The algebra of derivatives

Theorem 5.2.1. Let f, g: (a,b) → ℝ, differentiable on (a, b); then f ± g, f ⋅ g, and f/g (g ≠0) are differentiable on (a, b) and the following properties hold:

1. (f ± g)' = f' ± g'

2. (f ⋅ g)' = f' ⋅ g + f ⋅ g'

3. (1/g)' = −g/g².

4. (f / g)' = (f' ⋅ g −f ⋅ g')/g²

Proof.

1. Derivative of a sum. The derivative of the sum/difference of two differentiable functions equals the sum/difference of their derivatives.

   $$\begin{array}{rl}(f+g{)}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{[f(x+h)+g(x+h)]-[f(x)+g(x)]}{h}\\ & =\underset{h\to 0}{lim}[\frac{[f(x+h)-f(x)]}{h}+\frac{[g(x+h)-g(x)}{h}]\\ & =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}+\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\\ & =f^{\prime }(x)+g^{\prime }(x)\end{array}$$

   An analogous proof can be provided for the derivative of a difference.

2. Derivative of a product. We analyze the difference quotient

   $$\begin{array}{r}\frac{(f\cdot g)(x+h)-(f\cdot g)(x)}{h}=\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ & =\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ & =f(x+h)\left[\frac{g(x+h)-g(x)}{h}\right]+g(x)\left[\frac{f(x+h)-f(x)}{h}\right]\end{array}$$

   Here we have added and subtracted f(x + h)g(x) in the numerator and then regrouped the terms so as to display the difference quotients for f and g separately. Thus taking the limit lim_h→0 f(x + h) = f(x), and

   (f ⋅ g)'(x) = f(x)g'(x) + g(x)f'(x).

3. Since g is differentiable at x, it is also continuous at x by Theorem 5.2.1. Since g(x) ≠ 0, we know that the function (1/g)(x) is continuous at x. We have

   $$\begin{array}{rl}{\left(\frac{1}{g(x)}\right)}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{1}{h}[\frac{1}{g(x+h)}-\frac{1}{g(x)}]=\underset{h\to 0}{lim}\frac{g(x)-g(x+h)}{g(x+h)g(x)}\\ & =[-\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}]\cdot \left[\underset{h\to 0}{lim}\frac{1}{g(x+h)g(x)}\right]\\ & =-\frac{g^{\prime }(x)}{(g(x){)}^2}\end{array}$$

   where we used the fact that

   lim_h⟶0 1/g(x + h) = 1/g(x)

4. The quotient rule. Verify it by applying the rule for products and the reciprocal rule. □

The derivative of a composite function: The Chain Rule

Proposition 3.2 (Chain Rule) Let D,E ⊆ ℝ and f: D ⟶ ℝ, g: E ⟶ ℝ be functions such that f(D) ⊆ E. Suppose x is an interior point of E. If f is differentiable at x and g is differentible at f(c), then g ∘ f is differentiable at x and

Proof. We have

For the sake of simplifying the notation, let

then f(x + h) = y + k; from the continuity of f we have for h ⟶ 0 that k ⟶ 0. With the new notations

The derivative of g is:

$$g^{\prime }(y)=\underset{k\to 0}{lim}\frac{g(y+k)-g(y)}{k}$$

which can be written as well, for k ≠ 0:

$$\frac{g(y+k)-g(y)}{k}=g^{\prime }(y)+ϵ(k)$$

Where ε(k) indicates a quntity that goes to zero for k ⟶ 0. By multiplying both sides of the former by k, we obtain

which is perfeclty valid for k = 0 as well. Thus

dividing by h and observing that k/h → f'(x), we have proved the claim.  □

The Chain Rule has a particularly simple expression if we use the Leibniz notation for the derivative. Recall f ∘ g is defined as f(g(x)). The quantity f′(g(x)) is the derivative of f with x replaced by g; this can be written df/dg. As usual, g′(x) = dg/dx. Then the Chain Rule becomes

$$\frac{dw}{dx}=\frac{dw}{dy}\frac{dy}{dx}\text{.}$$

Examples . If w(x) = (sen x)², we have w '(x) = 2 sen x ⋅ cos x.

If w(x) = e^−x2 then w '(x) = e^−x2 ⋅ (−2x).

If w(x) = (sen cos x³)⁻² then w '(x) = −2(sen cos x³)⁻³ ⋅ cos cos x³ ⋅ (−sen x³) ⋅ 3x².   ■

Derivative of a periodic function

Let f: I ⟶ ℝ be a differentiable function, with period T, that is f(x + T) = f(x) ∀x ∈ I, then f' is also periodic with period T. We have if f is differentiable at x,

$$\begin{array}{rl}{f}^{\prime }(x+T)& =\underset{h\to <!--\; \to \; -->0}{lim}\frac{f(x+T+h)-<!--\; -\; -->f(x+T)}{h}\\ & =\underset{h\to <!--\; \to \; -->0}{lim}\frac{f(x+h)-<!--\; -\; -->f(x)}{h}\\ & =f^{\prime }(x).\end{array}$$

Let f: ℝ → ℝ be periodic with period T.

• The range of f is precisely f([0,T]); in particular, if f is continuous, the range of f is bounded.

• If f is differentiable, then f′ is periodic with period T.

Example 1. Verify whether f(x) = sin(x²) is a periodic function. Note that f(x) = sin(x²) is differentiable and f′(x)= 2x ⋅ cos(x²) which is unbounded. Therefore, f′ cannot be periodic by the first point, and hence f cannot be periodic by the second point.   ■

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