Problem 26

Question

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

Step-by-Step Solution

Verified

Answer

The derivative is $ \frac{dy}{d\theta} = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta}. $

1Step 1: Identify the function

The given function is $ y = \ln(\sec \theta + \tan \theta) $. Our goal is to differentiate this function with respect to $ \theta $.

2Step 2: Differentiate using the Chain Rule

To find the derivative of $ y $ with respect to $ \theta $, use the chain rule. The chain rule states that if $ y = \ln(u) $, then $ \frac{dy}{d\theta} = \frac{1}{u} \cdot \frac{du}{d\theta} $. Here, $ u = \sec \theta + \tan \theta $.

3Step 3: Differentiate $ u = \sec \theta + \tan \theta $

Differentiate $ u $ with respect to $ \theta $:1. The derivative of $ \sec \theta $ is $ \sec \theta \tan \theta $.2. The derivative of $ \tan \theta $ is $ \sec^2 \theta $.Thus, $ \frac{du}{d\theta} = \sec \theta \tan \theta + \sec^2 \theta $.

4Step 4: Substitute back into the chain rule expression

Substitute $ u $ and $ \frac{du}{d\theta} $ back into the chain rule formula:\[ \frac{dy}{d\theta} = \frac{1}{\sec \theta + \tan \theta} \cdot (\sec \theta \tan \theta + \sec^2 \theta). \]

5Step 5: Simplify the Expression

Simplify the expression:\[ \frac{dy}{d\theta} = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta}. \]

Key Concepts

Applying the Chain RuleTrigonometric Derivatives in ActionUnderstanding Derivatives of Logarithmic Functions

Applying the Chain Rule

The chain rule is essential in calculus for finding derivatives of composite functions, like our function involving both secant and tangent. To differentiate a composite function, imagine peeling an onion — you work from the outer layers to the innermost core. The chain rule states if you have a function composed of two or more functions, such as $ y = f(g(x)) $, then the derivative is:

$ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) $

For our specific problem, the outer function is the natural log or $ \ln(u) $ and the inner function is $ u = \sec \theta + \tan \theta $. By applying the chain rule, you first differentiate the outer function with respect to the inner function, giving $ \frac{1}{u} $. Now, find the derivative of the inner function $ u $ with respect to $ \theta $, putting it all together to get the complete derivative.

Trigonometric Derivatives in Action

Trigonometric derivatives are an integral part of calculus when working with functions involving trigonometric terms like secant and tangent. Knowing the basic derivatives:

The derivative of $ \sec \theta $ is $ \sec \theta \tan \theta $.
The derivative of $ \tan \theta $ is $ \sec^2 \theta $.

These derivatives help simplify the process. In the context of the function $ u = \sec \theta + \tan \theta $, you will differentiate each term separately:

For $ \sec \theta $, apply its derivative formula.
For $ \tan \theta $, use its specific derivative formula.

Putting them together, $ \frac{du}{d\theta} = \sec \theta \tan \theta + \sec^2 \theta $. This allows you to further use these results in the chain rule.

Understanding Derivatives of Logarithmic Functions

The derivative of logarithmic functions is a concept where understanding the natural log rules is crucial. For $ y = \ln(u) $, its derivative is characterized by:

$ \frac{dy}{du} = \frac{1}{u} $.

In this problem, $ u $ is not just a simple variable but a compound expression ($ \sec \theta + \tan \theta $).
When you take the derivative that incorporates the chain rule, you'll multiply $ \frac{1}{u} $ by the derivative of $ u $ with respect to $ \theta $, which we've identified through trigonometric derivatives. After substitution and simplification, the final expression becomes:

$ \frac{dy}{d\theta} = \frac{\sec \theta \tan \theta + \sec^2 \theta}{\sec \theta + \tan \theta} $.

This process illustrates the magnificence of coupling logarithmic and trigonometric derivatives with the chain rule to untangle complex function derivatives.

Problem 26

Problem 27

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