Problem 60 - Calculus Volume 1

Question

Use the definition of a derivative to find \(f^{\prime}(x)\). $$ f(x)=\sqrt{x-6} $$

Step-by-Step Solution

Answer

The derivative \( f'(x) = \frac{1}{2\sqrt{x-6}} \).

Step 1: Understanding the derivative definition

The definition of the derivative of a function \( f(x) \) is given by \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \). Our goal is to apply this definition to the function \( f(x) = \sqrt{x-6} \).

Step 2: Apply the function to the definition

Substitute \( f(x) = \sqrt{x-6} \) into the derivative formula. We have \( f(x+h) = \sqrt{(x+h)-6} = \sqrt{x + h - 6} \). Therefore, the formula becomes \( f'(x) = \lim_{h \to 0} \frac{\sqrt{x + h - 6} - \sqrt{x - 6}}{h} \).

Step 3: Simplify the numerator

To simplify the expression \( \sqrt{x + h - 6} - \sqrt{x - 6} \), multiply the numerator and the denominator by the conjugate, \( \sqrt{x + h - 6} + \sqrt{x - 6} \). This gives: \[ \frac{\sqrt{x + h - 6} - \sqrt{x - 6}}{h} \times \frac{\sqrt{x + h - 6} + \sqrt{x - 6}}{\sqrt{x + h - 6} + \sqrt{x - 6}} = \frac{(x + h - 6) - (x - 6)}{h(\sqrt{x + h - 6} + \sqrt{x - 6})} \].

Step 4: Simplify further

The expression simplifies to \( \frac{x + h - 6 - x + 6}{h(\sqrt{x + h - 6} + \sqrt{x - 6})} \), which becomes \( \frac{h}{h(\sqrt{x + h - 6} + \sqrt{x - 6})} \). Cancel \( h \) in the numerator and denominator to obtain \( \frac{1}{\sqrt{x + h - 6} + \sqrt{x - 6}} \).

Step 5: Find the limit as h approaches 0

Take the limit of \( \frac{1}{\sqrt{x + h - 6} + \sqrt{x - 6}} \) as \( h \to 0 \). Substituting \( h = 0 \) gives \( \frac{1}{2\sqrt{x-6}} \). Therefore, \( f'(x) = \frac{1}{2\sqrt{x-6}} \).

Key Concepts

Definition of Derivative

The derivative is a fundamental concept in calculus, essentially describing how a function changes at any given point. It can be thought of as the "instantaneous rate of change" or "slope" of a function at a particular point. The mathematical definition is given as follows:

• For a function \( f(x) \), the derivative \( f'(x) \) is defined as \( f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \).

In simpler terms, the derivative tells us how \( f \) is changing near \( x \). It addresses how small changes in \( x \) result in changes in \( f(x) \). This definition involves a limit, as we explore changes becoming infinitesimally small. Calculating a derivative through this definition can often involve some algebraic manipulation, especially when the function includes roots or more complex terms.

Limit Process

The limit process is crucial in finding derivatives using their definitions. In calculus, limits help us understand the behavior of functions as they approach a specific point, even if they don't actually reach it. Here, the limit \( \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \) describes how the function's average rate of change approaches a single value.
The concept of finding a "limit" means looking at what happens as \( h \), a small increment, gets closer and closer to zero. Instead of evaluating directly at zero, which often leads to indeterminate forms, we evaluate the expression at values arbitrarily close to zero. This process smoothens out abrupt changes and helps in determining the derivative accurately.
For our exercise, applying the limit process allows us to find the derivative of \( \sqrt{x-6} \) and involves simplifying under the limit framework by using algebraic techniques.

Simplifying Expressions

Simplifying expressions is an essential step when dealing with derivatives and limits. Often, the raw expressions obtained in derivatives can be complex or unwieldy. Simplification helps in making the problem manageable, especially before applying limits.

• In our exercise, initially substituting into the derivative formula gives a complicated fraction involving the difference of square roots.
• To simplify these, perform algebraic operations such as multiplying by conjugates or factoring out common terms.

The primary goal is to simplify both the numerator and the denominator such that the limit process can be applied without resulting in indeterminate forms like \( \frac{0}{0} \). This approach is crucial for calculating derivatives, as it lays the groundwork for applying calculus techniques efficiently.

Conjugate

The term "conjugate" refers to a specific algebraic technique that helps simplify expressions involving roots. When you have an expression with square roots, such as \( \sqrt{a} - \sqrt{b} \), multiplying by the conjugate \( \sqrt{a} + \sqrt{b} \) can help eliminate roots from the numerator.

• The product of \( \sqrt{a} - \sqrt{b} \) and \( \sqrt{a} + \sqrt{b} \) results in \( a - b \), a much simpler expression without square roots.
• This technique is used in our exercise to simplify the derivative expression before taking the limit.

Using conjugates is particularly useful in derivative calculations where direct substitution in limits might lead to indeterminate forms. By applying the conjugate, the expression becomes easier to evaluate as \( h \) approaches zero. This step is vital in reaching the final form of the derivative efficiently.