Problem 77 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that \(D_{x}|x|=|x| / x, x \neq 0 .\) Hint: Write \(|x|=\sqrt{x^{2}}\) and use the Chain Rule with \(u=x^{2}\).

Step-by-Step Solution

Answer

The derivative of \(|x|\) with respect to \(x\) for \(x \neq 0\) is \(\frac{|x|}{x}\).

Step 1: Rewrite |x| using square root

Recall from the hint that the absolute value function can be expressed as \[ |x| = \sqrt{x^2} \]. This is the first step in our solution to differentiate \( |x| \) with respect to \(x\).

Step 2: Differentiate using the Chain Rule

We can now differentiate \( |x| = \sqrt{x^2} \) using the chain rule. Let \( u = x^2 \). Then \( |x| = \sqrt{u} \).The derivative of \( \sqrt{u} \) with respect to \( u \) is \( \frac{1}{2\sqrt{u}} \). Applying the Chain Rule gives us:\[ \frac{d}{dx}(\sqrt{x^2}) = \frac{d}{du}(\sqrt{u}) \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot 2x = \frac{x}{\sqrt{x^2}} \].

Step 3: Simplify the expression

Since \( \sqrt{x^2} = |x| \), we simplify \( \frac{x}{\sqrt{x^2}} \) to:\[ \frac{x}{|x|} \].For \( x eq 0 \), this expression simplifies the derivative of \(|x|\), showing that \[ D_x |x| = \frac{|x|}{x} \].

Step 4: Conclusion of differentiation

We have shown through differentiation and simplification that the derivative of the function \( |x| \) with respect to \( x \), for \( x eq 0 \), is\[ \frac{|x|}{x} \]. This matches the desired result in the exercise.

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used for differentiating composite functions. When a function is nested within another, like an onion has layers, the chain rule allows us to find the derivative. Suppose we have a function like \( f(x) = \sqrt{x^2} \). Here, \( x^2 \) is inside the square root function, demonstrating a composite setup.

To apply the Chain Rule, follow these steps:

• Identify the outer function and the inner function. In our case, \( \sqrt{u} \) is the outer and \( u=x^2 \) is the inner function.
• Differentiating the outer function, \( \frac{d}{du}(\sqrt{u}) = \frac{1}{2\sqrt{u}} \).
• Differentiate the inner function, \( \frac{du}{dx} = 2x \).
• Multiply these derivatives as per the Chain Rule formula: \( \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \).

Applying this systematically lets us find the derivative of complex expressions efficiently.

Absolute Value Function

The Absolute Value Function, denoted as \( |x| \), measures the distance of a number \( x \) from zero on the number line, regardless of direction. By definition, \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \). It helps in understanding both positive and negative numbers uniformly.

Because of its piecewise nature, applying standard differentiation can be tricky. For calculus problems like this one, rewriting \( |x| \) as \( \sqrt{x^2} \) simplifies the process by utilizing known derivative rules. Using such manipulations is essential in transforming difficult math problems into manageable forms.

Calculus Problem Solving

Effective problem solving in calculus involves converting complex expressions into forms that are easier to work with. The problem \( D_x|x|=\frac{|x|}{x} \), hints at using the alternate expression \( |x| = \sqrt{x^2} \). While it seems complex, we can break it down logically.

Here's how you solve it:

• Understand the function and its components.
• Use derivative rules, like the Chain Rule, to transform and differentiate.
• Simplify the resulting expression to discern patterns that match a given solution.

Approaching calculus problems layer by layer, as shown here, builds deep comprehension and sharpens problem-solving skills.

Derivative of Square Root

In finding the derivative of a square root function, it's vital to understand how to differentiate \( y = \sqrt{u} \). This links back to our problem when \( |x| = \sqrt{x^2} \).

The steps to find the derivative are structured:

• Start with \( y = u^{1/2} \).
• Using the power rule, the derivative \( \frac{d}{du}(u^{1/2}) = \frac{1}{2}u^{-1/2} \), simplifies to \( \frac{1}{2\sqrt{u}} \)
• When \( u = x^2 \), substitute to apply the Chain Rule, getting \( \frac{1}{2\sqrt{x^2}} \cdot 2x = \frac{x}{\sqrt{x^2}} \)

Recognizing this pattern allows for swift differentiation of similar functions, proving again that mathematical patterns aid understanding and skill.