Problem 31 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=\theta \sin \left(\log _{7} \theta\right)\)

Step-by-Step Solution

Answer

The derivative is \(\sin(\log_7 \theta) + \frac{\cos(\log_7 \theta)}{\ln(7)}\)."

Step 1: Identify the rule to use

The function given is a product of two functions of \(\theta\): \(\theta\) and \(\sin(\log_7 \theta)\). Therefore, we will use the product rule to find the derivative. Recall the product rule: if \(u\) and \(v\) are functions of \(\theta\), then \(\frac{d}{d\theta}(uv) = u'v + uv'\).

Step 2: Differentiate the first function

The first function is \(u = \theta\). The derivative of \(\theta\) with respect to \(\theta\) is 1, so \(u' = 1\).

Step 3: Differentiate the second function

The second function is \(v = \sin(\log_7 \theta)\). Use the chain rule to differentiate: the outer function is \(\sin(x)\), and the inner function is \(\log_7 \theta\). First, find the derivative of \(\sin(x)\), \(\cos(x)\), and then multiply by the derivative of \(\log_7 \theta\), which is \(\frac{1}{\theta \ln(7)}\). Thus, the derivative \(v' = \cos(\log_7 \theta) \cdot \frac{1}{\theta \ln(7)}\).

Step 4: Apply the product rule

Using the product rule formula: \( u'v + uv'\). Substitute \(u = \theta\), \(u' = 1\), \(v = \sin(\log_7 \theta)\), and \(v' = \cos(\log_7 \theta) \cdot \frac{1}{\theta \ln(7)}\). Therefore, the derivative \(\frac{d}{d\theta} \left( \theta \sin(\log_7 \theta) \right) = 1 \cdot \sin(\log_7 \theta) + \theta \cdot \cos(\log_7 \theta) \cdot \frac{1}{\theta \ln(7)}\).

Step 5: Simplify the expression

Simplify the expression derived from applying the product rule: \(\sin(\log_7 \theta) + \frac{\cos(\log_7 \theta)}{\ln(7)}\). This is the final form of the derivative with respect to \(\theta\).

Key Concepts

Product Rule

When you're working with derivatives in calculus, you might come across functions that are products of two simpler functions. In these cases, the product rule is your best friend. It's all about how to differentiate a product of functions. Let's say you have two functions, \(u\) and \(v\). The product rule states that the derivative of their product \(uv\) is given by:

• \(\frac{d}{d\theta}(uv) = u'v + uv'\)

This means you differentiate one function while leaving the other one alone, then switch roles, and add these results. It's a powerful tool when functions are multiplied together.
In the original problem, where you have \(y = \theta \sin(\log_7 \theta)\), \(\theta\) is \(u\) and \(\sin(\log_7 \theta)\) is \(v\).

• You get \(u' = 1\) (since the derivative of \(\theta\) is 1).
• Then you find \(v'\), which involves the chain rule (we'll get to that next).

By applying the product rule, you simplify a complex expression step by step.

Chain Rule

The chain rule allows you to differentiate composite functions, where one function is inside another. It's like peeling an onion, one layer at a time. To use the chain rule, you differentiate the outer function and then multiply by the derivative of the inner function.

• If you have \(f(g(x))\), the chain rule tells you \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\).

In the original exercise, after identifying \(v = \sin(\log_7 \theta)\), the outer function \(\sin(x)\) differentiates to \(\cos(x)\). The inner function \(\log_7 \theta\) differentiates to \(\frac{1}{\theta \ln(7)}\).

• So, \(v' = \cos(\log_7 \theta) \cdot \frac{1}{\theta \ln(7)}\).

This shows how the chain rule breaks down these functions, making it easier to find complex derivatives.

Logarithmic Differentiation

Logarithmic differentiation is a technique especially helpful when dealing with products or quotients of functions, or functions raised to variable powers. This method leverages the properties of logarithms to simplify differentiation.
If you have a function such as \(y = f(x)^{g(x)}\), you take the natural log of both sides, use the properties of logs to simplify, and then differentiate. The key property used here is:

• \(\ln(u^v) = v \ln(u)\)

With the exercise's \(\log_7(x)\) being part of the derivative, the derivative is simplified by recalling that \(\frac{d}{d\theta} \log_b \theta = \frac{1}{\theta \ln(b)}\). Here, breaking the problem into separate parts using logs makes differentiation manageable, despite not being directly applied in this example.
Remember, logarithmic differentiation isn't always necessary, but it's good to know for handling tricky derivatives.