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8.3.4. Proposition. If f, gFa, then


Proof. Exercise. (Solution .)

The last two propositions prefigure the fact that differentiation is a linear operator; the next result will lead to Leibniz's rule for differentiating products.

8.3.5. Proposition. If φ, fFa, then


Proof. Problem.

Finally, we present a result which prepares the way for the chain rule.

8.3.6. Proposition. If fFa, gFf(a), and gfFa, then


Proof. Exercise. (Solution .)

8.3.7. Proposition. Let AR. A function f:AR is continuous at the point a in A if and only if fa is continuous at 0.

Proof. Problem.

8.3.8. Proposition. If f:UU1 is a bijection between subsets of R, then f0or each a in U the function fa:U-aU1-f(a) is invertible and


Proof. Problem.


We now have developed enough machinery to talk sensibly about differentiating real valued functions.

8.4.1. Definition. Let fFa. We say that f is differentiable at a if there exists a linear function which is tangent at 0 to fa. If such a function exists, it is called the differential of f at a and is denoted by dfa. (Don't be put off by the slightly complicated notation; dfa is just a member of L satisfying dfafa.) We denote by Da the family of all functions in Fa which are differentiable at a.

The next proposition justifies the use of the definite article which modifies ""differential“ in the preceding paragraph.

8.4.2. Proposition. Let fFa. If f is diffe rentiable at a, then its diffe rential is unique. (That is, there is at most one linear map tangent at 0 to fa.)

Proof. Proposition .

8.4.3. Example. It is instructive to examine the relationship between the differential of f at a, which we defined in , and the derivative of f at a as defined in beginning calculus. For fFa to be differentiable at a it is necessary that there be a linear function T:RR which is tangent at 0 to fa. According to there must exist a constant c such that Tx=cx for all x in R. For T to be tangent to fa, it must be the case that


that is,