Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). $$ g(x)=\frac{1}{\sqrt{3 x}} $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{-1}{2x\sqrt{3x}} \).

Step 1: Understand the Function

The function provided is \( g(x) = \frac{1}{\sqrt{3x}} \). To find its derivative using the definition of a derivative, you'll set up the expression that involves the limit.

Step 2: Set Up the Difference Quotient

Using the definition of a derivative, construct the difference quotient. This is: \[ \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{\frac{1}{\sqrt{3(x+h)}} - \frac{1}{\sqrt{3x}}}{h} \]

Step 3: Simplify the Difference

To simplify \( \frac{\frac{1}{\sqrt{3(x+h)}} - \frac{1}{\sqrt{3x}}}{h} \), find a common denominator: \[ \lim_{h \to 0} \frac{\sqrt{3x} - \sqrt{3(x+h)}}{h \sqrt{3x}\sqrt{3(x+h)}} \]

Step 4: Rationalize the Numerator

Multiply both the numerator and the denominator by the conjugate of the numerator, \( \sqrt{3x} + \sqrt{3(x+h)} \): \[ \lim_{h \to 0} \frac{(\sqrt{3x} - \sqrt{3(x+h)})(\sqrt{3x} + \sqrt{3(x+h)})}{h (\sqrt{3x} \sqrt{3(x+h)})(\sqrt{3x} + \sqrt{3(x+h)})} \]. This simplifies the numerator to: \[ 3x - 3(x+h) = -3h \]

Step 5: Simplify and Find the Limit

After simplification, the expression becomes: \[ \lim_{h \to 0} \frac{-3h}{h (\sqrt{3x} \sqrt{3(x+h)})(\sqrt{3x} + \sqrt{3(x+h)})} = \lim_{h \to 0} \frac{-3}{\sqrt{3x} \sqrt{3(x+h)}(\sqrt{3x} + \sqrt{3(x+h)})} \]As \(h\) goes to zero, \(\sqrt{3(x+h)} \to \sqrt{3x}\), so the expression simplifies further.

Step 6: Evaluate the Limit

The limit simplifies to: \[ \frac{-3}{\sqrt{3x} \cdot \sqrt{3x} \cdot 2\sqrt{3x}} = \frac{-3}{6x\sqrt{3x}} = \frac{-1}{2x\sqrt{3x}} \]. Therefore, the derivative of \( g(x) = \frac{1}{\sqrt{3x}} \) is \( g'(x) = \frac{-1}{2x\sqrt{3x}} \).

Key Concepts

Definition of Derivative

The derivative of a function represents the rate at which the function's value changes at a particular point. It provides us with the slope of the tangent line to the function's graph at that point. The formal definition of the derivative is given by the limit:

• \( f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)

This definition uses a limit process to calculate the slope as the distance between two points on a function, usually denoted by \( h \), becomes infinitesimally small. By substituting into this formula, we can determine the exact rate of change for any specific function.

Limit Process

The limit process is a crucial concept in calculus used to find derivatives. It solves the problem of finding the instantaneous rate of change at a specific point. To calculate a derivative, we take the limit of the difference quotient as \( h \) approaches zero:

• This seeks to find how the function behaves right next to a particular point, thus capturing the slope of the tangent line at that point.

Understanding this process involves imagining a series of secant lines, which are lines connecting two points on a curve, becoming the tangent line as the space between the points approaches zero. By using limits, we essentially "zoom in" on the curve.

Rationalizing Techniques

In calculus, rationalizing techniques often facilitate the simplification of expressions that involve square roots or other radical terms during limit processes. When working with a derivative involving roots, like in our example, rationalizing the numerator is key. This involves multiplying the expression by a conjugate to eliminate the square roots from the numerator:

• The conjugate of \( \sqrt{3x} - \sqrt{3(x+h)} \) is \( \sqrt{3x} + \sqrt{3(x+h)} \).

Multiplying both the numerator and the denominator by this conjugate allows us to convert the difference into a form that cancels out one of the variables \( h \) in the denominator, which reveals the slope at an infinitesimally close point.

Simplifying Expressions

Simplifying expressions is a significant step in the process of finding derivatives, especially with complex fractions and radicals. Once the numerator is rationalized, further simplification is needed to evaluate the limit accurately.

• After rationalizing, terms containing \( h \) in the numerator and denominator can often be canceled.
• Adjust the remaining expression to focus solely on constant terms and known values.

In our problem, we ended with an expression that involved \( h \) in the denominator. Simplifying meant removing \( h \) to evaluate the limit as \( h \to 0 \). This eventually allowed us to find the derivative \( g^{\prime}(x) = \frac{-1}{2x\sqrt{3x}} \), providing a clear depiction of how the original function's rate changes at any point \( x \).