Problem 22 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\frac{x \ln x}{1+\ln x} $$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{(\ln x)^2 + \ln x + 1}{(1 + \ln x)^2}.\)

Step 1: Identify the Derivative Technique Needed

The given function is a fraction, so we need to use the quotient rule to find the derivative. The quotient rule states if we have a function \(y = \frac{u}{v}\), then the derivative \(y'\) is \(\frac{u'v - uv'}{v^2}\).

Step 2: Apply the Quotient Rule

Given the function \(y = \frac{x \ln x}{1+\ln x}\), let \(u = x \ln x\) and \(v = 1 + \ln x\). First, find the derivatives \(u'\) and \(v'\).

Step 3: Differentiate \(u = x \ln x\)

To differentiate \(u = x \ln x\), apply the product rule: \(u' = x' \ln x + x (\ln x)' = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\).

Step 4: Differentiate \(v = 1 + \ln x\)

The derivative of \(v = 1 + \ln x\) is straightforward: \(v' = \frac{d}{dx}(1 + \ln x) = 0 + \frac{1}{x} = \frac{1}{x}\).

Step 5: Compute the Derivative Using the Quotient Rule

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the quotient rule formula: \( y' = \frac{(\ln x + 1)(1 + \ln x) - x \ln x \cdot \frac{1}{x}}{(1 + \ln x)^2}.\)

Step 6: Simplify the Derivative Expression

Simplify the numerator: \((\ln x + 1)(1 + \ln x) - \ln x =(\ln x)(1 + \ln x) + 1 + \ln x - \ln x = (\ln x)^2 + \ln x + 1 \). The final derivative is \( y' = \frac{(\ln x)^2 + \ln x + 1}{(1 + \ln x)^2}.\)

Key Concepts

Quotient Rule

When dealing with functions that are presented as a fraction of two other functions, the **quotient rule** becomes immensely useful. This rule specifically helps us find the derivative of a quotient. For a function in the form of \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \), the derivative \( y' \) is given by the formula:

• \( y' = \frac{u'v - uv'}{v^2} \)

Here's what these terms stand for:

• \( u' \): the derivative of the numerator \( u \)
• \( v' \): the derivative of the denominator \( v \)
• \( u'v \) and \( uv' \) : cross terms that combine both derivatives
• \( v^2 \): the square of the denominator

The formula emphasizes knowing both parts of the fraction completely, as both their derivatives are needed. When applying this rule to any function, ensure all parts—especially the differentiation—are calculated with precision. In our example, the function \( y = \frac{x \ln x}{1 + \ln x} \) requires careful attention to both functions involved.

Product Rule

In calculus, the **product rule** is crucial for differentiating functions that are products of two distinct expressions. When you have a function \( y = u(x) \cdot v(x) \), the derivative is given by:

• \( y' = u' v + u v' \)

Here, each term represents a product involving both functions:

• \( u' \): the derivative of the first function
• \( v \): the second function as is
• \( u \): the first function as is
• \( v' \): the derivative of the second function

By applying these steps, you can ensure no part of the function is missed. For differentiation like \( x \ln x \), as seen in our problem, this rule shows its importance by dividing the task into smaller, more manageable steps. Here, we treated \( x \ln x \) by identifying \( u = x \) and \( v = \ln x \). Differentiating both and applying the rule already simplifies the process significantly. Take each part as a building block to solve complex functions with ease.

Logarithmic Differentiation

**Logarithmic differentiation** is a handy technique when working with complex compositions, especially those involving logarithms. This becomes particularly efficient in cases where the function involves products, ratios, or exponents, which are challenging to differentiate directly.Here’s a step-by-step approach:

• First, take the natural logarithm, \( \ln \), of both sides of the equation to simplify the process.
• Next, differentiate implicitly with respect to \( x \). This means you differentiate both sides, keeping in mind that the derivative of \( \ln(u(x)) \) is \( \frac{u'(x)}{u(x)} \).
• Once differentiated, solve for \( y' \) by multiplying through, if needed.

By using this method, the differentiation becomes much more manageable, breaking down into simpler algebraic terms. In our exercise, logarithmic differentiation isn't directly applied but understanding it highlights why differentiating \( \ln x \) is straightforward, allowing focus to remain on more complex parts of the function. Remember, this method is a powerful alternative when standard rules become cumbersome.