Problem 39 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(33-54,\) find the derivative. $$ y=e^{x} \ln x $$

Step-by-Step Solution

Answer

The derivative of the function \(y = e^x \ln x\) is \(y' = e^x (\ln x + 1/x)\)

Step 1: Identify the Functions

In the function \(y = e^x \ln x\), there are two functions involved which are \(u = e^x\) and \(v = \ln x\).

Step 2: Apply the Product Rule

The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function. So, apply this rule to \(y = e^x \ln x\). It results in the first part of the solution: \(y' = u' v + u v'\).

Step 3: Find the Derivatives of Individual Functions

The derivative of \(u = e^x\) is \(u' = e^x\) and the derivative of \(v = \ln x\) is \(v' = 1/x\)

Step 4: Substitute Derivatives Back Into the Product Rule

Substitute the calculated derivatives into \(y' = u' v + u v'\) to find the derivative of the given function \(y = e^x \ln x\). By substituting it will become \(y' = e^x \ln x + e^x / x\)

Step 5: Simplify the Expression

Realize that \(y' = e^x (\ln x + 1/x) \)

Key Concepts

Product Rule

The product rule is a fundamental part of calculus used for finding the derivative of a product of two differentiable functions. Imagine you have two functions, say \( u(x) \) and \( v(x) \). The product rule helps differentiate their product, \( y = u(x)v(x) \). Simply put, the product rule formula is:

• \( (uv)' = u'v + uv' \)

The idea here is that instead of attempting to differentiate the entire product at once, you break it down:

• You differentiate one function at a time while keeping the other function constant.
• Then, you sum the results.
• Thus, this method helps when handling complex equations.

In our example, we have \( y = e^x \ln x \), where \( u = e^x \) and \( v = \ln x \). Applying the product rule gives \( y' = u'v + uv'\), which leads you directly to the next steps: finding the derivatives of \( u \) and \( v \).

Exponential Function

Exponential functions have a base number, commonly 'e', raised to the power of a variable, such as \( e^x \). These functions are essential in many areas due to their growth properties. Key points include:

• The function \( e^x \) itself has a special property where it is its own derivative.
• This makes differentiating it straightforward: \( \frac{d}{dx}[e^x] = e^x \).

By recognizing these properties:

• In our function, we observe \( u = e^x \), where finding the derivative \( u' \) is simply \( e^x \).
• This property often simplifies the differentiation process and can ease solving equations involving exponential growth or decay.

Natural Logarithm

The natural logarithm, represented by \( \ln(x) \), is the logarithm to the base 'e'. It is frequently used in calculus for converting multiplicative processes into additive ones, which simplifies calculation. Some important points are:

• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), given \( x > 0 \).
• It inversely describes the exponential function, providing the time needed to achieve a certain level of growth at a continuous, exponential rate.

In the problem \( y = e^x \ln x \), the component \( v = \ln x \) has a derivative \( v' = \frac{1}{x} \). Thus, when applying this to the product rule, helps simplify the computation immensely.

Differentiation

Differentiation is a process in calculus that finds the rate at which a function is changing at any given point. It is expressed as a derivative. The primary goal is to understand how a function behaves according to changes in its input. Some strategies include:

• Identifying components of a function that need differentiating individually with rules like the product or chain rule.
• Combining these derived values together to form the derivative of the larger, more complex function.
• Utilizing simplification techniques to make computation easier.

For the function \( y = e^x \ln x \), once the derivatives \( u' \) and \( v' \) are found, they are substituted back into the formula from the product rule, creating the simplified derivative: \( y' = e^x (\ln x + \frac{1}{x}) \). This final step illustrates the power of differentiation in breaking down complex expressions into more manageable parts.