Problem 39 - Calculus and its applications

Question

Differentiate each function $$ f(x)=\sqrt{\frac{x^{2}+x}{x^{2}-x}} $$

Step-by-Step Solution

Answer

The derivative of the function is \(-x^2 \cdot \frac{\sqrt{x^2 - x}}{\sqrt{x^2 + x} (x^2-x)^2}\).

Step 1: Set up the Function for Differentiation

We begin by expressing the function in a form suitable for applying differentiation rules. The function is given as: \[f(x) = rac{ extstyle extstyle ig(rac{x^2 + x}{x^2 - x}ig)^{1/2}}\] Notice this is a composite function and requires the use of the Chain Rule and the Quotient Rule for differentiation.

Step 2: Differentiate the Outer Function Using the Chain Rule

The Chain Rule states: if \(y = g(u)\) and \(u = h(x)\), then \(y' = g'(u) \cdot h'(x)\). In our case, \( g(u) = u^{1/2} \) where \( u = \frac{x^2 + x}{x^2 - x} \). The derivative of \(u^{1/2}\) is \( \frac{1}{2} u^{-1/2}\). Thus, we have: \[\frac{d}{du} \sqrt{u} = \frac{1}{2} u^{-1/2} = \frac{1}{2} \left(\frac{x^2 + x}{x^2 - x}\right)^{-1/2}\] We still need to find \(h'(x)\).

Step 3: Differentiate the Inner Function Using the Quotient Rule

The Quotient Rule states: for \( u = \frac{v}{w} \), \( u' = \frac{v'w - vw'}{w^2} \). Here, \[ v = x^2 + x \quad \text{and} \quad w = x^2 - x \] Differentiate both:\[ v' = 2x + 1 \quad \text{and} \quad w' = 2x - 1 \] Plug these into the Quotient Rule:\[ \frac{d}{dx} \left(\frac{x^2 + x}{x^2 - x}\right) = \frac{(2x + 1)(x^2 - x) - (x^2 + x)(2x - 1)}{(x^2 - x)^2} \]

Step 4: Simplify the Derivative of the Inner Function

Simplify the expression from the previous step:\[= \frac{(2x^3 + 1x^2 - 2x^2 - 1x) - (2x^3 - 1x^2 + 2x^2 - x)}{(x^2 - x)^2}\]\[= \frac{2x^3 + x^2 - 2x^2 - x - 2x^3 + x^2 - 2x^2 + x}{(x^2-x)^2}\]\[= \frac{0x^3 - 2x^2 + 0x}{(x^2-x)^2}\]\[= \frac{-2x^2}{(x^2-x)^2}\]

Step 5: Combine the Results from the Chain Rule and Quotient Rule

Now, substitute back into the original chain rule differentiation:\[f'(x) = \frac{1}{2} \left(\frac{x^2 + x}{x^2 - x}\right)^{-1/2} \cdot \frac{-2x^2}{(x^2-x)^2}\]Simplify by canceling constants and rearranging: \[f'(x) = -x^2 \left(\frac{x^2 - x}{x^2 + x}\right)^{1/2} \cdot \frac{1}{(x^2-x)^2}\]

Step 6: Final Simplification

Multiply out the power of the terms:\[f'(x) = -x^2 \cdot \frac{\sqrt{x^2 - x}}{\sqrt{x^2 + x} (x^2-x)^2}\]This is the most simplified form that captures the derivative of the original function after accounting for all composite and multiple rule applications.

Key Concepts

Chain Rule

The Chain Rule is an essential technique in calculus for differentiating composite functions. To understand it, think of a composite function as one function inside another. For example, in the function \( f(x) = \sqrt{\frac{x^2 + x}{x^2 - x}} \), the inner function \( u = \frac{x^2 + x}{x^2 - x} \) is within the outer function \( \sqrt{u} \). This setup is ideal for the Chain Rule. Here’s the simplicity of it: when you have a situation with a dependent inner and outer function, you take the derivative of the outer function with respect to the inner function (\( u \)) and then multiply by the derivative of the inner function with respect to \( x \).

• In formula terms, if \( y = g(h(x)) \), the derivative is \( y' = g'(h(x)) \cdot h'(x) \).
• This helps break down complex differentiation into manageable parts.

To apply this to our function, differentiate the outer part \( \sqrt{u} \) to get \( \frac{1}{2}u^{-1/2} \), then multiply it by the derivative of \( u \). It allows you to efficiently tackle the most intricate parts of differentiation tasks.

Quotient Rule

The Quotient Rule is necessary when you differentiate a fraction or a ratio of two functions. It answers the question: how does the ratio of two functions change as the input changes? The rule says that if you have a function \( u = \frac{v}{w} \), the derivative \( u' \) is given by:

• \[ u' = \frac{v'w - vw'}{w^2} \]
• This formula helps calculate the change in the quotient of \( v\) and \( w \).

Let's apply this to the function where \( v = x^2 + x \) and \( w = x^2 - x \). By first differentiating \( v \) and \( w \) individually, i.e., \( v' = 2x + 1 \) and \( w' = 2x - 1 \), we plug these values into the quotient formula. So, it becomes a matter of algebraic simplification: calculate \( (2x + 1)(x^2 - x) \) and \( (x^2 + x)(2x - 1) \), subtract them, and divide by \( (x^2 - x)^2 \). The result is the derivative of the inner function in our original problem.

Composite Function Differentiation

Composite functions, functions formed by combining two or more functions, are common in differentiation problems. They require special handling, primarily using the Chain Rule, to carefully determine their derivatives. The task becomes identifying the "layers" of functions involved and differentiating them in sequence. Consider our example function \( f(x) = \sqrt{\frac{x^2 + x}{x^2 - x}} \). It is composed of a square root function on the outside and a quotient on the inside. The process:

• Recognize each function's role and list them separately.
• Differentiate them, starting from the outermost to the innermost, and multiply as required by the Chain Rule.

For this example, use the Chain Rule to manage \( \sqrt{u} \) and employ the Quotient Rule for \( u = \frac{x^2 + x}{x^2 - x} \). The culmination of these results gives us the derivative of the original composite function. This structured approach dismantles complexity, layer by layer, making it straightforward once you understand each function’s role.