Problem 57 - Brief Applied Calculus

Question

Find the derivative of each function. $$ \frac{\sqrt{x}-1}{\sqrt{x}+1} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{x + 2\sqrt{x} + 1} \).

Step 1: Understand the Function Structure

The given function is a rational expression \( \frac{\sqrt{x}-1}{\sqrt{x}+1} \). It has a square root in both the numerator and the denominator, which suggests using the quotient rule to find its derivative.

Step 2: Identify Components for the Quotient Rule

For the quotient \( \frac{u}{v} \), the derivative is given by the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \). In our function, \( u = \sqrt{x} - 1 \) and \( v = \sqrt{x} + 1 \).

Step 3: Differentiate the Numerator

The numerator is \( u = \sqrt{x} - 1 \). Its derivative is \( \frac{du}{dx} = \frac{d}{dx}(x^{1/2} - 1) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

Step 4: Differentiate the Denominator

The denominator is \( v = \sqrt{x} + 1 \). Its derivative is \( \frac{dv}{dx} = \frac{d}{dx}(x^{1/2} + 1) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

Step 5: Apply the Quotient Rule

Using the quotient rule formula, substitute the derivatives into it: \(\frac{d}{dx} \left( \frac{\sqrt{x} - 1}{\sqrt{x} + 1} \right) = \frac{(\sqrt{x} + 1) \cdot \frac{1}{2\sqrt{x}} - (\sqrt{x} - 1) \cdot \frac{1}{2\sqrt{x}} }{(\sqrt{x} + 1)^2} \).

Step 6: Simplify the Numerator

Calculate the simplified form of the numerator: \( (\sqrt{x} + 1) \cdot \frac{1}{2\sqrt{x}} - (\sqrt{x} - 1) \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2} - (-\frac{1}{2}) = 1 \).

Step 7: Simplify the Denominator

The denominator of the expression \( (\sqrt{x} + 1)^2 \) becomes \( x + 2\sqrt{x} + 1 \).

Step 8: Final Derivative Expression

Combine the simplified numerator and denominator to get the final derivative: \( \frac{1}{x + 2\sqrt{x} + 1} \).

Step 9: Verify Process

Ensure each differentiation step followed derivative rules and confirm the simplification was accurate.

Key Concepts

Quotient Rule

When you have a function that is a ratio of two differentiable functions, you can use the Quotient Rule to find its derivative. The Quotient Rule is essential in calculus for handling these types of functions. The general formula looks like this:

• If you have a function of the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), then the derivative is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).

Use the Quotient Rule when differentiating a quotient to ensure accuracy. It's like performing a focused subtraction between two product derivatives, and then dividing by the square of the denominator. This particular rule simplifies the process of finding the derivative, as it accounts for both parts of a function. It keeps everything orderly.
In this exercise, \( u \) and \( v \) are both square root functions which make this rule particularly useful. With denominator and numerator being similar can make derivatives easier, often leading to beautiful simplifications.

Square Roots in Calculus

Square roots are common in mathematical equations and, in calculus, handling them requires special attention. When differentiating square root functions, rewriting them as exponential expressions is often helpful. Recall that \( \sqrt{x} = x^{1/2} \). This expression allows us to apply standard rules of calculus more easily, such as the power rule.

• For instance, the derivative of \( x^{1/2} \) is \( \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \).

Recognizing this conversion is a crucial step when you hit square roots during differentiation. It lets you apply the power rule without getting bogged down by radical signs. In our exercise, both the numerator and denominator include a square root, and differentiating them follows this method.
By breaking complex expressions into more manageable forms, it simplifies finding derivatives. This helps in understanding the function's behavior more thoroughly.

Rational Functions

Rational functions are expressions that form the quotient of two polynomials. In this exercise, however, we encounter a rational function that involves square roots. This adds a layer of complexity due to the square roots in both the numerator and denominator.

• Always remember: A rational function is expressed as \( \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials.
• Despite square roots, they maintain the basic form and can be manipulated similarly using specific rules, like the quotient rule and power rule for differentiation.

The goal when handling these functions is to find derivatives or to simplify them by eliminating radicals when possible. The challenge is to maintain accuracy throughout simplification, particularly when both parts of the function mirror each other in structure, leading often to neat results.
In our example, simplifying the square roots in the rational function ultimately leads to easier integration of the quotient rule. Understanding this is key to mastering calculus topics.