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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.21

In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ln(e^(θ)/(1+e^θ))

Verified step by step guidance
1
Rewrite the given function to simplify the expression inside the logarithm: \(y = \ln\left( \frac{e^{\theta}}{1 + e^{\theta}} \right)\).
Use the logarithm property \(\ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b)\) to separate the function into \(y = \ln(e^{\theta}) - \ln(1 + e^{\theta})\).
Simplify \(\ln(e^{\theta})\) using the property \(\ln(e^{x}) = x\), so the function becomes \(y = \theta - \ln(1 + e^{\theta})\).
Differentiate each term with respect to \(\theta\): the derivative of \(\theta\) is 1, and for \(-\ln(1 + e^{\theta})\), apply the chain rule.
For the second term, use the chain rule: \(\frac{d}{d\theta} \left[ -\ln(1 + e^{\theta}) \right] = - \frac{1}{1 + e^{\theta}} \cdot \frac{d}{d\theta} (1 + e^{\theta}) = - \frac{e^{\theta}}{1 + e^{\theta}}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of the Natural Logarithm Function

The derivative of ln(u), where u is a differentiable function of a variable, is (1/u) times the derivative of u. This rule allows us to differentiate logarithmic expressions by first identifying the inner function and then applying the chain rule.
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Chain Rule

The chain rule is used to differentiate composite functions. It states that the derivative of a function composed of another function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
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Simplification of Exponential Expressions

Simplifying expressions involving exponentials, such as e^θ/(1 + e^θ), helps in easier differentiation. Recognizing that e^θ/(1 + e^θ) can be rewritten or simplified aids in applying derivative rules more efficiently.
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