Problem 60 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=(\sin x)^{\ln x} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to find the derivative of the function \( y = (\sin x)^{\ln x} \). This expression is a combination of a variable base and a variable exponent. To differentiate such functions, we often use logarithmic differentiation.

Step 2: Using Logarithmic Differentiation

First, take the natural logarithm of both sides of the equation: \[ \ln y = \ln((\sin x)^{\ln x}) \] Using the logarithmic identity \( \ln(a^b) = b \ln a \), we can simplify the right side: \[ \ln y = \ln x \cdot \ln(\sin x) \].

Step 3: Differentiate Both Sides with Respect to x

Differentiate both sides with respect to \( x \): \[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\ln x \cdot \ln(\sin x)) \] This requires using the product rule \( u'v + uv' \), where \( u = \ln x \) and \( v = \ln(\sin x) \).

Step 4: Apply the Product Rule

Calculate \( u' \) and \( v' \): - \( u = \ln x \) gives \( u' = \frac{1}{x} \) - \( v = \ln(\sin x) \) gives \( v' = \frac{1}{\sin x} \cdot \cos x = \cot x \)Using the product rule: \[ \frac{d}{dx}(\ln x \cdot \ln(\sin x)) = \frac{1}{x} \cdot \ln(\sin x) + \ln x \cdot \cot x \].

Step 5: Solve for dy/dx

Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y \cdot \left( \frac{1}{x} \ln(\sin x) + \ln x \cdot \cot x \right) \] Recall that \( y = (\sin x)^{\ln x} \), substitute back: \[ \frac{dy}{dx} = (\sin x)^{\ln x} \cdot \left( \frac{1}{x} \ln(\sin x) + \ln x \cdot \cot x \right) \].

Step 6: Final Clean-Up (Optional)

You can express the derivative in a clearer format for final presentation or simplification, if desired, but it generally depends on further context or application constraints.

Key Concepts

Variable Exponent Differentiation

When you have a function where both the base and the exponent can vary, like in \((\sin x)^{\ln x}\), regular differentiation techniques might not be sufficient. This situation calls for logarithmic differentiation. Taking the natural logarithm of both sides is a common tactic. By doing this, you can bring the exponent down in front of the logarithm. - Start with \( y = (\sin x)^{\ln x} \)- Take the natural logarithm: \( \ln y = \ln((\sin x)^{\ln x}) \)- Simplify using \( \ln(a^b) = b \ln a \): \( \ln y = \ln x \cdot \ln(\sin x) \)Now, you have transformed the variable exponent function into something more manageable for differentiation.

Product Rule in Calculus

The product rule is a valuable tool when differentiating products of two functions. If you have a product like \( u(x)v(x) \), the derivative is \[ u'v + uv' \]. In the context of our problem, after applying logarithmic differentiation, we find ourselves with \( \ln x \cdot \ln(\sin x) \). Here, - \( u = \ln x \) and - \( v = \ln(\sin x) \).Thus, applying the product rule:- Derivative of \( u = \ln x \) is \( \frac{1}{x} \).- Derivative of \( v = \ln(\sin x) \) translates to \( \frac{1}{\sin x} \, \cos x = \cot x \).The full derivative is then \[ \frac{d}{dx}(\ln x \cdot \ln(\sin x)) = \frac{1}{x} \cdot \ln(\sin x) + \ln x \cdot \cot x \].

Trigonometric Functions Derivatives

Differentiating trigonometric functions can be tricky, but knowing the basic derivatives allows you to tackle complex expressions. In the exercise, we encounter \( \ln(\sin x) \), which involves the derivative of \( \sin x \).Here’s a brief guide:- The derivative of \( \sin x \) is \( \cos x \).- If you have a logarithm of a trigonometric function like \( \ln(\sin x) \), the derivative becomes a product following the chain rule: \( \frac{1}{\sin x} \cdot \cos x = \cot x \).These steps help you manage expressions where trigonometric functions and their derivatives interact with logarithms, providing a robust approach to calculus problems involving such mixed expressions.