Problem 31 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\ln (\sec (\ln \theta))$$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{d\theta} = \frac{\tan(\ln(\theta))}{\theta} \).

Step 1: Identify the outer function

The outer function is the natural logarithm function \( y = \ln(u) \), where \( u = \sec(\ln(\theta)) \). To find the derivative, use the chain rule, which states that \( \frac{dy}{d\theta} = \frac{dy}{du} \times \frac{du}{d\theta} \).

Step 2: Differentiate the natural logarithm

The derivative of \( y = \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Therefore, we have \( \frac{dy}{du} = \frac{1}{\sec(\ln(\theta))} \).

Step 3: Identify the inner function in \( u \)

The function \( u = \sec(\ln(\theta)) \) consists of another function inside: \( v = \ln(\theta) \). So, \( u = \sec(v) \). We need to find \( \frac{du}{dv} \) and then \( \frac{dv}{d\theta} \).

Step 4: Differentiate \( \sec(v) \)

The derivative of \( \sec(v) \) with respect to \( v \) is \( \sec(v)\tan(v) \). Thus, \( \frac{du}{dv} = \sec(\ln(\theta))\tan(\ln(\theta)) \).

Step 5: Differentiate \( \ln(\theta) \)

\( v = \ln(\theta) \) implies \( \frac{dv}{d\theta} = \frac{1}{\theta} \).

Step 6: Apply the chain rule

Compile the results using the chain rule. By substituting back, we have:\[\frac{dy}{d\theta} = \frac{dy}{du} \times \frac{du}{dv} \times \frac{dv}{d\theta} = \frac{1}{\sec(\ln(\theta))} \times \sec(\ln(\theta))\tan(\ln(\theta)) \times \frac{1}{\theta} = \frac{\tan(\ln(\theta))}{\theta}\].

Step 7: Conclusion: Final Expression

Therefore, the derivative of \( y \) with respect to \( \theta \) is \( \frac{\tan(\ln(\theta))}{\theta} \).

Key Concepts

Chain Rule

The Chain Rule is a crucial technique in calculus for finding derivatives of composite functions, which are functions made by putting one function inside another. In the context of our exercise, we are working with a nested or composite function:

• An outer function: the natural logarithm, \( y = \ln(u) \).
• An inner function: \( u = \sec(\ln(\theta)) \).

To find the derivative of such a composite function, the chain rule suggests that we take the derivative of the outer function and multiply it by the derivative of the inner function. This approach is like peeling away the layers of an onion: first handling the outside, and then progressing inward. In practice, this means if you have a function \( y = f(g(x)) \), the derivative is found as \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
For our specific problem, this requires multiple applications of the chain rule. We first differentiate the outermost function, \( \ln(u) \), then tackle the derivative of \( \sec(\ln(\theta)) \), bringing all the derivatives together through multiplication.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental concept in calculus. It is the inverse of the exponential function \( e^x \), relying heavily on its base value \( e \), which is approximately 2.71828. When it comes to derivatives, the derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). The natural logarithm is particularly useful for transforming multiplicative relationships into additive ones, making complex calculations more manageable.
In our problem, the outer function \( y = \ln(u) \) required us to use this logarithmic property. The derivative of the natural logarithm with respect to its argument, \( u \), resulted in a straightforward expression, \( \frac{1}{u} \). In our case, \( u \) was \( \sec(\ln(\theta)) \), and substituting back, we have \( \frac{1}{\sec(\ln(\theta))} \), which begins our journey of unpacking this composite function's derivative.

Trigonometric Derivatives

Trigonometric functions, such as sine, cosine, and secant, have well-defined derivatives that are essential for calculus. Each trigonometric function has a specific pattern of derivative, which can simplify the process of finding complex derivatives when they arise within other functions.

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \(-\sin(x) \).
• For \( \sec(x) \), which appears in our exercise, the derivative is \( \sec(x) \tan(x) \).

To solve our problem, we differentiated \( \sec(\ln(\theta)) \), using the known derivative of \( \sec(v) \) with respect to \( v \), which is \( \sec(v) \tan(v) \). When \( v = \ln(\theta) \), this becomes \( \sec(\ln(\theta)) \tan(\ln(\theta)) \). This step is fundamental to solving the derivative of our composite function and showcases the significance of understanding trigonometric derivatives.

Logarithmic Differentiation

Logarithmic Differentiation is a technique often used when dealing with products, quotients, or powers that are tricky to differentiate directly. The main idea is to take the natural logarithm of both sides of an equation, transforming complex algebraic forms into simpler ones before differentiating. This leverages the simpler properties of logarithms to make the differentiation process more straightforward.
Though logarithmic differentiation itself is not explicitly used in the step-by-step solution, it's akin to the kind of simplifications we see in our problem, especially when managing complex derivative chains. With \( y = \ln(\sec(\ln(\theta))) \), understanding how to juggle multiple layers of functions or how to break them down iteratively mirrors the mental steps invoked by logarithmic differentiation. This highlights its utility for similar problems, where untangling nested expressions can be key to achieving a clean, solvable differential equation.