Problem 57 - 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. $$ f(x)=\ln \left(x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{2}{x} \) using both methods.

Step 1: Applying the Chain Rule

First, identify the outer function and the inner function. Here, the outer function is \( g(u) = \ln(u) \) and the inner function is \( u(x) = x^2 \). According to the chain rule, the derivative \( f'(x) \) can be found by differentiating the outer function with respect to the inner function and then multiplying by the derivative of the inner function. This gives: \[ f'(x) = \frac{d}{dx}[g(u(x))] = g'(u(x)) \cdot u'(x) \] Since \( g(u) = \ln(u) \), \( g'(u) = \frac{1}{u} \). The derivative of \( u(x) = x^2 \) is \( u'(x) = 2x \). Therefore, \( f'(x) = \frac{1}{x^2} \cdot 2x \). Simplifying gives: \[ f'(x) = \frac{2x}{x^2} = \frac{2}{x} \]

Step 2: Simplifying the Function and Differentiating

Simplify \( f(x) = \ln(x^2) \) using logarithmic identities. The property \( \ln(a^b) = b\ln(a) \) allows us to rewrite \( \ln(x^2) \) as \( 2\ln(x) \).Now differentiate \( f(x) = 2\ln(x) \). The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so: \[ f'(x) = 2 \cdot \frac{1}{x} = \frac{2}{x} \]

Step 3: Verification

Compare the results from Step 1 and Step 2. Both methods resulted in the derivative \( f'(x) = \frac{2}{x} \). This confirms consistency across both approaches and verifies the result.

Key Concepts

Understanding the Chain Rule in Calculus

When working with derivatives in calculus, the Chain Rule is a fundamental tool that helps us differentiate composite functions. A composite function is essentially a function within another function, like nesting dolls. For example, if we have a function like \( f(x) = \ln(x^2) \), we can see that it's a combination of the outer function \( g(u) = \ln u \) and the inner function \( u(x) = x^2 \).

The Chain Rule tells us that the derivative of a composite function \( f'(x) \) can be determined by first finding the derivative of the outer function with respect to the inner function, and then multiplying it by the derivative of the inner function itself. Think of it as peeling away layers, working from the outside in.

• The derivative of the outer function \( g(u) = \ln(u) \) is \( g'(u) = \frac{1}{u} \).
• The derivative of the inner function \( u(x) = x^2 \) is \( u'(x) = 2x \).

Using the Chain Rule, we multiply these two results to get \( f'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \).

This rule simplifies the differentiation process for complex functions and is especially powerful when dealing with functions expressed in layers.

Logarithmic Differentiation Made Easy

Logarithmic Differentiation is a powerful technique often used when dealing with products, quotients, or exponential functions. It's particularly useful when simplifying the differentiation of a complex function like \( f(x) = \ln(x^2) \).

By utilizing properties of logarithms and simplifying the function first, we make the differentiation process much easier. In the case of \( \ln(x^2) \), we apply the property \( \ln(a^b) = b\ln(a) \), which simplifies \( \ln(x^2) \) to \( 2\ln(x) \). This reformulation transforms a potentially tricky expression into something far simpler to differentiate.

Once simplified to \( 2\ln(x) \), the differentiation becomes straightforward. The derivative of \( \ln(x) \) is \( \frac{1}{x} \), so for the function \( 2\ln(x) \), the derivative becomes:

• \( f'(x) = 2 \cdot \frac{1}{x} = \frac{2}{x} \).

This demonstrates how logarithmic identities not only simplify the problem but also provide a clean and efficient path to the solution. Notably, this technique often helps to avoid more complex multi-step processes.

Simplification Techniques in Calculus

Simplification in calculus is not just about making expressions shorter or neater; it's about making them manageable and solvable. When you simplify before differentiating, you often reduce the complexity of your calculations, leading to quicker and more accurate results. In the given problem, before differentiation, we simplified \( f(x) = \ln(x^2) \) to \( 2\ln(x) \).

One main simplification strategy is using properties of logarithms, such as \( \ln(a^b) = b\ln(a) \). This approach not only simplifies the expression but helps avoid potential errors during the differentiation process.

Moreover, simplification is crucial when verifying your results. After obtaining the derivatives using different methods (Chain Rule and Logarithmic Differentiation), simplified forms make it easy to compare and verify if they match, ensuring consistency. This ultimately builds confidence in your solution's accuracy.

• Simplification reduces complexity.
• Makes verification straightforward.
• Avoids unnecessary errors.

In summary, simplification in calculus is like clearing the path to a solution, ensuring it's straight, accurate, and efficient.