Problem 67 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=(\ln x)^{x} $$

Step-by-Step Solution

Answer

The derivative of \((\ln x)^x\) is \((\ln x)^x \cdot \left( \frac{1}{\ln x} + \ln(\ln x) \right)\)."

Step 1: Take the natural logarithm on both sides

To use logarithmic differentiation, start by taking the natural logarithm of both sides of the function: \[ y = (\ln x)^x \]Applying the natural log, we get:\[ \ln y = \ln((\ln x)^x) \]Using the property of logarithms, this becomes:\[ \ln y = x \cdot \ln(\ln x) \]

Step 2: Differentiate both sides with respect to x

Now that we have simplified the equation, differentiate both sides with respect to \(x\): For the left side, use implicit differentiation:\[ \frac{1}{y} \cdot \frac{dy}{dx} \]The right side uses the product rule since it is in the form \(u \cdot v\) where \(u = x\) and \(v = \ln(\ln x)\):\[ \frac{d}{dx}[x \cdot \ln(\ln x)] = x \cdot \frac{d}{dx}[\ln(\ln x)] + \ln(\ln x) \cdot \frac{d}{dx}[x] \]Simplify each term:- \(\frac{d}{dx}[\ln(\ln x)] = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}\)- The derivative of \(x\) is 1.So the differentiated right side is:\[ x \cdot \frac{1}{x \ln x} + \ln(\ln x) = \frac{1}{\ln x} + \ln(\ln x) \]

Step 3: Solve for dy/dx

Now equate the derivatives and solve for \(\frac{dy}{dx}\):\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{\ln x} + \ln(\ln x) \]Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \cdot \left( \frac{1}{\ln x} + \ln(\ln x) \right) \]Since \(y = (\ln x)^x\), substitute back:\[ \frac{dy}{dx} = (\ln x)^x \cdot \left( \frac{1}{\ln x} + \ln(\ln x) \right) \]

Key Concepts

First Derivative

Finding the first derivative of a function is all about determining how the function changes as its inputs change. Put simply, it tells you the slope of the tangent line at any given point on a curve. This is fundamental in calculus and helps in understanding the behavior of functions by identifying where functions increase, decrease, or have horizontal tangents.

In the given exercise, we are tasked with finding the first derivative of the function \( f(x) = (\ln x)^x \). This particular function is quite complex because it's a power of a logarithmic expression. Therefore, using standard differentiation rules isn't straightforward, and logarithmic differentiation becomes a powerful tool.

By taking the natural logarithm of both sides, we reformulate the problem into a more manageable form. This step simplifies the differentiation process and undoes the complexity posed by the expression \((\ln x)^x\). Once you have taken the logarithm, the next step involves applying implicit differentiation and the product rule to find the derivative.

Implicit Differentiation

Implicit differentiation is a method used when dealing with functions where the dependent variable is not isolated on one side of the equation. In simpler terms, it helps differentiate equations that are not explicitly solved for one variable in terms of another.

The given exercise involves an implicit form initially, after taking the logarithm: \( \ln y = x \cdot \ln(\ln x) \). Since \( y \) is not by itself on one side, implicit differentiation lets us find \( \frac{dy}{dx} \) by viewing \( y \) as a function of \( x \).

To differentiate \( \ln y \) with respect to \( x \), the chain rule is employed, giving us \( \frac{1}{y} \cdot \frac{dy}{dx} \). The rest of the differentiation process on the right side makes use of other rules, such as the product rule, to fully establish the derivative relationship.

Product Rule

The product rule is a fundamental rule in calculus that allows us to differentiate functions that are products of two functions. If you have two functions \( u \) and \( v \), their product \( u \cdot v \) is differentiated by taking \( u'v + uv' \), where \( u' \) and \( v' \) represent their respective derivatives.

In the exercise, after taking the logarithm, we differentiate the right side: \( x \cdot \ln(\ln x) \). This expression is a classic example where the product rule is applicable. Here, \( u = x \) and \( v = \ln(\ln x) \). By applying the product rule, we differentiate each function separately and then appropriately apply the formula.

The first part of the derivative involves multiplying \( x \) by the derivative of \( \ln(\ln x) \), and the second part involves multiplying \( \ln(\ln x) \) by the derivative of \( x \). Derivatives are simplified, leading us to an expression ready to be solved and equated to the implicit derivative of \( y \). This results in connecting the dots to find \( \frac{dy}{dx} \).