Question

Derivatives Find and simplify the derivative of the following functions. \(y=\frac{x-a}{\sqrt{x}-\sqrt{a}},\) where \(a\) is a positive constant

Short answer

Answer: The simplified derivative of the function is \(\frac{dy}{dx} = \frac{x+a-2\sqrt{ax}}{2\sqrt{x}(\sqrt{x}-\sqrt{a})^2}\).

Step-by-step solution

Apply the Quotient Rule for Differentiation

For a function \(y=\frac{f(x)}{g(x)}\), the derivative is given by $$\frac{dy}{dx} = \frac{g(x)\frac{df(x)}{dx} - f(x)\frac{dg(x)}{dx}}{[g(x)]^2}.$$ In our case, \(f(x)=x-a\) and \(g(x)=\sqrt{x}-\sqrt{a}.\) So we first need to find \(\frac{df(x)}{dx}\) and \(\frac{dg(x)}{dx}\). For \(\frac{df(x)}{dx}\), we differentiate \(f(x)=x-a\) with respect to x: $$\frac{df(x)}{dx}=1$$ For \(\frac{dg(x)}{dx}\), we differentiate \(g(x)=\sqrt{x}-\sqrt{a}\) with respect to x. Note that \(\sqrt{x}=x^{1/2}\): $$\frac{dg(x)}{dx}=\frac{d}{dx}(x^{1/2}-a^{1/2})=\frac{1}{2}(x^{-1/2})=\frac{1}{2\sqrt{x}}$$ Now we can apply the quotient rule with \(f(x)=x-a\), \(g(x)=\sqrt{x}-\sqrt{a}\), \(\frac{df(x)}{dx}=1\), \(\frac{dg(x)}{dx}=\frac{1}{2\sqrt{x}}\).

Substitute into Quotient Rule Formula and Simplify

Using the expressions we obtained in step 1, we can substitute and compute the derivative as follows: \begin{align*} \frac{dy}{dx} &= \frac{[\sqrt{x}-\sqrt{a}](1)-(x-a)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x}-\sqrt{a})^2}\\ &= \frac{\sqrt{x}-\sqrt{a}-\frac{x-a}{2\sqrt{x}}}{(\sqrt{x}-\sqrt{a})^2} \end{align*} To simplify, we can multiply both the numerator and denominator by \(2\sqrt{x}\) to eliminate the fraction: $$\frac{dy}{dx} = \frac{(2\sqrt{x}-2\sqrt{a})\sqrt{x}-(x-a)}{2\sqrt{x}(\sqrt{x}-\sqrt{a})^2}$$

Combine and Simplify

Now let's combine and simplify the numerator: \begin{align*} (2\sqrt{x}-2\sqrt{a})\sqrt{x}-(x-a) &= (2\sqrt{x}\sqrt{x}-2\sqrt{a}\sqrt{x})-(x-a) \\ &= (2x-2\sqrt{ax})-(x-a) \\ &= x+a-2\sqrt{ax} \end{align*} Therefore, the derivative of the given function, after simplifying, is: $$\frac{dy}{dx} = \frac{x+a-2\sqrt{ax}}{2\sqrt{x}(\sqrt{x}-\sqrt{a})^2}$$

Key concepts

Understanding the Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of a function that is the quotient of two smaller functions. In simpler terms, it's a way to differentiate when you have one function divided by another. To use the quotient rule, for a function defined as \( y = \frac{f(x)}{g(x)} \), the derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2} \)

Breaking this down, \( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \) respectively. This formula helps to methodically differentiate across numerators and denominators.
In our problem, we applied the quotient rule to \( y = \frac{x-a}{\sqrt{x}-\sqrt{a}} \), identifying \( f(x) = x-a \) and \( g(x) = \sqrt{x} - \sqrt{a} \). By calculating their derivatives (\( f'(x) = 1 \) and \( g'(x) = \frac{1}{2\sqrt{x}} \)), we were able to substitute these into the quotient rule.

The Art of Simplification

Simplification is a crucial part of calculus problem solving, especially after applying the quotient rule. Once you have your derivative expression, it's important to simplify it so that it's easier to understand and use in further calculations.
In the exercise provided, simplification involved dealing with complex fractions. By multiplying both the numerator and the denominator by \( 2\sqrt{x} \), we were able to eliminate a fraction within the numerator. This is a common technique in simplification: multiplying by a term that cleans up the expression without affecting its value.

• The expression \( \frac{\sqrt{x}-\sqrt{a} - \frac{x-a}{2\sqrt{x}}}{(\sqrt{x}-\sqrt{a})^2} \) required combining terms in the numerator to solve for \( x + a - 2\sqrt{ax} \).

Simplifying expressions makes working with derivatives much smoother and showcases the beauty of clean, concise calculus.

The Process of Differentiation

Differentiation is a fundamental concept in calculus that refers to finding the rate at which a function is changing at any given point. This process is key for determining the derivative of a function.
To differentiate a function, you follow a set procedure based on rules such as the product rule, chain rule, and in this case, the quotient rule. By finding the derivative of both the numerator \( f(x) = x-a \) and the denominator \( g(x) = \sqrt{x} - \sqrt{a} \), the quotient rule could easily be applied.

• For \( f(x) = x-a \), the differentiation is straightforward, \( f'(x) = 1 \).
• For \( g(x) = \sqrt{x} - \sqrt{a} \), recognizing \( \sqrt{x} \) as \( x^{1/2} \) allows us to differentiate to get \( g'(x) = \frac{1}{2\sqrt{x}} \).

Understanding differentiation gives insight into how functions behave and is instrumental in problem-solving across many scientific fields.

Strategic Calculus Problem Solving

Solving calculus problems like these requires a strategic approach, mixing different skills and concepts for a correct and simplified answer. Calculus problem solving distills down to following a methodical procedure:

• First, identify the components of your function. Here, \( f(x) = x-a \) and \( g(x) = \sqrt{x} - \sqrt{a} \).
• Next, use appropriate differentiation rules, such as the quotient rule in this scenario, to find derivatives of these components.
• Simplify your findings to a form that is practical for interpretation or further use.

This structured approach helps prevent errors and achieves a concise final result. By applying these techniques consistently, calculus problems become manageable and less daunting. Each step builds on the previous, thus solidifying your understanding of fundamental calculus concepts. Whether you are working on a simple derivative or a complex integral, strategic problem-solving skills remain crucial.