Differentiability and the Theorems Behind the Rules
Define the derivative as a limit, prove differentiability implies continuity, state and prove Rolle's theorem using the extreme value theorem, and derive the mean value theorem from Rolle's theorem.
What a learner can do afterwards
- Prove that differentiability at a point forces continuity there
- Give a continuous function that is not differentiable and say where the definition fails
- Prove Rolle's theorem using the extreme value theorem, then prove the mean value theorem from Rolle's theorem
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The derivative at a point a is the limit of the secant slope as x approaches a. If f is differentiable at a, write f(x) minus f(a) as the difference quotient times x minus a. The quotient tends to the finite number f'(a), while x minus a tends to 0. Thus f(x) tends to f(a), so f is continuous at a. This limit is taken at the single point a, so differentiability at a says nothing about differentiability at nearby points.
Continuity does not imply differentiability. For f(x) = |x minus 2| at 2, the left secant slopes tend to -1 and the right slopes tend to 1. The derivative limit has no single value. A cusp can make slopes diverge, and a vertical tangent does not give a finite derivative. Check the two-sided difference quotient rather than relying on a graph's smooth look.
Rolle's theorem assumes continuity on [a, b], differentiability on (a, b), and f(a) = f(b). It concludes f'(c) = 0 at some interior c. The extreme value theorem gives a maximum and a minimum on [a, b]. If both occurred only at endpoints, both would equal f(a), so f would be constant and any interior c works. Otherwise an extremum sits at an interior c. At an interior maximum, right difference quotients are at most 0 and left ones at least 0, so f'(c) = 0. A minimum works the same way.
The mean value theorem keeps Rolle's hypotheses except f(a) = f(b): continuity on [a, b] and differentiability only on (a, b). Let m = (f(b) minus f(a)) divided by (b minus a) and g(x) = f(x) minus f(a) minus m times (x minus a). Then g(a) = g(b) = 0, so Rolle gives c with g'(c) = 0, that is f'(c) = m. For f(x) = x cubed on [-1, 2], m = 3 and 3c squared = 3, so c = 1 or -1. Only c = 1 is interior.
Differentiability gives continuity, Rolle's theorem uses an interior extremum, and subtracting the secant line turns Rolle's theorem into the mean value theorem.
Lesson, questions and flashcards are original text written for this dataset.
The rewritten reading was written for this dataset. Reading notes summarise three Khan Academy AP Calculus AB articles. The videos are Khan Academy lessons on YouTube. Questions are original to this dataset, and flashcards were drafted from collected fact candidates.
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