Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Implicit Differentiation
Problem 3.R.58
Textbook Question
Evaluate and simplify y'.
sin x cos(y−1) = 1/2

1
First, identify the given equation: \( y' \cdot \sin(x) \cdot \cos(y - 1) = \frac{1}{2} \). Here, \( y' \) represents the derivative of \( y \) with respect to \( x \).
To isolate \( y' \), divide both sides of the equation by \( \sin(x) \cdot \cos(y - 1) \). This gives: \( y' = \frac{1}{2} \div (\sin(x) \cdot \cos(y - 1)) \).
Simplify the expression for \( y' \) by rewriting the division as a multiplication by the reciprocal: \( y' = \frac{1}{2} \cdot \frac{1}{\sin(x) \cdot \cos(y - 1)} \).
Combine the fractions to express \( y' \) as a single fraction: \( y' = \frac{1}{2 \cdot \sin(x) \cdot \cos(y - 1)} \).
The expression for \( y' \) is now simplified. Ensure that the domain restrictions for \( \sin(x) \) and \( \cos(y - 1) \) are considered, as they cannot be zero.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent variable is not isolated on one side. In this case, we have a relationship between x and y given by sin x cos(y−1) = 1/2. By differentiating both sides with respect to x, we can find dy/dx, which represents y'.
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Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When differentiating terms involving y, such as cos(y−1), we apply the chain rule to account for the derivative of the inner function (y−1) with respect to x, which involves dy/dx. This is crucial for correctly finding y'.
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Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables. In this problem, understanding the properties of sine and cosine is essential, especially since we are working with sin x and cos(y−1). These identities can help simplify the expression after differentiation and aid in solving for y'.
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