3.3 Differentiation Rules

Big idea

Computing every derivative from the limit definition is correct and unsustainable. The definition is the foundation, but running it on a polynomial of degree six means expanding a binomial to the sixth power for the sake of a formula you could have written down in ten seconds. This section replaces the limit with a small set of rules, each proved once from the definition and then used forever.

The rules are organized by how the function was built. A function assembled by adding pieces differentiates piece by piece. A function assembled by multiplying does not: the derivative of a product is not the product of the derivatives, and the correct statement has two terms because when both factors move, the output changes for two reasons at once. Division is handled by a companion rule with a similar shape and a minus sign that matters.

Underneath everything is the power rule, which handles $x^n$ in one stroke. Proved first for positive integers by expanding $(x+h)^n$, it turns out to hold for negative and fractional exponents too, which is why roots and reciprocals need no special treatment once you rewrite them as powers. The rewriting step is worth building as a habit: $\sqrt{x}$ becomes $x^{1/2}$ and $1/x^4$ becomes $x^{-4}$ before any rule is applied.

With these in hand, differentiating an algebraic expression becomes structural rather than computational. You look at the outermost operation - is this a sum, a product, a quotient? - pick the matching rule, and recurse on the pieces. That habit is what makes the chain rule, two sections later, feel like one more entry in the same list rather than a new subject.

Decoder

The derivative of a product is the first factor times the derivative of the second, plus the second factor times the derivative of the first; the derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the denominator squared.

Both statements are lists of instructions, and in both the striking feature is that the answer has two terms where the original had one. The reason is that a product changes for two independent reasons. Think of a rectangle whose width and height both grow: the new area exceeds the old by a strip along the top, a strip along the side, and a tiny corner square. The two strips are the two terms of the product rule; the corner is second order in the increment and vanishes in the limit.

The quotient rule is not symmetric, and its asymmetry is where the errors live. The numerator’s derivative comes first and carries the plus sign; the denominator’s derivative comes second and carries the minus. Swap them and you get the negative of the right answer, which is easy to miss because it looks structurally correct. One quick audit: apply the rule to $x/x^2$ and confirm you get $-1/x^2$, the derivative of $1/x$.

The classic mistake with both rules is to expect them to distribute the way the sum rule does. Only addition and subtraction pass through differentiation untouched, along with multiplication by a constant. Multiplication of two varying quantities never does.

Definitions and results

Constant rule. If $f(x) = c$ for all $x$, then $f'(x) = 0$. A horizontal graph has zero slope everywhere.

Power rule. For any real exponent $n$,

$$ \frac{d}{dx}x^{n} = n x^{\,n-1} $$

In particular $\frac{d}{dx}x = 1$, $\frac{d}{dx}x^{-1} = -x^{-2}$, and $\frac{d}{dx}x^{1/2} = \frac{1}{2}x^{-1/2}$.

Constant multiple rule. $\frac{d}{dx}\big[c f(x)\big] = c f'(x)$. Constants ride along outside the derivative.

Sum and difference rules. $\frac{d}{dx}\big[f(x) \pm g(x)\big] = f'(x) \pm g'(x)$. Together with the previous two rules this differentiates any polynomial term by term.

Product rule. If $f$ and $g$ are differentiable at $x$, then

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

Quotient rule. If $f$ and $g$ are differentiable at $x$ and $g(x) \ne 0$, then

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

Three factors. Applying the product rule twice gives a pattern that generalizes: the derivative of $fgh$ is $f'gh + fg'h + fgh'$, one term per factor, each with exactly one factor differentiated.

Rewrite before differentiating. A quotient with a single term in the denominator, such as $\frac{3x^2 + 1}{x}$, is faster split into $3x + x^{-1}$ and handled with the power rule. Use the quotient rule when the denominator genuinely has several terms.

Tangent lines from the rules. With the derivative in hand as a formula, horizontal tangents are the solutions of $f'(x) = 0$, and the tangent line at any point is still $y = f(a) + f'(a)(x-a)$.

Worked examples

A polynomial and a rewrite

Differentiate $f(x) = 5x^4 - 3x^2 + 7x - 2$. The sum, constant multiple, constant, and power rules do all of it term by term:

$$ f'(x) = 20x^3 - 6x + 7 $$

Now $g(x) = \dfrac{4}{x^3} + \sqrt{x}$. Rewrite first as $g(x) = 4x^{-3} + x^{1/2}$, then apply the power rule to each piece:

$$ g'(x) = -12x^{-4} + \tfrac{1}{2}x^{-1/2} = -\frac{12}{x^4} + \frac{1}{2\sqrt{x}} $$

A product, checked by expanding

Differentiate $y = (2x^2 + 1)(x^3 - 4x)$. Taking $f = 2x^2+1$ and $g = x^3 - 4x$, so $f' = 4x$ and $g' = 3x^2 - 4$:

$$ \frac{dy}{dx} = 4x(x^3 - 4x) + (2x^2+1)(3x^2-4) = 4x^4 - 16x^2 + 6x^4 - 8x^2 + 3x^2 - 4 $$

Collecting terms gives $10x^4 - 21x^2 - 4$. Because both factors are polynomials, this one can be checked independently: multiplying out first gives $y = 2x^5 - 7x^3 - 4x$, whose term-by-term derivative is $10x^4 - 21x^2 - 4$. The two routes agree, which is the reassurance worth having the first few times you use the rule.

A genuine quotient

Differentiate $y = \dfrac{3x - 1}{x^2 + 2}$. The denominator has two terms, so the quotient rule earns its place. With $f = 3x-1$, $f' = 3$, $g = x^2+2$, $g' = 2x$:

$$ \frac{dy}{dx} = \frac{3(x^2+2) - (3x-1)(2x)}{(x^2+2)^2} = \frac{3x^2 + 6 - 6x^2 + 2x}{(x^2+2)^2} = \frac{-3x^2 + 2x + 6}{(x^2+2)^2} $$

At $x = 0$ the formula gives $6/4 = 3/2$. A numerical check supports it: $y(0) = -1/2$ and $y(0.01) \approx -0.48498$, and the difference over $0.01$ is about $1.50$.

Locating horizontal tangents

Where is the tangent to $y = \dfrac{x^2}{x+1}$ horizontal? Differentiate with the quotient rule:

$$ \frac{dy}{dx} = \frac{2x(x+1) - x^2(1)}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2} = \frac{x(x+2)}{(x+1)^2} $$

A fraction is zero exactly when its numerator is zero and its denominator is not, so the horizontal tangents occur at $x = 0$ and $x = -2$. The points are $(0, 0)$ and $(-2, -4)$, and $x = -1$ is excluded from the domain rather than being a third solution. The derivative is positive for $x > 0$ and for $x < -2$, and negative between them, which matches a curve that falls from $(-2,-4)$ toward the asymptote and climbs again after the origin.

Practice

Start with the rules that apply term by term: constants, powers, constant multiples, sums and differences. Rewrite roots and reciprocals as powers before you begin.

Next, products. Identify the two factors, differentiate each separately, and assemble the two terms before simplifying.

Last, quotients. Watch the order in the numerator and remember to square the denominator, and check whether the problem is faster as a rewrite.

Quiz

Six items across the power and sum rules, the product rule, the quotient rule, and tangent lines built from them.