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Topos Uranos
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Differentiation rules

Summary

This lesson derives from the Carathéodory characterization the rules that make it possible to differentiate without returning each time to the difference quotient: the derivative of a linear combination, of a product, of a reciprocal and of a quotient. Each rule is obtained in the same way: the increment of the new function is written as a product with x−ax - a, whose other factor is continuous at aa, and the value of that factor at aa is the derivative. With the rules one differentiates powers with an integer exponent, polynomials and rational functions. The lesson ends with piecewise-defined functions: inside each piece the formula of the piece governs, and at the junction points the one-sided derivatives decide, which are not always the derivatives of the formulas.

Learning objectives

  1. Prove with the Carathéodory characterization the rules for a linear combination, a product, a reciprocal and a quotient.
  2. Differentiate powers with an integer exponent, polynomials and rational functions, and specify the domain of each derivative.
  3. Decide with the one-sided derivatives whether a piecewise-defined function is differentiable at a junction point.
  4. Distinguish sufficient hypotheses from necessary ones in the rules, and recognize when the derivatives of the formulas of the pieces do not give the derivative of the function.

The derivative depends only on what happens near the point

The lesson on the derivative as a limit computed each derivative with its difference quotient. For a function such as x3−2xx2+1\frac{x^3 - 2x}{x^2 + 1}, that computation is long; it is better, instead, to know how the derivative behaves under the operations. All the rules of this lesson rest on the Carathéodory characterization, from the same lesson: ff is differentiable at aa, with f′(a)=Lf'(a) = L, if and only if there exists a function φ\varphi, continuous at aa, with φ(a)=L\varphi(a) = L, such that f(x)−f(a)=φ(x)(x−a)f(x) - f(a) = \varphi(x)(x - a) for every xx in the domain. Therefore, to prove that a function is differentiable at aa it suffices to write its increment as a product with x−ax - a whose other factor is continuous at aa; the value of that factor at aa is the derivative. The theorem on operations with continuous functions, from the lesson on limit and continuity, takes care of the continuity of the factor.

First a question of domains must be settled. Two functions may have different domains, and the quotient fg\frac{f}{g} is defined only where gg does not vanish. The lemma below says that the derivative does not change when the domain is reduced, provided the point remains an accumulation point, and that it depends only on the values near the point.

LemmaLocal character of the derivative

Let ff be a function defined on AA, aa an element of AA, BB a subset of AA that contains aa, and FF the restriction of ff to BB.

  • If aa is an accumulation point of BB and ff is differentiable at aa, then FF is differentiable at aa, and F′(a)=f′(a)F'(a) = f'(a).
  • If aa is an accumulation point of AA and there exists a δ>0\delta > 0 such that every xx in AA with ∣x−a∣<δ\abs{x - a} < \delta belongs to BB, then ff is differentiable at aa if and only if FF is, and in that case F′(a)=f′(a)F'(a) = f'(a).

The first part is the lemma on restriction from the algebra of limits; the second, that the definition of the limit only looks at nearby points.

Proof

  1. x∈B∖{a}→F(x)−F(a)x−a=f(x)−f(a)x−ax \in B \setminus \{a\} \rightarrow \frac{F(x) - F(a)}{x - a} = \resaltar{\frac{f(x) - f(a)}{x - a}}

    The difference quotient of FF is the restriction to B∖{a}B \setminus \{a\} of that of ff. If aa is an accumulation point of BB, the lemma on the limit of a restriction, from the lesson on the algebra of limits, gives the same limit.

  2. x∈A∧0<∣x−a∣<min⁡{δ,δ′}→∣f(x)−f(a)x−a−L∣<εx \in A \wedge 0 < \abs{x - a} < \min\{\delta, \delta'\} \rightarrow \resaltar{\abs{\frac{f(x) - f(a)}{x - a} - L} < \varepsilon}

    In the second case, every point of AA other than aa and at a distance less than δ\delta lies in BB: since aa is an accumulation point of AA, it is also one of BB, and the first part gives one implication. For the other, if F′(a)=LF'(a) = L, let ε>0\varepsilon > 0 be given and let δ′\delta' be a response for the quotient of FF; the smaller of δ\delta and δ′\delta' is a response for the quotient of ff, because the two quotients coincide at the xx it covers.

From the second part it follows that two functions defined on AA that coincide at the points of AA near aa are both differentiable at aa, with the same derivative, or neither is: the two have the same restriction to those points. And from the first, that if ff and gg have different domains, both containing the point aa, one can work on the intersection of the domains, provided aa is an accumulation point of it. That is why, in what follows, ff and gg are assumed to be defined on one and the same set AA.