Differentiation Rules

01The power rule

The workhorse of the whole course:

$$\frac{d}{dx}\,x^n = n\,x^{n-1} \qquad \text{for any real } n$$

Bring the exponent down, knock the exponent down by one. So $\frac{d}{dx}x^7 = 7x^6$, $\frac{d}{dx}x = 1$, and $\frac{d}{dx}c = 0$ for any constant. The rule's real power comes from fractional and negative exponents: $\sqrt{x} = x^{1/2}$ differentiates to $\frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$, and $\frac{1}{x^3} = x^{-3}$ differentiates to $-3x^{-4}$. The single most valuable habit in this module: rewrite roots and reciprocals as powers before differentiating.

02Linearity: constants and sums

Derivatives pass through constant multiples and sums untouched:

$$\frac{d}{dx}\left[c\,f(x)\right] = c\,f'(x) \qquad\qquad \frac{d}{dx}\left[f(x) \pm g(x)\right] = f'(x) \pm g'(x)$$

Together with the power rule this dispatches every polynomial term by term: $\frac{d}{dx}(5x^4 - 3x^2 + 2x - 9) = 20x^3 - 6x + 2$. Note what linearity does not cover — products and quotients. $\frac{d}{dx}[f \cdot g]$ is emphatically not $f' \cdot g'$, which is the single most common wrong answer in Calc 1. The next two rules exist precisely because of that.

03The product rule

$$\frac{d}{dx}\left[f(x)\,g(x)\right] = f'(x)\,g(x) + f(x)\,g'(x)$$

In words: derivative of the first times the second, plus the first times the derivative of the second. Example: for $y = x^2(x^3 - 2x)$,

$$y' = 2x\,(x^3 - 2x) + x^2\,(3x^2 - 2) = 5x^4 - 6x^2$$

(You could expand first and get the same thing — a good self-check while learning. Soon the factors won't expand, and the rule becomes mandatory.)

04The quotient rule

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{g(x)^2}$$

The mnemonic that sticks: "low d-high minus high d-low, over low squared." Order matters in the numerator — the subtraction makes it non-symmetric, and swapping the terms flips the sign. Example: $y = \dfrac{2x+1}{x-3}$ gives

$$y' = \frac{2(x-3) - (2x+1)(1)}{(x-3)^2} = \frac{-7}{(x-3)^2}$$

Before reaching for the quotient rule, check whether the fraction simplifies: $\frac{x^3 + 2x}{x} = x^2 + 2$ is a power-rule problem in disguise.

05The chain rule

The most important rule in calculus, for composite functions:

$$\frac{d}{dx}\,f\big(g(x)\big) = f'\big(g(x)\big)\cdot g'(x)$$

In words: derivative of the outside (leaving the inside alone), times the derivative of the inside. For $y = (3x^2 + 5)^8$: the outside is $u^8$, the inside is $u = 3x^2 + 5$, so

$$y' = 8(3x^2+5)^7 \cdot 6x = 48x\,(3x^2+5)^7$$

This is exactly the inside/outside decomposition drilled in Modules 01's composition problems. In Leibniz notation the rule reads $\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$ — the $du$'s "cancel," which is why the notation is beloved. Nested compositions chain further: peel one layer at a time, outermost first.

06Combining the rules

Real problems stack rules. For $y = x^3\sqrt{2x + 1}$: product rule on the outside, chain rule inside the root:

$$y' = 3x^2\,(2x+1)^{1/2} + x^3\cdot\tfrac{1}{2}(2x+1)^{-1/2}\cdot 2 = 3x^2\sqrt{2x+1} + \frac{x^3}{\sqrt{2x+1}}$$

Strategy: identify the outermost structure first (is the whole thing a product? a quotient? a power of something?), apply that rule, and recurse inward. Writing $u = $ (inside) explicitly is not a crutch; it's what careful differentiation looks like.

07Higher-order derivatives

Differentiating $f'$ again gives the second derivative $f''(x)$, also written $\frac{d^2y}{dx^2}$; then $f'''$, $f^{(4)}$, and so on. Physically: if $s(t)$ is position, $s'$ is velocity and $s''$ is acceleration. Geometrically $f''$ measures how the slope itself is changing — the concavity story of Module 08. Each differentiation of a polynomial drops its degree by one, so a degree-$n$ polynomial has $f^{(n+1)} = 0$ identically.

08Rewriting before differentiating

The best students do algebra first so the calculus is easy. Three habitual rewrites: (1) roots and reciprocals to powers, $\frac{4}{\sqrt[3]{x}} = 4x^{-1/3}$; (2) split fractions with single-term denominators, $\frac{x^2 + 3x - 1}{x} = x + 3 - x^{-1}$, avoiding the quotient rule entirely; (3) expand cheap products, $(x+2)(x-5) = x^2 - 3x - 10$, when expansion is faster than the product rule. One minute of rewriting routinely saves five minutes of rule-stacking — and several sign errors.