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 Integrals3h 25m
3. Techniques of Differentiation
The Chain Rule
Problem 3.57d
Textbook Question
Suppose that functions ƒ(x) and g(x) and their first derivatives have the following values at x = 0 and x = 1.
x ƒ(x) g(x) ƒ'(x) g'(x)
0 1 1 -3 1/2
1 3 5 1/2 -4
Find the first derivatives of the following combinations at the given value of x.
d. ƒ(g(x)), x = 0

1
To find the derivative of the composition of two functions, ƒ(g(x)), we use the chain rule. The chain rule states that the derivative of a composite function ƒ(g(x)) is ƒ'(g(x)) * g'(x).
First, identify the values needed from the table: at x = 0, we have g(x) = 1 and g'(x) = 1/2.
Next, evaluate ƒ'(g(x)) at x = 0. Since g(x) = 1 when x = 0, we need ƒ'(1). From the table, ƒ'(1) = 1/2.
Now, apply the chain rule: the derivative of ƒ(g(x)) at x = 0 is ƒ'(g(0)) * g'(0). Substitute the values: ƒ'(1) * g'(0) = (1/2) * (1/2).
Finally, multiply the values obtained: (1/2) * (1/2). This gives the derivative of the composite function at x = 0.
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