KindiakMath

The Trifecta of Differentiation

In differentiation, three results are used repeatedly to establish a slew of commonly-used derivatives: the chain rule, the product rule, and the quotient rule.

The chain rule is the key result that establishes the other two, which we have proven before, and will re-state for clarification.

Theorem 1 (Chain Rule). Let f,g: \mathbb R \to \mathbb R be functions. Suppose f is differentiable at c and g is differentiable at f(c). Then g \circ f is differentiable at c with derivative (g \circ f)'(c) = g'(f(c))f'(c).

With the chain rule, we obtain the product rule as a corollary.

Theorem 2 (Product Rule). Let f,g be functions differentiable at some c. Then the product fg defined by

(fg)(x) := f(x)g(x)

is differentiable at c with derivative

(fg)'(c) = f'(c)g(c) + g'(c)f(c).

Proof. Apply linearity and the chain rule on the function

fg = \frac 12 ((f + g)^2 - f^2 - g^2).

The quotient rule then follows as an immediate corollary of both the chain rule and the product rule via f/g = f \cdot 1/g.

Theorem 3 (Quotient Rule). Let f,g be functions differentiable at some c. If g(c) neq 0, then the quotient f/g defined by

\displaystyle \left( \frac fg \right)(x) := \frac{f(x)}{g(x)}

is differentiable at c with derivative

\displaystyle \left( \frac fg \right)'(x) := \frac{f'(c) g(c) - f(c) g'(c)}{{g(c)}^2}.

These properties, coupled with the starting points \sin' = \cos and \exp' = \exp, suffice to prove most commonly used derivatives in high school and even in freshmen calculus.

To properly define the functions \sin and \exp, however, requires more thought. We will explore them in real analysis.

For now, let’s use the quotient rule to prove that \tan' = \sec^2.

Theorem 4. We have \tan' = \sec^2.

Proof. Writing \tan(x) = \sin(x)/{\cos(x)},

\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm d x}(\tan(x)) &= \frac{\mathrm d}{\mathrm d x} \left( \frac{\sin(x)}{\cos(x)} \right) \\ &= \frac{\cos(x) \cdot \cos(x) - \sin(x) \cdot (-{\sin(x)})}{\cos^2(x)} \\ &= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}\\ &= \frac{1}{\cos^2(x)} \\ &= \sec^2(x). \end{aligned}

—Joel Kindiak, 20 Oct 24, 1504H

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