Problem 51 - Contemporary Precalculus

Question

Find the rule of the derivative of the function \(f .\) [The difference quotients of these functions were found and simplified in Exercises \(57-60 \text { of Section } 5.1 .]\) $$f(x)=\sqrt{x+1}$$

Step-by-Step Solution

Answer

Answer: The derivative of the function \(f(x) = \sqrt{x+1}\) is \(f'(x) = \frac{1}{2\sqrt{x+1}}\).

Step 1: Write the given function and the derivative definition

We have a function \(f(x) = \sqrt{x+1}\), and we want to find the rule of its derivative using the definition: $$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$

Step 2: Compute \(f(x+h)\) and \(f(x)\)

To compute \(f(x+h)\), replace \(x\) with \((x+h)\) in the given function: $$f(x+h) = \sqrt{(x+h)+1} = \sqrt{x+h+1}$$ The expression for \(f(x)\) is already given: $$f(x) = \sqrt{x+1}$$

Step 3: Compute \(f(x+h) - f(x)\)

Now we will find the expression \(f(x+h) - f(x)\): $$f(x+h) - f(x) = \sqrt{x+h+1} - \sqrt{x+1}$$

Step 4: Simplify the difference quotient

We need to simplify the expression \(\frac{f(x+h) - f(x)}{h}\): $$\frac{f(x+h) - f(x)}{h} = \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h}$$ We can rationalize the numerator by multiplying the expression by the conjugate of the numerator, which is \(\sqrt{x+h+1} + \sqrt{x+1}\): $$\frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \cdot \frac{\sqrt{x+h+1} + \sqrt{x+1}}{\sqrt{x+h+1} + \sqrt{x+1}} = \frac{(x+h+1)-(x+1)}{h(\sqrt{x+h+1} + \sqrt{x+1})}$$ This simplifies to: $$\frac{h}{h(\sqrt{x+h+1} + \sqrt{x+1})} = \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}$$

Step 5: Evaluate the limit as \(h\) approaches 0

Now we can evaluate the limit as \(h\) approaches 0: $$f'(x) = \lim_{h\to 0} \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}}$$ By substituting \(h=0\) in the limit, we get: $$f'(x) = \frac{1}{\sqrt{x+1} + \sqrt{x+1}} = \frac{1}{2\sqrt{x+1}}$$ Therefore, the rule of the derivative of the function \(f(x) = \sqrt{x+1}\) is: $$f'(x) = \frac{1}{2\sqrt{x+1}}$$

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept when calculating the derivative of a function. To understand it, think of it as a way to calculate the 'slope' at a specific point on a curve. This slope helps us figure out how a function is changing at any point.

In our exercise, we're given the function \( f(x) = \sqrt{x+1} \). The difference quotient involves creating a new function \( f(x + h) \) to replace \( x \) with \( x + h \), where \( h \) is a very small number close to zero.

• In the given solution, \( f(x+h) = \sqrt{x+h+1} \).
• We then compute \( f(x+h) - f(x) \) to find how much the function changes as \( h \) approaches zero.
• This is expressed as \( \frac{f(x+h) - f(x)}{h} \), which is the difference quotient.

The goal is to simplify this expression to eventually determine the derivative of the function.

Rationalization

Rationalization is a technique we use to simplify expressions, particularly when dealing with square roots. In this task, we used rationalization to simplify the difference quotient.

Our initial task was simplifying \( \frac{\sqrt{x+h+1} - \sqrt{x+1}}{h} \). To do this, we multiply by the conjugate \( \sqrt{x+h+1} + \sqrt{x+1} \), which helps eliminate the square roots in the numerator.

• The conjugate changes \( a - b \) into \( a + b \), where \( a \) and \( b \) are the original square root terms.
• This multiplication involves the formula \((a - b)(a + b) = a^2 - b^2 \), resulting in a simplified numerator.

In our situation, \((x+h+1)-(x+1)\) simplified the numerator to just \( h \), making further simplification possible.

Limit Evaluation

In calculus, evaluating a limit allows us to observe the behavior of a function as it approaches a specific point. In derivatives, this limit helps determine the slope at a single point.

After rationalizing and simplifying the difference quotient, \( \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} \), we need to evaluate it as \( h \) tends to 0.

• Substitute \( h = 0 \) into the expression. This makes it evaluate to \( \frac{1}{\sqrt{x+1} + \sqrt{x+1}} \).
• Simplification gives us \( \frac{1}{2\sqrt{x+1}} \), representing the derivative \( f'(x) \).

This derivative tells us the rate of change of the original function at any point \( x \). When \( h \) approaches 0, it allows for precise calculation at specific values of \( x \).

Function Simplification

Function simplification is the process of making complex algebraic expressions easier to work with. It helps achieve a form where calculations, like limit evaluating, become straightforward.

During the exercise, we dealt with simplifying the rationalizing step and then simplifying through limit evaluation. Simplifying functions is essential, especially in calculus, as it leads to more straightforward and sometimes more insightful answers.

• The rationalization step simplified our expression to \( \frac{1}{\sqrt{x+h+1} + \sqrt{x+1}} \), making the limit evaluation clear and manageable.
• This final expression \( \frac{1}{2\sqrt{x+1}} \) was achieved by simplifying the initial complicated expression into a useful, simple form of the derivative.

Each simplification step contributes to solving the problem efficiently and understanding the changes in \( f(x) \). Simplified forms make it easier to interpret and apply derivatives in problem-solving scenarios.