Problem 59 - Calculus Single Variable

Question

Evaluate the derivative \(f^{\prime}\) of the given function \(f\) in two ways. First, apply the Chain Rule to \(f(x)\) without simplifying \(f(x)\) in advance. Second, simplify \(f(x)\), and then differentiate the simplified expression. Verify that the two expressions are equal. $$ f(x)=\ln (1 / \sqrt{x}) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \ln(1 / \sqrt{x}) \) is \( -\frac{1}{2x} \).

Step 1: Identify the Function Structure

The given function is \( f(x) = \ln(1/\sqrt{x}) \). Notice that this expression can be rewritten as \( \ln((x^{-1/2})) \). This suggests a composition of functions. We have an inner function \( g(x) = x^{-1/2} \) and an outer function \( h(u) = \ln(u) \) where \( u = g(x) \).

Step 2: Apply the Chain Rule to Find the Derivative

To differentiate \( f(x) = \ln(1/\sqrt{x}) \) using the Chain Rule, find the derivatives of the outer and inner functions: 1. Derivative of the outer function, \( h(u) = \ln(u) \), is \( h'(u) = 1/u \).2. Derivative of the inner function, \( g(x) = x^{-1/2} \), is \( g'(x) = -\frac{1}{2}x^{-3/2} \).Now apply the chain rule: \[ f'(x) = h'(g(x)) \cdot g'(x) = \frac{1}{g(x)} \cdot g'(x) = \frac{1}{x^{-1/2}} \cdot \left(-\frac{1}{2}x^{-3/2}\right) = -\frac{1}{2}x^{-1} = -\frac{1}{2} \cdot \frac{1}{x}. \]

Step 3: Simplify the Function

Simplify the function \( f(x) = \ln(1/\sqrt{x}) \) as follows:\[ f(x) = \ln(x^{-1/2}) = -\frac{1}{2}\ln(x). \]

Step 4: Differentiate the Simplified Expression

Differentiate the simplified expression \( f(x) = -\frac{1}{2} \ln(x) \):The derivative of \( \ln(x) \) is \( 1/x \).Using the constant multiple rule, the derivative of \( -\frac{1}{2}\ln(x) \) is:\[ f'(x) = -\frac{1}{2} \cdot \frac{1}{x} = -\frac{1}{2x}. \]

Step 5: Verify the Equality of Derivatives

Both methods to differentiate the function \( f(x) \) resulted in the same derivative:\[ f'(x) = -\frac{1}{2x}. \]This confirms the two expressions for the derivative are equal.

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus used to differentiate composite functions. When you have a function composed of an outer function and an inner function, like in our case, the Chain Rule helps us find the derivative more systematically.
By identifying the outer and inner functions, we express the problem in simpler parts:- Outer function: In our example, it's the natural logarithm, denoted as \( h(u) = \ln(u) \).- Inner function: Here, it's \( g(x) = x^{-1/2} \).
The Chain Rule states that the derivative of \( f(x) = h(g(x)) \) is given by \( f'(x) = h'(g(x)) \cdot g'(x) \). To derive \( f(x) = \ln(1/\sqrt{x}) \), we first find \( h'(u) = 1/u \) and \( g'(x) = -\frac{1}{2}x^{-3/2} \), and then plug them into the rule. This reveals how interconnected differentiations of separate composed components lead to the overall function's derivative.

Function Simplification

Simplifying a function before differentiation can make the process easier and less error-prone. With the function \( f(x) = \ln(1/\sqrt{x}) \), we recognized that rewriting can simplify our calculations:
1. Express \( 1/\sqrt{x} \) as \( x^{-1/2} \).2. Rewrite the entire function as \( f(x) = \ln(x^{-1/2}) \).
Further expansion shows that \( \ln(x^{-1/2}) \) can be simplified to \(-\frac{1}{2}\ln(x) \). This step makes differentiation straightforward because the resulting expression is a simple logarithmic form, which many students find easier to differentiate than a composite expression. Simplification helps double-check work and ensures a derived solution may be verified through straightforward calculation.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithmic function that uses the base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. It's a crucial concept in calculus because of its unique properties that simplify the differentiation process. One important property of the natural logarithm is its derivative: - The derivative of \( \ln(x) \) is \( 1/x \).
In our problem, \( \ln(x) \) is part of an expression \( \ln(x^{-1/2}) \). Knowing the derivative of the natural logarithm is essential for applying differentiation techniques effectively, especially when combined with real constants or coefficients, as seen through the simplified expression \( -\frac{1}{2}\ln(x) \). The natural logarithm makes calculating exponentials manageable and is prominently used in calculus for continuous growth processes.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function, representing the rate of change. Using different techniques helps us handle various function forms. Let's look at some key differentiation techniques employed in the problem:

• Chain Rule: When dealing with composite functions, as explained earlier, the Chain Rule is indispensable for breaking down derivatives into manageable parts.
• Power Rule: This rule simplifies differentiation of powers of \( x \); the derivative of \( x^n \) is \( nx^{n-1} \). It’s crucial when tackling expressions like \( x^{-1/2} \).
• Constant Multiplication Rule: If a function is multiplied by a constant, its derivative is the constant multiplied by the derivative of the function. This principle was applied directly to the scaled expression \(-\frac{1}{2}\ln(x) \).

Mastering these techniques takes practice and allows students to tackle increasingly complex calculus problems. Each method builds on foundational principles to provide comprehensive solutions.