Derivative Flashcards
For a function f to be differentiable at x=a, the function has to be..
If f is differentiable at a point x0, then f must also be continuous at x0. In particular, any differentiable function must be continuous at every point in its domain.
The converse does not hold.
A continuous function need not be differentiable. True or False?
True. A continuous function need not be differentiable. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Example of of a function which is continous but not differentiable.
An ordinary cusp on the cubic curve (semicubical parabola) x3 – y2 = 0, which is equivalent to the multivalued function f(x) = ± x3/2. This relation is continuous, but is not differentiable at the cusp.
Rolle’s theorem
If a real-valued function ƒ is continuous on aclosed interval [a, b], differentiable on the open interval(a, b), and ƒ(a) = ƒ(b), then there exists a c in the open interval (a, b) such that f’**(c) = 0.
Indian mathematician Bhāskara II (1114–1185) is credited with knowledge of Rolle’s theorem.[1]
Mean Value Theorem
For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.
Proof of Mean Value Theorem
* Let g(x) = f(x) − rx, where r is a constant. * Since f is continuous on [a, b] and differentiable on (a, b), the same is true for g. * Choose r so that g satisfies the conditions of Rolle’s theorem. * Since g is differentiable and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,
Which theorem do we use to prove the Mean Value Theorem
Rolle’s theorem
If f is differentiable at x = a then
The limit h approaching 0, (f(a+h) - f(a))/h
Product rule
(f*g)’ = f’*g + g’f
Proof of product rule
- Let h(x) = f(x) g(x),
- suppose that f and g are each differentiable at x
- We want to prove that h is differentiable at x and that its derivative h’(x) is given by f’(x) g(x) + f(x) g’(x).
- Add f(x)g(x+deltax) - f(x) g(x+deltax) to the numerator to permit its factoring
- Properties of limits are used.
What is (f/g)’?
(g*f’ - f*g’)/g2
(f(g(x))’
f’(g(x)) ( g’(x)
(f-1(x))’
- Let y = f-(x)
- Then f-1(x))’ = 1/f’(y)
f(g(x)
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Why should we be more careful than usual in our statement of the differentiability result for inverse functions?
- The differentiation formula for the inverse function involves division by f ‘(f -1(x)).
- We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity.
- f(x) = xa.*
- f”(x) = ?*
axa-1