Problem 58 - 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. $$ f(x)=\ln \left(x e^{x}\right) $$

Step-by-Step Solution

Answer

The derivative \(f'(x) = \frac{1}{x} + 1\) for both methods.

Step 1: Apply the Chain Rule

We have the function \(f(x) = \ln(x e^x)\). The outer function is \(g(u) = \ln(u)\), where \(u = x e^x\). The derivative of \(g(u)\) is \(g'(u) = \frac{1}{u}\). The inner function \(u(x) = x e^x\), has a derivative \(u'(x) = e^x + x e^x\) (using the product rule). Applying the chain rule gives:\[ f'(x) = \frac{1}{u} imes u'(x) = \frac{1}{x e^x} imes (e^x + x e^x) \] Simplifying the expression:\( f'(x) = \frac{e^x + x e^x}{x e^x}\)\.

Step 2: Simplify and Differentiate the Simplified Expression

Let's simplify the original function \(f(x) = \ln(x e^x)\). Using the log properties, this becomes \(f(x) = \ln(x) + \ln(e^x)\), which simplifies to \(f(x) = \ln(x) + x\) since \(\ln(e^x) = x\).Now, differentiate \(f(x) = \ln(x) + x\):The derivative of \(f(x)\) is:\[ f'(x) = \frac{1}{x} + 1 \]

Step 3: Verify the Equality of the Two Derivatives

Now, compare the results from the chain rule and the simplified method.From Step 1:\[ f'(x) = \frac{e^x + x e^x}{x e^x} = \frac{e^x(1 + x)}{x e^x} = \frac{1 + x}{x} \]From Step 2:\[ f'(x) = \frac{1}{x} + 1 \]Both expressions simplify to the same thing:\( f'(x) = \frac{1}{x} + 1 \).Thus, the derivatives are indeed equal.

Key Concepts

Product Rule

The Product Rule is a fundamental technique in calculus used to find the derivative of a product of two functions. When we have two functions, say \( u(x) \) and \( v(x) \), their derivative when multiplied together is not just the product of their derivatives. Instead, the Product Rule tells us:

• \( (uv)' = u'v + uv' \)

This formula is essential for correctly differentiating products of functions.
In the original exercise, we apply the Product Rule to the term \( x e^x \). Here, \( u(x) = x \) and \( v(x) = e^x \).

• First, find \( u'(x) = 1 \) and \( v'(x) = e^x \).
• Apply the rule: \( u'v + uv' = 1 \times e^x + x \times e^x = e^x + xe^x \).

The Product Rule allows us to correctly differentiate this product, which is a crucial step in applying the Chain Rule to the given function in the original exercise.

Logarithmic Differentiation

Logarithmic Differentiation is a powerful strategy for differentiating complex functions, particularly useful when dealing with functions involving products, quotients, or even powers. By using the properties of logarithms, we can simplify our differentiation process.
In the context of the given function \( f(x) = \ln(x e^x) \), taking the natural log enables us to break it into the sum of logs:

• \( \ln(x e^x) = \ln(x) + \ln(e^x) \)
• Using the property \( \ln(e^x) = x \), this simplifies to \( \ln(x) + x \).

By rewriting the original complex logarithmic expression, we transform it into an easier form to differentiate.
After simplifying with log properties, differentiating becomes straightforward:

• \( \frac{d}{dx}[\ln(x) + x] = \frac{1}{x} + 1 \)

Logarithmic Differentiation not only simplifies the function before differentiation, but confirms that the simpler derivative aligns with the one calculated using the Chain Rule, demonstrating the equality of different methods.

Function Simplification

Function Simplification plays a pivotal role in calculus, as it helps in reducing complex functions into more manageable forms. This is particularly advantageous when calculating derivatives, as seen in our exercise where simplifying \( f(x) = \ln(x e^x) \) was necessary for method comparison.
Simplifying involves using mathematical identities and properties to break the function into simpler parts:

• For \( f(x) = \ln(x e^x) \), we use the logarithmic property: \( \ln(a \cdot b) = \ln(a) + \ln(b) \).
• This turns \( \ln(x e^x) \) into \( \ln(x) + \ln(e^x) \).
• Knowing \( \ln(e^x) = x \), the function simplifies to \( \ln(x) + x \).

After simplification, the function becomes easier to handle, allowing us to use basic differentiation rules rather than complex techniques.
In our exercise, this allowed us to see directly that the derivative \( \frac{1}{x} + 1 \) matches the result we obtained using the Chain Rule, proving both perspectives are valid. Simplifying functions like this is a strategic move, not just for derivatives, but for solving many calculus problems efficiently.