In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
19. y = (sech θ)(1-ln(sech θ))

Verified step by step guidance

1. Identify the function to differentiate: \(y = (\text{sech} \ \theta)(1 - \ln(\text{sech} \ \theta))\). Notice this is a product of two functions of \(\theta\).
2. Apply the product rule for derivatives: If \(y = u(\theta) \cdot v(\theta)\), then \(\frac{dy}{d\theta} = u'(\theta) v(\theta) + u(\theta) v'(\theta)\), where \(u(\theta) = \text{sech} \ \theta\) and \(v(\theta) = 1 - \ln(\text{sech} \ \theta)\).
3. Find the derivative of \(u(\theta) = \text{sech} \ \theta\). Recall that \(\frac{d}{d\theta} \text{sech} \ \theta = -\text{sech} \ \theta \tanh \ \theta\).
4. Find the derivative of \(v(\theta) = 1 - \ln(\text{sech} \ \theta)\). Use the chain rule: \(\frac{d}{d\theta} \ln(\text{sech} \ \theta) = \frac{1}{\text{sech} \ \theta} \cdot \frac{d}{d\theta} \text{sech} \ \theta\). Substitute the derivative of \(\text{sech} \ \theta\) from the previous step.
5. Combine all parts using the product rule formula: \(\frac{dy}{d\theta} = u'(\theta) v(\theta) + u(\theta) v'(\theta)\), then simplify the expression as much as possible.

Key Concepts

Derivative of Hyperbolic Functions

Hyperbolic functions like sech(θ) have specific derivatives; for example, the derivative of sech(θ) is -sech(θ) tanh(θ). Understanding these derivatives is essential to differentiate expressions involving hyperbolic functions correctly.

Product Rule

The product rule is used to differentiate the product of two functions. It states that (fg)' = f'g + fg', meaning you differentiate each function separately and combine the results appropriately.

Chain Rule

The chain rule helps differentiate composite functions, such as ln(sech(θ)). It requires differentiating the outer function and multiplying by the derivative of the inner function, enabling correct handling of nested expressions.