Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\). $$ y=(\ln x)^{\ln x} $$

Short answer

\(\frac{dy}{dx} = (\ln x)^{\ln x} \left( \frac{1 + \ln (\ln x)}{x} \right)\)

Step-by-step solution

Take Natural Logarithm on Both Sides

We start by applying the natural logarithm to both sides of the equation to make differentiation easier. \[\ln y = \ln((\ln x)^{\ln x})\]Using the properties of logarithms, we can simplify this as:\[\ln y = (\ln x) \cdot \ln(\ln x)\]

Differentiate Both Sides

Next, we differentiate both sides with respect to \(x\). Recall that when differentiating \(\ln y\), we use the chain rule:\[\frac{d}{dx}(\ln y) = \frac{1}{y}\frac{dy}{dx}\]On the right side, we use the product rule to differentiate \((\ln x) \cdot \ln(\ln x)\):\[\frac{d}{dx}(\ln x) \cdot \ln(\ln x) = (\ln x)\left(\frac{d}{dx}(\ln(\ln x))\right) + \ln(\ln x) \cdot \left(\frac{1}{x}\right)\]

Differentiate \(\ln(\ln x)\) Using Chain Rule

To differentiate \(\ln(\ln x)\), we use the chain rule:\[\frac{d}{dx}(\ln(\ln x)) = \frac{1}{\ln x} \cdot \frac{1}{x}\]Substitute this back into the product differentiation:\[(\ln x) \cdot \frac{1}{\ln x \cdot x} + \ln(\ln x) \cdot \frac{1}{x} = \frac{1}{x} + \frac{\ln(\ln x)}{x}\]

Solve for \(\frac{dy}{dx}\)

Substitute the result from Step 3 into the derivative of \(\ln y\):\[\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + \frac{\ln (\ln x)}{x}\]Multiply through by \(y\) to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y \left( \frac{1 + \ln (\ln x)}{x} \right)\]Substitute \(y = (\ln x)^{\ln x}\) back in:\[\frac{dy}{dx} = (\ln x)^{\ln x} \left( \frac{1 + \ln (\ln x)}{x} \right)\]

Final Answer

The derivative of the original function \( y = (\ln x)^{\ln x} \) with respect to \(x\) is:\[\frac{dy}{dx} = (\ln x)^{\ln x} \left( \frac{1 + \ln (\ln x)}{x} \right)\]

Key concepts

Product Rule

One of the most useful techniques in calculus for finding derivatives of products of two functions is the product rule. If you have two functions, say \( u(x) \) and \( v(x) \), their product is denoted as \( u(x) \cdot v(x) \). The product rule states that the derivative of this product is given by:

• \((uv)' = u'v + uv'\)

This means you differentiate the first function and multiply it by the second function plus the first function times the derivative of the second function.
This formula is crucial especially when differentiating expressions where two separate functions are multiplied together, such as in step 2 of our solution. The product rule allows each function to be processed separately, making complex differentiation manageable.

Chain Rule

The chain rule is the go-to tool when dealing with the derivative of a composite function. This rule lets you differentiate functions that are nested inside other functions. Suppose you have two functions, \( f(x) \) and \( g(x) \), and you want to differentiate a composite function \( f(g(x)) \). The chain rule tells us:

• \((f(g(x)))' = f'(g(x)) \cdot g'(x)\)

This approach simplifies the process by breaking down a complex function into smaller, more manageable parts.
In step 3, the chain rule was used to successfully differentiate \( \ln(\ln x) \). This involves differentiating the outer function (\( \ln(u) \)) and multiplying by the derivative of the inner function \( u = \ln(x) \).
By using the chain rule, you're effectively peeling off layers of functions, which is extremely helpful in obtaining derivatives of nested functions.

Natural Logarithm

The natural logarithm, denoted \( \ln(x) \), is a key mathematical function often used in calculus. It is the logarithm to the base \( e \), where \( e \) is approximately 2.718. The natural logarithm has unique properties, such as:

• \( \ln(xy) = \ln(x) + \ln(y) \)
• \( \ln(x/y) = \ln(x) - \ln(y) \)
• \( \ln(x^a) = a \cdot \ln(x) \)

These properties become particularly useful when simplifying expressions before differentiation.
In step 1 of our solution, we used \( \ln(y) = (\ln x) \cdot \ln(\ln x) \) to simplify the problem, making the differentiation more straightforward.
Understanding natural logarithms helps to handle exponents and transforms a hard-to-differentiate function into one that is linear and much easier to work with.

Derivative

The derivative is a fundamental concept in calculus representing the rate of change of a function. In simple terms, it tells you how fast or slow a function is changing at any point. The notation \( \frac{dy}{dx} \) is often used, where \( y \) is the function of \( x \).
To find a derivative, you usually follow derivative rules like the power rule, product rule, chain rule, and others. These rules guide you in taking the derivative of various types of functions effectively.
In our solution, to find \( \frac{dy}{dx} \), logarithmic differentiation was used. This process involves taking the natural log of both sides and differentiating with respect to \( x \). This technique is especially helpful when dealing with complex expressions like products or powers involving logarithms or other functions.
Derivatives open up countless possibilities not only in pure mathematics but also in real-world applications, from physics to economics.