Calculus Differentiability and the Chain Rule October 9, 2018
Definitions
(a) Let f be a function f defined on the interval I = (s, t). f is said to be differentiable at a point x ∈ I if the limit
f0(x) = lim
h→0
f (x + h) − f (x)
h exists.
(b) A function f is said to be differentiable on the interval I = (s, t) if f is differentiable at each point in I.
Remark f is differentiable on an interval I = (s, t) if and only if for each x ∈ I there exists l = l(x) ∈ R such that the function
o(h) = f (x + h) − f (x) − l h for all x + h in I satisfies that
h→0lim o(h)
h = 0.
Here, the function o(h) is read as ”little o of h” and f0(x) = l(x) = l if l exists.
Proof
=⇒ Suppose that f is differentiable at some point x ∈ I, i.e.
f0(x) = lim
h→0
f (x + h) − f (x)
h exists.
By setting
l = f0(x) and o(h) = f (x + h) − f (x) − l h for x + h ∈ I, we have
h→0lim o(h)
h = lim
h→0
f (x + h) − f (x) − l h
h = lim
h→0
f (x + h) − f (x)
h − f0(x) = 0.
⇐= Let x ∈ I. Suppose that l = l(x) is a real number such that the function o(h) = f (x + h) − f (x) − l h for all x + h in I
satisfies that
lim
h→0
o(h) h = 0.
Then we have
0 = lim
h→0
o(h)
h = lim
h→0
f (x + h) − f (x)
h − l.
Hence,
lim
h→0
f (x + h) − f (x)
h = l exists.
By setting f0(x) = l, we have shown that f is differentiable at x ∈ I.
Calculus Differentiability and the Chain Rule (Continued) October 9, 2018
Theorem (the Chain Rule) Let f be differentiable for all x ∈ I = (s, t) and let g be differen- tiable for all y ∈ J = (u, v). Suppose that
Range of f = f (I) = {f (x) | x ∈ I} ⊆ J = Domain of g.
Then g ◦ f is differentiable for all x ∈ I with
(g ◦ f )0(x) = g0(f (x)) f0(x).
Proof For each x ∈ I, since f is differentiable at x, the function
o(h) = f (x + h) − f (x) − f0(x) h defined for sufficiently small h satisfies that
h→0lim o(h)
h = 0.
Also since f (I) ⊆ J, the range of f is a subset of the domain of g, y = f (x) ∈ J and since g is differentiable at f (x), the function
o(k) = g(f (x) + k) − g(f (x)) − g0(f (x)) k for sufficiently small k satisfies that
lim
k→0
o(k) k = 0.
By setting k = f (x + h) − f (x) and using that lim
h→0k = 0 by the continuity of f at x and lim
h→0
k
h = f0(x), we have
h→0lim
g ◦ f (x + h) − g ◦ f (x) − g0(f (x)) f0(x) h h
= lim
h→0
g(f (x + h)) − g(f (x)) − g0(f (x)) f0(x) h h
= lim
h→0
g(f (x) + f (x + h) − f (x)) − g(f (x)) − g0(f (x))f (x + h) − f (x) − o(h)
h
= lim
h→0
g(f (x) + k) − g(f (x)) − g0(f (x))k − o(h)
h
= lim
h→0
g(f (x) + k) − g(f (x)) − g0(f (x)) k k
k h + lim
h→0
o(h) h
= 0.
Hence g ◦ f is differentiable for all x ∈ I with
(g ◦ f )0(x) = g0(f (x)) f0(x) for all x ∈ I.
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