Derivatives

Given a function f, assume that f′(a) = m, where a is fixed. We put

ϵ(h) = \frac{f(a + h) − f(a)}{h} − m for h ≠ 0

in order to arrange ϵ(h) → 0 as h → 0. When we multiply up, we get

f(a + h) = f(a) + mh + hϵ(h)  (1)

On the right, we have the constant term f(a), then a constant multiple of h, then an error term which is small compared to h.

Definition 2 The function f is differentiable at a, with derivative f′(a) = m, if and only if equation (1) holds with ϵ(h) → 0 as h → 0.

In equation (1) we no longer have any division by h, and it remains valid for h = 0, whatever value we choose for ϵ(0). The reasonable choice is ϵ(0) = 0, to make the function ϵ continuous at h = 0. This form of the definition leads to cleaner proofs of the standard rules.

Theorem 3 If f is differentiable at a, it is continuous at a.

Proof We must show that f(a+h) → f(a) as h → 0. From equation (1) we have

f(a + h) → f(a) + 0 + 0 = f(a) as h → 0.

Constant multiples If we multiply equation (1) by a constant c, we get

cf(a + h) = cf(a) + cmh + chϵ(h)

to which Definition 2 applies immediately. We write cm = cf′(a).

Theorem 4 Given a constant c, if f is differentiable at a, then so is cf, with derivative cf′(a).

Sums and Differences Assume also that g′(a) = p, so that like equation (1) we have

g(a + h) = g(a) + ph + hθ(h)  (5)

with θ(h) → 0 as h → 0. If we add (or subtract) equations (1) and (5), we see that

f(a+h) ± g(a+h) = f(a) ± g(a) + (m±p)h + h{ϵ(h)±θ(h)}

which has the form prescribed by Definition 2 for showing that f ± g is differentiable.

Theorem 6 If f and g are differentiable at a, then so is f ± g, with derivative f′(a) ± g′(a).

The Product Rule We can just as well multiply equations (1) and (5). There are nine terms, which we collect as we go,

\begin{matrix}f(a+h)g(a+h) = & f(a)g(a) + \{f(a)p+mg(a)\}h \\ h\{f(a)θ(h) + mph + mhθ(h) + ϵ(h)g(a) + ϵ(h)ph + hϵ(h)θ(h)\} \end{matrix}

Again, Definition 2 applies, and we plug in p = g′(a) and m = f′(a).

Theorem 7 If f and g are differentiable at a, then so is fg, with derivative

f′(a)g(a) + f(a)g′(a)

The Chain Rule Suppose that u = g(x) and y = f(u), so that y = f(u) = f(g(x)). Assume first that g′(a) = p, so that we have by equation (5)

\begin{aligned}g(a + h) & = g(a) + ph + hθ(h) \\ & = b + k \end{aligned}

where we write b = g(a) and k = h(p + θ(h)) in preparation for the next step, which is to expand f(g(a+h)) = f(b+k). We note that k → 0 as h → 0.

Assume that f′(b) = q, so that like equation (1),

\begin{aligned}f(g(a+h)) = f(b+k) & = f(b) + qk + kϵ(k) \\ & = f(b) + qph + h\{qθ(h) + pϵ(k) + θ(h)ϵ(k)\} \end{aligned}

where ϵ(k) → 0 as k → 0 and we substituted for k in some places. Once again, Definition 2 applies, and the derivative of f ∘ g is qp. (It is possible to have k = 0 exactly, but this is not a problem if we define ϵ(0) = 0 as suggested, to make ϵ continous.) We plug in q = f′(b) = f′(g(a)) and p = g′(a).

Theorem 8 If g is differentiable at a and f is differentiable at g(a), then the composite f ∘ g is differentiable at a, with derivative f′(g(a)) g′(a).

Reciprocals We apply Theorem 8 with f(u) = 1/u, for which f′(u) = −1/u^{2}.

Theorem 9 If f is differentiable at a, then so is 1/f, with derivative \frac{− f′(a)}{f(a)^{2}}

(provided that f(a) ≠ 0).

The Quotient Rule

Theorem 10 If u and v are differentiable at a, then so is \frac{u}{v}, with derivative

\frac{v(a)u′(a) − u(a)v′(a)}{v(a)^{2}}

provided v(a) ≠ 0.

Proof We write u/v = u.(1/v) and apply the product rule Theorem 7 and Theorem 9, to get

u′(a)\frac{1}{v(a)} + u(a) \frac{− v′(a)}{v(a)^{2}}

which can be rearranged as stated.