Problem 16 - Calculus

Question

Find the derivatives of the functions $$ e^{\ln \left(x^{2}\right)}+\ln \left(e^{x^{2}}\right) $$

Step-by-Step Solution

Answer

The derivative is \(4x\).

Step 1: Simplify the Expression

The given function is \( e^{\ln(x^2)} + \ln(e^{x^2}) \). Let's simplify it: \( e^{\ln(x^2)} = x^2 \) (because \( e^{\ln(a)} = a \)) and \( \ln(e^{x^2}) = x^2 \) (because \( \ln(e^a) = a \)). Thus, the simplified expression is \( x^2 + x^2 = 2x^2 \).

Step 2: Find the Derivative

The function has simplified to \( 2x^2 \). To find the derivative of \( 2x^2 \), apply the power rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). Therefore, \( \frac{d}{dx}(2x^2) = 2 \cdot 2x^{2-1} = 4x \).

Key Concepts

Power Rule

The power rule is a fundamental technique in calculus used to easily find the derivative of functions in the form of a power. In essence, this rule states that if you have a function such as \( f(x) = x^n \), the derivative \( f'(x) \) is \( n \times x^{n-1} \). The idea is to bring down the exponent as a coefficient and then reduce the exponent by one.

For example, in the given exercise, after simplifying the expression to \( 2x^2 \), we apply the power rule to find the derivative. Here, the function is represented as \( 2 \times x^2 \). The coefficient \( 2 \) remains unaffected by the differentiation, as constants multiply through, while the power rule is applied to \( x^2 \).

• Bring the exponent 2 down to multiply with the existing coefficient 2, giving \( 4 \).
• Then, reduce the exponent 2 by one, resulting in \( x^{2-1} \) which simplifies to \( x^1 \) or simply \( x \).

This combination results in the derivative \( 4x \), effectively using the power rule to reach the solution. It greatly simplifies the process of finding derivatives for polynomial expressions, making it an essential tool in calculus.

Simplification

Simplification is a crucial step before taking derivatives, as it makes the expression easier to manage and reduces the potential for errors. Consider the original function in the exercise: \( e^{\ln(x^2)} + \ln(e^{x^2}) \). At first glance, it looks a bit complicated, but the properties of exponents and logarithms allow for straightforward simplification.

Let's break it down:

• \( e^{\ln(x^2)} \) simplifies to \( x^2 \) because \( e \) and \( \ln \) are inverse functions, and the composition \( e^{\ln(a)} \) reduces to \( a \). Thus, \( a = x^2 \) in this instance.
• Similarly, \( \ln(e^{x^2}) \) simplifies directly to \( x^2 \), as \( \ln \) is the inverse of the exponential function and cancels it out, leaving just the exponent.

Once these simplifications are made, the expression becomes \( 2x^2 \). This simplified form is much easier to work with when finding the derivative, applying rules like the power rule quickly and efficiently. Simplifying complex expressions allows for smooth and accurate calculations, turning daunting problems into manageable ones.

Exponential Functions

Exponential functions are essential in both mathematics and real-world applications, representing growth and decay processes. However, they can sometimes appear tricky, especially when combined with logarithms or in more complex expressions.

In the original problem, you encounter exponential expressions like \( e^{\ln(x^2)} \) and logarithmic expressions \( \ln(e^{x^2}) \). Understanding how these functions interact is key. The exponential function \( e^x \) and the natural logarithm \( \ln(x) \) are inverse operations. This means that when you have a situation like \( e^{\ln(a)} \), you can simplify it directly to \( a \). Similarly, the expression \( \ln(e^a) \) simplifies to \( a \).

Points to remember:

• Exponential and logarithmic functions undo each other: \( e^{\ln(x)} = x \) and \( \ln(e^x) = x \).
• These properties allow for the simplification of expressions, especially when derivatives are involved.

Being comfortable with these transformations is crucial when differentiating or simplifying more complex mathematical problems. They help to clean up expressions, making the derivative process more straightforward and efficient.