Calculus I · A Toolkit for Slopes

Sums, differences, and constant multiples

In Chapter 3 you found $\tfrac{d}{dx}(3x^2 - 4x + 1) = 6x - 4$ — you differentiated term by term, and you predicted the rule in the Looking-ahead problems. Here it is, and it’s immediate from the definition (the difference quotient of a sum is the sum of the difference quotients):

Sum / Constant-Multiple Rules. $\displaystyle \frac{d}{dx}\big[f(x) + g(x)\big] = f'(x) + g'(x), \qquad \frac{d}{dx}\big[c\,f(x)\big] = c\,f'(x).$

Together with the power rule, these differentiate any polynomial instantly. For example, $\tfrac{d}{dx}(5x^4 - 2x^3 + 7x - 9) = 20x^3 - 6x^2 + 7$ — bring each exponent down, subtract one, drop the constant. What took a page in Chapter 3 now takes a glance.

The product rule (and why it is not what you’d guess)

Here intuition fails, so we derive carefully. You might guess that the derivative of a product is the product of the derivatives — that $(fg)' = f'g'$. It is not, and the derivation shows why. Start from the definition and use one clever move: add and subtract the same term, $f(a+h)g(a)$, to split the quotient into two recognizable pieces.

$$\frac{f(a+h)g(a+h) - f(a)g(a)}{h} = \frac{f(a+h)g(a+h) - f(a+h)g(a)}{h} + \frac{f(a+h)g(a) - f(a)g(a)}{h}$$ $$= f(a+h)\cdot\frac{g(a+h)-g(a)}{h} + g(a)\cdot\frac{f(a+h)-f(a)}{h}.$$

Now let $h \to 0$. The two difference quotients become $g'(a)$ and $f'(a)$; and $f(a+h) \to f(a)$ because $f$ is continuous (which, by Chapter 3, follows from $f$ being differentiable). So:

The Product Rule. $\displaystyle \frac{d}{dx}\big[f(x)\,g(x)\big] = f'(x)\,g(x) + f(x)\,g'(x).$

In words: derivative of the first times the second, plus the first times the derivative of the second. The extra term — the one the naive guess misses — is exactly the cross-term that the add-and-subtract trick exposed. Example: $\tfrac{d}{dx}\big[x^2(x^3+1)\big] = (2x)(x^3+1) + (x^2)(3x^2) = 2x^4 + 2x + 3x^4 = 5x^4 + 2x$. (Check: expanding first gives $x^5 + x^2$, whose derivative is $5x^4 + 2x$. They agree.)

The chain rule: differentiating a function inside a function

This is the one that matters most, because most real functions are compositions — a function applied to the output of another function, like $(x^2+1)^5$ (raise-to-the-fifth applied to $x^2+1$) or $\sqrt{x^2+1}$ or $e^{x^2}$. The chain rule tells you how to differentiate them, and Leibniz’s notation makes it almost think for you. Write $y = f(u)$ where $u = g(x)$ — an outer function $f$ and an inner function $g$. Then:

The Chain Rule. $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}, \qquad\text{or}\qquad \frac{d}{dx}\,f(g(x)) = f'(g(x))\cdot g'(x).$

The idea behind it is beautifully simple: rates of change multiply through a chain. If $y$ changes $\tfrac{dy}{du}$ times as fast as $u$, and $u$ changes $\tfrac{du}{dx}$ times as fast as $x$, then $y$ changes $\tfrac{dy}{du}\cdot\tfrac{du}{dx}$ times as fast as $x$ — the same way gear ratios multiply. (The honest derivation writes $\tfrac{\Delta y}{\Delta x} = \tfrac{\Delta y}{\Delta u}\cdot\tfrac{\Delta u}{\Delta x}$ and takes limits; a careful proof handles the case $\Delta u = 0$ separately, which is the Diamond problem.) In practice: differentiate the outer function, leaving the inside alone, then multiply by the derivative of the inside.

• $\dfrac{d}{dx}(x^2+1)^5 = 5(x^2+1)^4 \cdot \underbrace{2x}_{\text{inside}'} = 10x(x^2+1)^4.$
• $\dfrac{d}{dx}\sqrt{x^2+1} = \dfrac{1}{2\sqrt{x^2+1}}\cdot 2x = \dfrac{x}{\sqrt{x^2+1}}.$
• $\dfrac{d}{dx}e^{x^2} = e^{x^2}\cdot 2x$ (using the exponential’s derivative, below).

A glimpse of why machine learning runs on this rule

A neural network is a tower of functions — the input passes through layer after layer, each one a function of the layer below: $f\big(g\big(h(\cdots)\big)\big)$, sometimes hundreds deep. To train it, you need the derivative of the final error with respect to every knob in every layer — and you get it by the chain rule, multiplying the rates of change layer by layer, working from the output back toward the input. That backward sweep has a name — backpropagation — and it is nothing more exotic than the chain rule, applied across an enormous composition, a few billion times per second. The rule you’re learning on $(x^2+1)^5$ is the literal engine of modern AI.

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