Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(D_{\theta} \sqrt{\log _{10}\left(3^{\theta^{2}-\theta}\right)}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{\ln(3)(2\theta-1)}{2\ln(10)\sqrt{\log_{10}(3^{\theta^2-\theta})}} \).

Step 1: Understand the Problem

We need to find the derivative of the expression \( \sqrt{\log_{10}(3^{\theta^2-\theta})} \) with respect to \( \theta \). This involves understanding that we have a composite function, including a square root, a logarithm, and an exponential function.

Step 2: Apply Chain Rule

Since this is a composite function, we'll use the chain rule. The chain rule states that if you have a composite function \( f(g(x)) \), then its derivative is \( f'(g(x)) \cdot g'(x) \). In our problem, this will involve taking the derivative of the outer function first, which is the square root.

Step 3: Differentiate the Outer Function

The outside function is \( f(u) = \sqrt{u} \). Its derivative is \( \frac{d}{du} \sqrt{u} = \frac{1}{2\sqrt{u}} \). Thus, the derivative of \( \sqrt{\log_{10}(3^{\theta^2-\theta})} \) with respect to its inner function is \( \frac{1}{2\sqrt{\log_{10}(3^{\theta^2-\theta})}} \).

Step 4: Differentiate the Inner Function

The inner function is \( g(\theta) = \log_{10}(3^{\theta^2-\theta}) \). To differentiate this, use the change of base formula for logarithms: \( \log_{10}(x) = \frac{\ln(x)}{\ln(10)} \). So, \( \log_{10}(3^{x}) = \frac{1}{\ln(10)} \ln(3^{x}) \).

Step 5: Differentiate the Logarithmic Function

Differentiate \( \ln(3^{\theta^2-\theta}) \) using the chain rule. The derivative of \( \ln(3^{x}) \) is \( \frac{1}{3^x} \cdot 3^x \cdot \ln(3) \cdot x' = \ln(3) \cdot x' \). Here, \( x = \theta^2-\theta \), so finding \( x' \), we get \( 2\theta-1 \). Thus, the derivative of the logarithmic part is \( \ln(3)(2\theta-1) \).

Step 6: Combine All the Derivatives

Combine the results from the previous steps to get the full derivative: \[ \frac{1}{2\sqrt{\log_{10}(3^{\theta^2-\theta})}} \cdot \frac{\ln(3)(2\theta-1)}{\ln(10)} \] This is the derivative of the given expression \( D_{\theta} \sqrt{\log_{10}(3^{\theta^2-\theta})} \).

Step 7: Final Simplification

If needed, simplify the expression further, but in many cases, leaving it in the derived form is acceptable depending on the context. For now, we consider the expression as simplified.

Key Concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When you have a function within another function, for instance, such as the function inside the square root in our exercise, the chain rule simplifies the differentiation process. Essentially, it allows us to break down the derivative of the complex function into more manageable parts.

• Composite Functions: Example functions could be \( f(g(x)) \) where \( f \) is the outer function and \( g(x) \) is the inner function.
• Applying the Rule: To differentiate, you first find the derivative of \( f \) as if \( g \) were a variable. Then multiply it by the derivative of \( g \), i.e., \( f'(g(x)) \) times \( g'(x) \).

The chain rule is essential because it provides a system for tackling complex derivatives by turning them into sequences of simpler derivatives.

Logarithmic Differentiation

Logarithmic differentiation turns functions with products, quotients, or exponents into more solvable expressions. Instead of making our problem more complicated, it can simplify differentiation, which is especially helpful with exponential and logarithmic functions like in our exercise.

• Basic Idea: Take the natural log of both sides of an equation \( y = f(x) \), so you have \( \ln(y) = \ln(f(x)) \).
• Rules Used: Utilize properties of logs to simplify, especially products and powers: \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(a^b) = b \ln(a) \).
• Derive Each Part: Differentiate both sides with respect to \( x \). Remember to apply the derivative rules, such as the chain rule again, as needed.

This technique helps especially when dealing with variable exponents, as seen with \( 3^{\theta^2-\theta} \), allowing us to turn complicated products and powers into sums and products which are easier to differentiate.

Composite Function

In calculus, a composite function comes into play when one function is applied inside another. This is exactly what happened in our given problem, which included a square root function, a logarithm, and an exponential function all wrapped into one.

• Structure: Composite functions are structured as \( (f \circ g)(x) = f(g(x)) \) where \( g(x) \) is the inner function and \( f \) is the outer function. Each needs its own attention during differentiation.
• Example: In our exercise, the square root is \( f \) and \( \log_{10}(3^{\theta^2-\theta}) \) is \( g(x) \).

Working with composite functions demands understanding each layer, so we apply appropriate rules like the chain rule to differentiate efficiently.

Differentiation Techniques

Differentiating functions involves various methods depending on the kind of function we're dealing with. Being familiar with differentiation techniques provides the flexibility needed to handle complex expressions like the one given in our problem.

• Product Rule: Useful when differentiating products of two functions.
• Quotient Rule: Essential when you have a division of two functions to differentiate.
• Chain Rule: As mentioned earlier, indispensable for composite functions.
• Power Rule and Logarithmic Differentiation: These simplify differentiation, especially useful in exponential functions and natural logs.

By mastering these techniques, tackling any differentiation exercise becomes approachable, as various rules like these lay the foundation for solving derivatives.