Problem 60 - Calculus Volume 1

Question

For the following exercises, use the definition of a derivative to find \(f^{\prime}(x)\) . $$f(x)=\sqrt{x-6}$$

Step-by-Step Solution

Answer

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

Step 1: Write the Definition of the Derivative

The derivative of a function \(f(x)\) using the limit definition is given by: \[f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\]

Step 2: Substitute the Given Function

Substitute \(f(x) = \sqrt{x-6}\) into the derivative definition to get: \[f'(x) = \lim_{{h \to 0}} \frac{\sqrt{(x+h)-6} - \sqrt{x-6}}{h}\]

Step 3: Simplify the Expression Using Conjugates

Multiply the numerator and the denominator by the conjugate of the numerator: \(\sqrt{(x+h)-6} + \sqrt{x-6}\). This helps to simplify the expression. \[f'(x) = \lim_{{h \to 0}} \frac{(\sqrt{(x+h)-6} - \sqrt{x-6})(\sqrt{(x+h)-6} + \sqrt{x-6})}{h(\sqrt{(x+h)-6} + \sqrt{x-6})}\]

Step 4: Expand and Simplify

The numerator becomes a difference of squares which simplifies to: \[(x+h-6) - (x-6) = h\] which simplifies the function to: \[f'(x) = \lim_{{h \to 0}} \frac{h}{h(\sqrt{(x+h)-6} + \sqrt{x-6})}\]

Step 5: Cancel Out the Terms and Evaluate Limit

Cancel \(h\) in the numerator and denominator: \[f'(x) = \lim_{{h \to 0}} \frac{1}{\sqrt{(x+h)-6} + \sqrt{x-6}}\] Evaluate the limit as \(h \to 0\): \[f'(x) = \frac{1}{2\sqrt{x-6}}\]

Step 6: Write the Derivative

The derivative is written as: \[f'(x) = \frac{1}{2\sqrt{x-6}}\]

Key Concepts

Limit Definition of Derivative

To understand the derivative of a function, we use the limit definition. The derivative of a function at a point, denoted as \(f'(x)\), is essentially the slope of the tangent line to the function at that point. Mathematically, it is expressed as: \[f'(x) = \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h}\] This equation allows us to find how the function changes as we make tiny increments, \(h\), from any point \(x\). Consider \(h\) as a very small number that approaches zero. Substituting a specific function into this equation allows us to calculate its derivative by carefully evaluating the expression within the limit as \(h\) gets exceedingly small.

Simplifying Expressions Using Conjugates

When dealing with functions involving square roots, simplifying expressions can be quite challenging. One effective technique is multiplying by a conjugate. The conjugate of a binomial \(a - b\) is \(a + b\). By multiplying the numerator and denominator by this conjugate, we can simplify complex terms. Applying this to our example with square roots:

• Given: \(\sqrt{(x+h)-6} - \sqrt{x-6}\)
• Conjugate: \(\sqrt{(x+h)-6} + \sqrt{x-6}\)

By multiplying, we use the difference of squares identity, \(a^2-b^2 = (a-b)(a+b)\), which removes the square roots. This process is crucial for simplifying the original expression so that we can evaluate the limit more easily.

Evaluating Limits

With the expression simplified, the next step is to evaluate the limit. Initially, we have: \[\frac{h}{h(\sqrt{(x+h)-6} + \sqrt{x-6})}\] Here, \(h\) can be canceled from the numerator and the denominator, simplifying the expression further. We are then left with: \[\lim_{{h \to 0}} \frac{1}{\sqrt{(x+h)-6} + \sqrt{x-6}}\] Consider what happens as \(h\) approaches zero. Any terms involving \(h\) dwindly steadily close to zero, and the limit can now be evaluated without obstacles. As \(h\) disappears, the expression becomes smoother, leading us to the final result. This limit represents the instantaneous rate of change or the slope of the tangent line at point \(x\).

Function Differentiation

Function differentiation allows us to compute the derivative of a function, which informs us about the function's behavior. In simpler terms, it reveals how a function evolves at any given point. The finalized expression from our limit evaluation in the previous section becomes the derivative of the function. For the function \(f(x)=\sqrt{x-6}\), after our step-by-step process: \[f'(x) = \frac{1}{2\sqrt{x-6}}\] This result is the derivative. Knowing \(f'(x)\) allows us to predict how \(f(x)\) will change with small changes in \(x\). It serves as a fundamental concept for calculus, helping you solve problems involving rates of change, tangents, and optimizations. Embrace function differentiation to unpack the "why" and "how" functions vary in mathematical beauty.