Problem 73 - Calculus Single Variable

Question

Find \(f^{\prime}(x)\) for the given function \(f\). $$ f(x)=6 \sqrt{\sqrt{2 x-1}+\sqrt{2 x+1}} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{3}{\sqrt{\sqrt{2x-1} + \sqrt{2x+1}}} \left( \frac{1}{\sqrt{2x-1}} + \frac{1}{\sqrt{2x+1}} \right) \).

Step 1: Understand the Problem

We need to find the derivative of the function \( f(x) = 6 \sqrt{\sqrt{2x-1} + \sqrt{2x+1}} \). Our approach will involve using the chain rule multiple times due to the nested functions.

Step 2: Identify the Outer Function

The function \( f(x) \) can be written as \( u = \sqrt{\sqrt{2x-1} + \sqrt{2x+1}} \). The outermost operation on \( f(x) \) is multiplication by 6 and taking the square root of \( u \). Use the chain rule to find the derivative of the form \( g(u) = \sqrt{u} \), i.e., \( g'(u) = \frac{1}{2\sqrt{u}} \).

Step 3: Derivative of the Inner Function

Next, we need to find the derivative of the inner function: \( u = \sqrt{2x-1} + \sqrt{2x+1} \). The derivative with respect to \( x \) is \( u' = \frac{1}{2\sqrt{2x-1}} \cdot 2 + \frac{1}{2\sqrt{2x+1}} \cdot 2 = \frac{1}{\sqrt{2x-1}} + \frac{1}{\sqrt{2x+1}} \).

Step 4: Apply the Chain Rule

Applying the chain rule, the derivative of \( f(x) = 6 \sqrt{u} \) is:\[ f'(x) = 6 \cdot \left( \frac{1}{2\sqrt{u}} \right) \cdot u' = \frac{3}{\sqrt{u}} \cdot \left( \frac{1}{\sqrt{2x-1}} + \frac{1}{\sqrt{2x+1}} \right), \]where \( u = \sqrt{2x-1} + \sqrt{2x+1} \).

Step 5: Substitute Back and Simplify

Substitute \( u = \sqrt{2x-1} + \sqrt{2x+1} \) back into the derivative expression:\[ f'(x) = \frac{3}{\sqrt{\sqrt{2x-1} + \sqrt{2x+1}}} \left( \frac{1}{\sqrt{2x-1}} + \frac{1}{\sqrt{2x+1}} \right). \]

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. It's like peeling an onion, layer by layer, to find the rate of change from the inside out. When applying the chain rule, we distinguish between the "outer" function and the "inner" function.
- The outer function is the last operation performed. In our case, it is taking the square root of something.- The inner functions are under the "hood" of the outer operation. They need individual differentiation before anything else. In the exercise, the outermost operation on the function is multiplying by 6 and taking the square root.By applying the chain rule, we find the derivative of the outer function and multiply it by the derivative of the inner function.
Always remember:

• Identify the nested structure of the function.
• Differentiate the outer part first.
• Use the chain rule formula: if you have a function of a function, such as \( y = g(h(x)) \), the derivative is \( y' = g'(h(x)) \times h'(x) \).

This approach is crucial in tackling complex functions, and it builds the foundation for understanding derivatives of nested functions.

Nested Functions

Nested functions appear when you have one function inside another, creating multiple layers of operations. Think of it like Russian nesting dolls; each doll represents a function inside another.
For our exercise, the innermost functions are \( \sqrt{2x - 1} \) and \( \sqrt{2x + 1} \). Together, they form another layer: \( u = \sqrt{2x-1} + \sqrt{2x+1} \). Then, the entire expression is wrapped in yet another function, which is the square root. This multi-layer nesting requires us to carefully track and differentiate each layer step-by-step.When dealing with nested functions:

• Break down the function into recognizable layers or pieces.
• Differentiate from the outermost layer inward using the chain rule technique for each distinct level.
• Track substitutions and its impact on each level to avoid errors.

Understanding the structure of nested functions helps identify what needs differentiating first when tackling complex calculus problems.

Square Root Function

The square root function, symbolized as \( \sqrt{x} \) or \( x^{1/2} \), is a specific and common type of root function. Its derivative is found using basic differentiation rules and is crucial in the chain rule application.Here's how it works:

• The derivative of a simple square root function \( f(x) = \sqrt{x} \) is \( f'(x) = \frac{1}{2\sqrt{x}} \).

In complex functions, like our given \( f(x) = 6 \sqrt{\sqrt{2x-1} + \sqrt{2x+1}} \), you repeat twice the differentiation for the square roots during the chain rule application.
Remember:

• The square root function's behavior can drastically alter the appearance of derivative equations. Pay attention to fractional powers.
• Square roots translate into derivatives that frequently feature the denominator with another square root.
• “Chains” of square roots require you to differentiate each root step-by-step until you reach the inner bulk of the expression.

Mastering the square root's derivative simplifies working through equations and prepares you for numerous problems you'd encounter in calculus.