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
3. Techniques of Differentiation
The Chain Rule
Problem 96c
Textbook Question
Composition containing sin x Suppose f is differentiable on [−2,2] with f′(0)=3 and f′(1)=5. Let g(x)=f(sin x). Evaluate the following expressions.
c. g'(π)
![](/channels/images/assetPage/verifiedSolution.png)
1
Step 1: Recognize that g(x) is a composition of functions, where g(x) = f(sin x). To find g'(x), we need to use the chain rule for differentiation.
Step 2: The chain rule states that if you have a composition of functions h(x) = f(u(x)), then the derivative h'(x) = f'(u(x)) * u'(x).
Step 3: In this problem, u(x) = sin x, so u'(x) = cos x. Therefore, g'(x) = f'(sin x) * cos x.
Step 4: We need to evaluate g'(π). First, find sin(π), which is 0. So, g'(π) = f'(sin(π)) * cos(π).
Step 5: Substitute the known values: f'(0) = 3 and cos(π) = -1. Therefore, g'(π) = f'(0) * (-1).
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