Labware - MA35 Multivariable Calculus - Single Variable Calculus
 MA35 Labs 1 » Single Variable Calculus Contents1.2 Differentiation 1.2.1 Derivatives 1.2.2 Critical Points 1.2.4 Chain Rule 1.2.5 Differentiability 1.4 Integration Search

Differentiability

Text

According to the definition, a function f of one variable is said to be differentiable at a point x0 in its domain if there exists a function f '(x) such that

f'(x0) = limh --> 0 [f(x0 + h) - f(x0)]/h

where h is a real number. This is equivalent to

limh --> 0 [f(x0 + h) - f(x0) - f'(x_0)*h]/h = 0.

This condition can be written in terms of epsilons and deltas as well. We say that f is differentiable at x0 if for any ε > 0 there exists a &delta > 0 such that

[f(x_0 + h) - f(x_0) - f'(x_0)h]/h < ε or equivalently,

[ f '(x0 - ε) h < f(x0 + h) - f(x0) < (f '(x0) + ε) h

whenever |h| < &delta. This definition is similar in form to the definition of continuity. Recall that in the geometrical interpretation of continuity the challenge was to find a small enough &delta-disc domain of x0 such that the graph of the function over the domain would lie between two horizontal bars a distance ε above and below f(x0). In the geometrical interpretation of differentiability, as seen in the inequality above, the challenge is to find a &delta-disc domain, centered at x0, that is small enough so that the graph of f over the domain lies between lines with slopes (f'(x0) + ε and (f'(x0) - ε).

Demos

 Differentiability This demo shows two blue lines through a point (x0, f(x0)), one with a slope ε greater than that of the tangent line and one with a slope ε less than that of the tangent line. If the function is differentiable at x0, you should always be able to choose a &delta (controlled by the hotspot) small enough that the portion of the curve between x0 - &delta and x0 + &delta lies inside the blue lines.

Exercises

• 1. Use the demo to test whether the following functions are differentiable at the points given:
• f(x) = x^2 at x = 5
• f(x) = tan(x) at x = π/4
• f(x) = |x| at x = 0
• f(x) = |x| at x = 1
• 2. Use the demonstrations for continuity and differentiability in one variable to give an argument for why differentiability at a point implies continuity at a point.