- 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
Product and Quotient Rules
Problem 3.57b
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.
b. ƒ(x)g²(x), x = 0

1
To find the derivative of the product ƒ(x)g²(x), we will use the product rule. The product rule states that if you have two functions u(x) and v(x), the derivative of their product is given by (u(x)v(x))' = u'(x)v(x) + u(x)v'(x).
In this case, let u(x) = ƒ(x) and v(x) = g²(x). We need to find the derivative of v(x) = g²(x) using the chain rule. The chain rule states that if you have a composite function h(x) = f(g(x)), then h'(x) = f'(g(x))g'(x).
Apply the chain rule to find the derivative of g²(x): v'(x) = 2g(x)g'(x).
Now, apply the product rule: (ƒ(x)g²(x))' = ƒ'(x)g²(x) + ƒ(x)(2g(x)g'(x)).
Substitute the given values at x = 0: ƒ(x) = 1, g(x) = 1, ƒ'(x) = -3, and g'(x) = 1/2 into the expression from the product rule to find the derivative at x = 0.
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