Problem 9 - Single Variable Calculus

Question

Use the definition of the derivative to find the derivative of the function. What is its domain? \(f(x)=\sqrt{x+1}\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\sqrt{x+1}\) is given by \(f'(x)=\frac{1}{2\sqrt{x+1}}\). The domain of the derivative is \((-1, \infty)\) or \(x>-1\).

Step 1: Write down the definition of the derivative

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

Step 2: Substitute the function into the definition

Substitute the function \(f(x)=\sqrt{x+1}\) into the definition of the derivative, to find the difference quotient: \[\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{(x+h)+1}-\sqrt{x+1}}{h}\]

Step 3: Simplify the difference quotient

In order to simplify the difference quotient, we'll multiply the numerator and the denominator by the conjugate of the numerator: \[\frac{\sqrt{(x+h)+1}-\sqrt{x+1}}{h} \cdot \frac{\sqrt{(x+h)+1}+\sqrt{x+1}}{\sqrt{(x+h)+1}+\sqrt{x+1}}\] This will give us: \[\frac{((x+h)+1)-(x+1)}{h(\sqrt{(x+h)+1}+\sqrt{x+1})}\]

Step 4: Further simplify the difference quotient

Now, simplify the numerator by canceling out the terms: \[\frac{h}{h(\sqrt{(x+h)+1}+\sqrt{x+1})}\] Notice that \(h\) can be canceled from the numerator and the denominator, which leaves us with: \[\frac{1}{\sqrt{(x+h)+1}+\sqrt{x+1}}\]

Step 5: Evaluate the limit

Now, we want to find the limit as \(h\) approaches \(0\): \[f'(x)=\lim_{h\to 0}\frac{1}{\sqrt{(x+h)+1}+\sqrt{x+1}}\] As \(h\) approaches \(0\), the expression becomes: \[f'(x)=\frac{1}{\sqrt{x+1}+\sqrt{x+1}}\]

Step 6: Simplify and find the derivative

The final step in finding the derivative is to simplify the expression: \[f'(x)=\frac{1}{2\sqrt{x+1}}\]

Step 7: Find the domain of the derivative

To find the domain of the derivative, we need to determine the values of \(x\) for which \(f'(x)\) is defined. Since we cannot have the square root of a negative number or a division by zero, we need to find the values of \(x\) that make the denominator nonzero and the radicand non-negative: \[2\sqrt{x+1}>0\]\[\Rightarrow\sqrt{x+1}>0\] This means that \(x+1>0\), and therefore, \(x>-1\). So, the domain of the derivative is \((-1, \infty)\), which we can write as \(x>-1\).

Key Concepts

Difference Quotient

The difference quotient is a crucial concept when discussing derivatives. It provides a way to approximate the slope of the tangent line to a curve at a certain point. Imagine you are measuring the slope of a hill at a tiny section; this is similar to finding the difference quotient.
The formula for the difference quotient is:

• \[\frac{f(x+h)-f(x)}{h}\]

Here, \(h\) represents a small change in \(x\), and \(f(x+h)\) is the function value at \(x+h\). By calculating this fraction, you find how much the function \(f(x)\) changes per unit \(h\).
It's like calculating an average rate of change over an interval, giving a snapshot of the function's behavior. This foundational tool leads into defining the derivative itself.

Limit

Limits are essential when moving from the difference quotient to finding the exact derivative. When we use limits, we take our approximation of change and make it precise.
The definition of a derivative involves taking the limit of the difference quotient as \(h\) approaches zero:

• \[\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\]

This approach ensures the slope calculated is accurate at a single point, revealing the instantaneous rate of change.
Understanding limits is vital for calculus as they help solve problems where straightforward methods break down. Think of limits as a tool that zooms into the tiniest details of a function's behavior.

Domain of a Function

The domain of a function denotes all the permissible input values \(x\) for which the function results in a real number. When calculating derivatives, knowing the domain helps you understand where the function and its derivative make sense.
For the function \(f(x) = \sqrt{x+1}\), the expression under the square root must be non-negative.

• Therefore, \(x+1\geq0\)
• Which implies \(x\geq-1\)

For the derivative \(f'(x)\), further considerations such as avoiding division by zero are essential. Hence, ensuring the domain also respects the needs of the derivative is crucial. This gives us a domain for the derivative: \((-1, \infty)\).
By understanding domains, calculations remain valid, and complex functions are correctly interpreted in their respective intervals.

Simplifying Expressions

Simplifying expressions allows us to manipulate complex function outputs or derivatives into easier, more understandable forms.
During the derivative process, you often encounter intricate expressions when working through the difference quotient. By using algebraic techniques like multiplying by a conjugate, you can simplify these expressions into more manageable terms.

• For example, \[\frac{\sqrt{(x+h)+1} - \sqrt{x+1}}{h}\] can be burdened with complexity.
• Multiplying by the conjugate gives \[\frac{((x+h)+1) - (x+1)}{h(\sqrt{(x+h)+1} + \sqrt{x+1})}\]

This process, as demonstrated, helps remove problematic forms like zero in the denominator or complex radicals, leading to clearer derivatives.
Mastering simplification is vital for approaching calculus problems more easily and accurately, ensuring less error-prone solutions.