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Find the derivative of each function. $$ f(x)=\ln e^{2 x} $$

Short Answer

Expert verified
The derivative of \( f(x) = \ln e^{2x} \) is 2.

Step by step solution

01

Simplify the Function

First, recognize that the function can be simplified using logarithmic properties. The given function is \( f(x) = \ln e^{2x} \). Using the property \( \ln a^b = b \ln a \), this simplifies to \( f(x) = 2x \) because \( \ln e = 1 \).
02

Differentiate the Simplified Function

Now that we have the simplified function \( f(x) = 2x \), we can differentiate it with respect to \( x \). The derivative of \( 2x \) with respect to \( x \) is simply the coefficient of \( x \), which is \( 2 \).
03

Conclude the Derivative

Since we've found that the derivative of the simplified function \( f(x) = 2x \) is \( 2 \), the derivative of the original function \( \ln e^{2x} \) is also \( 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Logarithmic Differentiation
Logarithmic differentiation is a powerful derivative technique, especially useful with functions involving logarithms or products and quotients of functions. It leverages the properties of logarithms to simplify the differentiation process. When using logarithmic differentiation, the first step often involves taking the natural log of both sides of an equation. This can simplify expressions by turning multiplication into addition and exponentiation into multiplication. For instance, the function we encountered, \(f(x) = \ln e^{2x}\), can be simplified using the logarithmic identity \(\ln a^b = b \ln a\). Given that \(\ln e = 1\), simplifying \(\ln e^{2x}\) directly yields \(2x\). By understanding and applying these properties, differentiation becomes straightforward, leading us effortlessly to the derivative.
Simplifying Expressions
Simplifying expressions is a crucial step in solving calculus problems efficiently. It involves the process of rewriting an expression in a simpler or more manageable form without changing its value. In the exercise, the function \(f(x) = \ln e^{2x}\) is simplified using the basic logarithmic identity \(\ln a^b = b \ln a\). This helps us realize that \(\ln e^{2x} = 2x\) because of \(\ln e = 1\). Simplifying expressions before differentiation can help clarify the problem and make finding the derivative easier. It's especially important when dealing with complex functions as it reduces the likelihood of errors. Remember, simpler forms often reveal straightforward pathways to solutions, as seen here by converting a logarithmic expression to a linear one.
Calculus Problem Solving
The process of solving calculus problems often involves both strategic thinking and a firm grasp of mathematical fundamentals. Here's a step-by-step approach to tackle such problems:
  • Identify the type of function you are dealing with, such as polynomials, exponential functions, or logarithmic functions.
  • Simplify the function if possible. This can involve algebraic manipulations or applying identities like \(\ln a^b = b \ln a\).
  • Once simplified, proceed to find the derivative using basic derivative rules (like the power rule or constant rule).
In this problem, recognizing that simplifying \(f(x) = \ln e^{2x}\) to \(2x\) let us swiftly find that the derivative \(f'(x) = 2\). Approaching calculus with a clear method helps manage the complexity and develops problem-solving skills, making it easier to solve similar problems in the future. Understanding both the underlying concepts and the procedural techniques is key to mastering calculus.

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Most popular questions from this chapter

Sales The weekly sales (in thousands) of a new product are predicted to be \(S(x)=1000-900 e^{-0.1 x}\) after \(x\) weeks. Find the rate of change of sales after: a. 1 week. b. 10 weeks.

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