The Quotient Rule is used to find the derivative of a function that is the quotient of two differentiable functions.
If you have a function \( u = \frac{a}{b} \), the Quotient Rule states that the derivative \( \frac{du}{dx} \) is given by:

• \( \frac{du}{dx} = \frac{a' \cdot b - a \cdot b'}{b^2} \)

Here, \( a' \) and \( b' \) are the derivatives of \( a \) and \( b \) respectively. This method helps us efficiently tackle derivatives of fractions.
In the example provided, \( u = \frac{x-\pi}{x-2} \):

• We identify \( a = x-\pi \) and \( b = x-2 \)
• Derivatives are \( a' = 1 \) and \( b' = 1 \)
• Applies as \( \frac{du}{dx} = \frac{(1) \cdot (x-2) - (x-\pi) \cdot (1)}{(x-2)^2} \)
• Simplifies to \( \frac{du}{dx} = \frac{\pi-2}{(x-2)^2} \)

The Quotient Rule, despite having some steps, is a practical tool in calculus to manage the derivatives of functions like ratios of two expressions.

Exponent Properties in Calculus

Exponents are a fundamental concept in mathematics, and understanding their properties is crucial, especially when dealing with calculus problems.
Some key exponent properties that are frequently applied include:

• The power of a power property: \((a^m)^n = a^{m \times n}\)
• Negative exponents: \(a^{-n} = \frac{1}{a^n}\)
• Fractional exponents: \(a^{m/n} = \sqrt[n]{a^m}\)

In calculus, these properties allow us to manipulate expressions into more differentiable forms.
For the initial function given as \( y = \left(\frac{x-2}{x-\pi}\right)^{-3} \), recognizing that a negative exponent implies a reciprocal was key:

• Rewriting \( y = \left(\frac{x-\pi}{x-2}\right)^{3} \)
• This simplification made the differentiation process much more straightforward

Having a solid grip on exponent properties can make solving calculus problems more intuitive and manageable, as seen in this exercise.