Problem 81 - Calculus and its applications

Question

Differentiate. $$ g(x)=e^{x} \ln x^{2} $$

Step-by-Step Solution

Answer

The derivative of \( g(x) = e^x \ln(x^2) \) is \( g'(x) = \frac{2e^x}{x} + 2e^x \ln(x) \).

Step 1: Identify the Functions

We have two functions: an exponential function and a logarithmic function. Specifically, the function is a product of these two: \[ g(x) = e^{x} \cdot \ln(x^2) \]

Step 2: Recall the Product Rule

The Product Rule for differentiation states that if you have two functions multiplied together, \( u(x) \) and \( v(x) \), then \[ (uv)' = u'v + uv' \]

Step 3: Assign and Differentiate Each Function

Assign \( u(x) = e^x \) and \( v(x) = \ln(x^2) \). Now find their derivatives:- Differentiate \( u(x) = e^x \): \( u'(x) = e^x \) - Differentiate \( v(x) = \ln(x^2) \) using the chain rule: first, recall \( \ln(x^2) = 2\ln(x) \), then \( v'(x) = \frac{2}{x} \)

Step 4: Apply the Product Rule

Using the derivatives found in Step 3, apply the Product Rule: \[ g'(x) = u'v + uv' = e^x \cdot \ln(x^2)' + \ln(x^2) \cdot e^x \] Substitute the derivatives:\[ g'(x) = e^x \cdot \frac{2}{x} + 2\ln(x) \cdot e^x \]

Step 5: Simplify the Expression

Combine and simplify the terms:\[ g'(x) = \frac{2e^x}{x} + 2e^x \ln(x) \]This is the final differentiated form of the function \( g(x) = e^x \ln(x^2) \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which focuses on finding the rate at which a function changes at any given point. This rate is what we call the derivative. In essence, differentiation helps us understand how a function behaves, especially when dealing with changes in variables.

For example, consider a function where two different mathematical expressions are multiplied together. When you need to find the derivative of such a product, you would use the **Product Rule**. The Product Rule is applied when differentiating functions that are multiplied together. To use it, assign each part of the product to a variable and find their individual derivatives.

• If you have two functions, say \( u(x) \) and \( v(x) \), the Product Rule states that the derivative of their product is \((uv)' = u'v + uv'\).
• This rule helps break down complex expressions into simpler, more manageable parts that can each be differentiated separately.

By understanding differentiation and applying the Product Rule, you can tackle a wide variety of problems in calculus.

Exponential Function

An exponential function is a mathematical function of the form \( f(x) = a^x \), where \( a \) is a constant and \( x \) is a variable. In most scenarios, the base \( a \) is Euler's number \( e \), which is approximately equal to 2.71828. Exponential functions are highly significant in various fields such as biology, economics, and physics because they describe rapid growth or decay processes.

In our exercise, we have an exponential function within the expression, specifically \( e^x \). Here are some key points about this type of function:

• The derivative of \( e^x \) is quite simple, as it equals itself: \( \frac{d}{dx}[e^x] = e^x \).
• Exponential functions have the special property of their rate of change being proportional to the value of the function itself.
• This property makes them unique and particularly useful in modeling exponential growth or decay scenarios.

In our differentiation process, using the product rule, remembering that the derivative of an exponential function like \( e^x \) is still \( e^x \) helps simplify calculations.

Logarithmic Function

Logarithmic functions are the inverse of exponential functions. The logarithm with base \( e \) is called the natural logarithm, represented as \( \ln(x) \). Logarithms are crucial for simplifying expressions and solving equations involving exponential functions. They have properties that make calculations easier, especially when dealing with large numbers.

In the given exercise, the expression \( \ln(x^2) \) is part of the differentiated function. It is simplified as \( 2\ln(x) \), utilizing the logarithmic identity \( \ln(a^b) = b\ln(a) \).

• When differentiating a natural logarithm, the rule is \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \).
• In the context of the function \( \ln(x^2) \), we apply the chain rule, yielding \( \frac{d}{dx}[\ln(x^2)] = \frac{2}{x} \).
• This differentiation rule helps in finding how quickly a logarithmic function changes, which is particularly useful in calculus.

Understanding the synergy of exponential and logarithmic functions is essential in calculus, as they often occur together in complex expressions, requiring the application of differentiation techniques like the Product Rule to solve.