Question

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

Short answer

The derivative is \( \frac{d y}{d x} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \cdot \ln x \right) \).

Step-by-step solution

Take the Natural Logarithm of Both Sides

We begin by taking the natural logarithm of both sides of the equation to simplify the differentiation process. This utilizes the property of logarithms that simplifies powers:\[\ln(y) = \ln(x^{\cot x})\]Using the property \( \ln(a^b) = b \cdot \ln(a) \), we rewrite it as:\[\ln(y) = \cot x \cdot \ln x\]

Differentiate Both Sides with Respect to x

Differentiate both sides with respect to \(x\). We apply the chain rule to \(\ln(y)\) and the product rule to \(\cot x \cdot \ln x\):\[\frac{1}{y} \cdot \frac{d y}{d x} = \frac{d}{d x} (\cot x \cdot \ln x)\]Using the product rule:\[\frac{d}{d x} (\cot x \cdot \ln x) = \cot x \cdot \frac{d}{d x}(\ln x) + \ln x \cdot \frac{d}{d x}(\cot x)\]

Compute the Derivatives

Calculate each derivative:1. \( \frac{d}{d x}(\ln x) = \frac{1}{x} \)2. \( \frac{d}{d x}(\cot x) = -\csc^2 x \)Substituting these into our expression, we get:\[\cot x \cdot \frac{1}{x} + \ln x \cdot (-\csc^2 x) = \frac{\cot x}{x} - \csc^2 x \cdot \ln x\]

Solve for \(\frac{d y}{d x}\)

Substitute the expression from Step 3 back into our differentiated equation from Step 2:\[\frac{1}{y} \cdot \frac{d y}{d x} = \frac{\cot x}{x} - \csc^2 x \cdot \ln x\]Multiply both sides by \(y\) to isolate \(\frac{d y}{d x}\):\[\frac{d y}{d x} = y \left( \frac{\cot x}{x} - \csc^2 x \cdot \ln x \right)\]Remember that \(y = x^{\cot x}\), so:\[\frac{d y}{d x} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \cdot \ln x \right)\]

Final Result

Summarizing, the final result for the derivative of \(y\) with respect to \(x\) is:\[\frac{d y}{d x} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \cdot \ln x \right)\]

Key concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a special type of logarithm where the base is \( e \), an irrational constant approximately equal to 2.71828. In calculus and mathematical analysis, the natural logarithm is used extensively because of its unique mathematical properties, especially when dealing with growth and rates of change.
In our exercise, the natural logarithm simplifies the differentiation process of complex functions like \( y = x^{\cot x} \).
Why use natural logarithms? Here are a few reasons:

• Simplification: The natural logarithm can convert a power function into a product, making differentiation manageable using the property \( \ln(a^b) = b \cdot \ln(a) \).
• Relation to exponential functions: \( e^x \) and \( \ln \) are inverse operations, which is useful for solving, simplifying and analyzing equations.
• Chain rule applications: The derivative of \( \ln u \) requires the chain rule, which is a core concept for this exercise.

Understanding natural logarithms equips students with an invaluable tool for handling a wide array of mathematical problems.

Product Rule

The product rule is a fundamental technique in calculus for dealing with derivatives of functions that are products of two or more functions. If you have two functions, \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product \( u(x) \cdot v(x) \) is:\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x)\]In our problem, after taking the natural logarithm, we encounter \( \ln(y) = \cot x \cdot \ln x \).
To differentiate \( \cot x \cdot \ln x \), we must use the product rule:

• Choose functions: Here, \( u(x) = \cot x \) and \( v(x) = \ln x \).
• Differentiate separately: Find \( u'(x) \) and \( v'(x) \) individually.
• Apply the rule: Combine using \( u'(x)v(x) + u(x)v'(x) \).

By applying the product rule, we can handle complex expressions involving products of functions more efficiently.

Chain Rule

The chain rule in calculus is essential for finding the derivative of composite functions. When you have a function nested inside another, like \( f(g(x)) \), the derivative is found using the chain rule:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]In this exercise, the chain rule is primarily applied when differentiating \( \ln(y) \). Here, \( y \) is a function of \( x \), so \( \ln(y) \) creates a composite function.
To differentiate \( \ln(y) \), we use the chain rule:

• Identify outer and inner functions: \( f(u) = \ln(u) \) and \( u(x) = y \).
• Differentiate: \( \frac{d}{dy}[\ln(y)] = \frac{1}{y} \) and multiply by \( \frac{dy}{dx} \) to get \( \frac{1}{y} \cdot \frac{dy}{dx} \).

Thus, applying the chain rule allows us to find the derivative of more intricate functions, making it a versatile tool in calculus.

Differentiation

Differentiation is a core concept in calculus, representing the process of finding the derivative, which is the rate of change or slope of a function. To differentiate a function means to find how it changes with respect to a variable, typically \( x \).
In the given exercise, differentiation is used in the context of logarithmic differentiation, a method that simplifies finding derivatives of complex expressions by transforming them into simpler forms using logarithms.
Key steps and approaches of differentiation include:

• Identify functions: Understand the structure and form of functions involved.
• Choose methods: Apply appropriate rules such as the product, quotient, or chain rules, depending on the function.
• Compute derivatives: Perform the actual mathematical operations to find \( \frac{dy}{dx} \).

By mastering the techniques of differentiation, one can solve a variety of intricate mathematical problems efficiently.