Calculus I · A Toolkit for Slopes

The quotient rule (which the chain rule hands us)

Now that we have the product and chain rules, the quotient rule is almost free. Write a quotient as a product with a negative power: $\tfrac{f}{g} = f\cdot g^{-1}$. The derivative of $g^{-1}$ is, by power-rule-plus-chain, $-g^{-2}g'$. So the product rule gives $f'g^{-1} + f(-g^{-2}g') = \tfrac{f'}{g} - \tfrac{fg'}{g^2}$, and combining over $g^2$:

The Quotient Rule. $\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\big(g(x)\big)^2}.$

In words: bottom times derivative of top, minus top times derivative of bottom, all over the bottom squared. (The order matters — the minus sign makes it non-symmetric.) Example: $\tfrac{d}{dx}\tfrac{x}{x^2+1} = \tfrac{(1)(x^2+1) - (x)(2x)}{(x^2+1)^2} = \tfrac{1 - x^2}{(x^2+1)^2}$.

The trigonometric derivatives (where Chapter 2 finally pays off)

We promised, back in Chapter 2, that the two strange limits we proved there — $\tfrac{\sin h}{h} \to 1$ and $\tfrac{\cos h - 1}{h} \to 0$ — were the tools that would unlock the derivatives of sine and cosine. Here is that promise kept, and it’s worth savoring because it’s the whole reason those limits earned a place in the book.

Differentiate $\sin x$ from the definition. The key is the angle-addition formula $\sin(x+h) = \sin x\cos h + \cos x\sin h$:

$$\frac{\sin(x+h) - \sin x}{h} = \frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}.$$

Now let $h \to 0$. The first fraction goes to $0$ and the second goes to $1$ — exactly the two Chapter 2 limits — so the whole thing becomes $\sin x \cdot 0 + \cos x \cdot 1$:

$$\boxed{\;\frac{d}{dx}\sin x = \cos x\;}$$

The identical computation with $\cos(x+h) = \cos x\cos h - \sin x\sin h$ gives

$$\boxed{\;\frac{d}{dx}\cos x = -\sin x\;}$$

(the minus sign because cosine is falling as it leaves $0$). And $\tan x = \tfrac{\sin x}{\cos x}$ now yields to the quotient rule: $\tfrac{d}{dx}\tan x = \tfrac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \tfrac{\cos^2 x + \sin^2 x}{\cos^2 x} = \tfrac{1}{\cos^2 x} = \sec^2 x$. These are the derivatives that Chapter 7 used to build the Taylor series of $\sin$ and $\cos$ (and, through them, Euler’s formula) — now derived from the ground up.

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