Vehicular stopping distance Based on data from the U.S. Bureau of Public Roads, a model for the total stopping distance of a moving car in terms of its speed is s = 1.1v + 0.054v², where s is measured in ft and v in mph. The linear term 1.1v models the distance the car travels during the time the driver perceives a need to stop until the brakes are applied, and the quadratic term 0.054v² models the additional braking distance once they are applied. Find ds/dv at v = 35 and v = 70 mph, and interpret the meaning of the derivative.
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 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
- 12. Techniques of Integration7h 39m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
3. Techniques of Differentiation
Basic Rules of Differentiation
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Find the indicated derivative.
d/dt(−5x)
A
0
B
x
C
t
D
−5

1
Identify the function to differentiate: \(-5x\).
Recognize that the derivative is with respect to \(t\), but the function is in terms of \(x\).
Since \(-5x\) is a constant with respect to \(t\), its derivative with respect to \(t\) is 0.
Apply the constant rule of differentiation: \(\frac{d}{dt}(c) = 0\) where \(c\) is a constant.
Conclude that the derivative \(\frac{d}{dt}(-5x) = 0\) because \(-5x\) does not depend on \(t\).
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Basic Rules of Differentiation practice set
