Basic Differentiation Rules

Differentiating from the definition works, but evaluating $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ by hand for every new function is hopeless. This topic builds the algebra of derivatives: a short list of rules, each proved from the definition, that lets you differentiate any polynomial, any rational function, and any product or quotient of functions whose derivatives you already know. Everything that comes later in the course — the chain rule, implicit differentiation, related rates, curve sketching, optimisation — assumes you can apply these rules quickly and without algebra errors.

Standing assumptions

Unless stated otherwise, $f$ and $g$ are real functions and $a$ is an interior point of their domains. "$f$ is differentiable at $a$" means the limit $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ exists as a finite real number. Recall two facts used repeatedly below: differentiability at $a$ implies continuity at $a$, and a derivative statement is always local — it says nothing at points outside the domain.

Linearity of differentiation

Theorem (constant multiple and sum rules). If $f$ and $g$ are differentiable at $a$ and $c\in\mathbb{R}$ is a constant, then $cf$ and $f+g$ are differentiable at $a$ and $(cf)'(a)=c\,f'(a),\qquad (f+g)'(a)=f'(a)+g'(a).$

Derivation sketch. $\dfrac{cf(a+h)-cf(a)}{h}=c\cdot\dfrac{f(a+h)-f(a)}{h}$ and $\dfrac{(f+g)(a+h)-(f+g)(a)}{h}=\dfrac{f(a+h)-f(a)}{h}+\dfrac{g(a+h)-g(a)}{h}$; now apply the limit laws for constant multiples and sums, which are legitimate exactly because both limits are assumed to exist.

Combining the two and using $f-g=f+(-1)g$ gives the general form: for constants $c_1,\dots,c_n$ and functions differentiable at $a$, $\left(\sum_{k=1}^{n}c_k f_k\right)'(a)=\sum_{k=1}^{n}c_k f_k'(a).$

Caution. Linearity covers sums and constant multiples only. There is no rule $(fg)'=f'g'$ and no rule $(f/g)'=f'/g'$; both are false (see Common mistakes).

The power rule

Theorem (power rule). For a fixed exponent $r$, $\dfrac{d}{dx}x^{r}=r\,x^{r-1}$, at every $x$ for which both $x^{r}$ and $x^{r-1}$ make sense. The precise statement depends on the kind of exponent:

The third column lists where the formula is guaranteed; the parenthetical remarks record the extra points that are easy to lose. For example $\frac{d}{dx}x^{5/3}=\frac53x^{2/3}$ is valid at $x=0$ as well (both sides are $0$ there), whereas $\frac{d}{dx}x^{2/3}=\frac23x^{-1/3}$ genuinely fails at $x=0$, where the graph has a cusp.

Derivation sketch for a positive integer $n$. Using $b^n-a^n=(b-a)\left(b^{n-1}+b^{n-2}a+\cdots+a^{n-1}\right)$ with $b=x+h$, $a=x$, $\frac{(x+h)^n-x^n}{h}=(x+h)^{n-1}+(x+h)^{n-2}x+\cdots+x^{n-1}\ \xrightarrow[h\to 0]{}\ \underbrace{x^{n-1}+\cdots+x^{n-1}}_{n\ \text{terms}}=n\,x^{n-1}.$

Edge cases you must name.

• $\sqrt{x}=x^{1/2}$ has domain $[0,\infty)$ but $\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}$ only for $x>0$. At $x=0$ the one-sided difference quotient is $\frac{\sqrt{h}}{h}=\frac{1}{\sqrt h}\to+\infty$, so $\sqrt{x}$ is not differentiable at $0$.
• $x^{1/3}$ is defined for all real $x$, but $\frac{d}{dx}x^{1/3}=\frac13 x^{-2/3}$ requires $x\ne 0$; at $x=0$ the graph has a vertical tangent and no derivative.
• The power rule needs a variable base and a constant exponent. It does not apply to $2^{x}$, and $\frac{d}{dx}2^{x}\ne x\,2^{x-1}$.

Useful special cases: $\dfrac{d}{dx}\left(\dfrac1x\right)=-\dfrac{1}{x^2}$, $\dfrac{d}{dx}\left(\dfrac{1}{x^n}\right)=-\dfrac{n}{x^{n+1}}$, $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt x}$, $\dfrac{d}{dx}\left(\dfrac{1}{\sqrt x}\right)=-\dfrac{1}{2x^{3/2}}$.

The product rule

Theorem. If $f$ and $g$ are differentiable at $a$, then $fg$ is differentiable at $a$ and $(fg)'(a)=f'(a)\,g(a)+f(a)\,g'(a).$

Derivation sketch. Add and subtract $f(a)g(a+h)$ in the numerator:

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

Let $h\to0$. The two difference quotients tend to $f'(a)$ and $g'(a)$ by hypothesis, and $g(a+h)\to g(a)$ because $g$, being differentiable at $a$, is continuous at $a$. That continuity step is exactly where the hypothesis on $g$ is used.

For three factors, apply the rule twice: $(fgh)'=f'gh+fg'h+fgh'$ — differentiate one factor at a time and leave the others alone.

The quotient rule

Theorem. If $f$ and $g$ are differentiable at $a$ and $g(a)\ne 0$, then $f/g$ is differentiable at $a$ and $\left(\frac{f}{g}\right)'(a)=\frac{f'(a)\,g(a)-f(a)\,g'(a)}{\big(g(a)\big)^{2}}.$ The hypothesis $g(a)\ne0$ is essential: it also guarantees, by continuity of $g$, that $g\ne0$ on a whole interval around $a$, so the quotient is defined near $a$.

Derivation sketch.

$$\begin{aligned} \frac1h\left(\frac{f(a+h)}{g(a+h)}-\frac{f(a)}{g(a)}\right)&=\frac{f(a+h)g(a)-f(a)g(a+h)}{h\,g(a+h)g(a)}\\ &=\frac{g(a)\big[f(a+h)-f(a)\big]-f(a)\big[g(a+h)-g(a)\big]}{h\,g(a+h)g(a)}\\ &=\frac{\dfrac{f(a+h)-f(a)}{h}\,g(a)-f(a)\,\dfrac{g(a+h)-g(a)}{h}}{g(a+h)g(a)}, \end{aligned}$$

where the second line came from adding and subtracting $f(a)g(a)$ in the numerator. Letting $h\to0$ and using $g(a+h)\to g(a)\ne0$ gives the formula.

Taking $f\equiv 1$ gives the reciprocal rule: $\left(\dfrac{1}{g}\right)'=-\dfrac{g'}{g^{2}}$ wherever $g\ne0$.

Order matters in the numerator: it is $(\text{derivative of top})(\text{bottom})-(\text{top})(\text{derivative of bottom})$. Swapping the two terms flips every sign.

Polynomials and rational functions

A polynomial $P(x)=c_nx^n+\cdots+c_1x+c_0$ is differentiable on all of $\mathbb{R}$, with $P'(x)=nc_nx^{n-1}+(n-1)c_{n-1}x^{n-2}+\cdots+c_1,$ a polynomial of degree $n-1$ when $n\ge1$ and $c_n\ne0$. A rational function $R=P/Q$ is differentiable at every point of its domain (every $x$ with $Q(x)\ne0$), and $R'=\frac{P'Q-PQ'}{Q^{2}}$ is again a rational function whose denominator is $Q^2$. Useful check on degrees: if $\deg P=p\ge1$ and $\deg Q=q\ge1$, then $P'Q$ and $PQ'$ each have degree $p+q-1$, so $P'Q-PQ'$ has degree at most $p+q-1$. Writing $P=ax^{p}+\cdots$ and $Q=bx^{q}+\cdots$ with $ab\ne0$, the coefficient of $x^{p+q-1}$ in $P'Q-PQ'$ is $pab-qab=(p-q)ab$, so the leading terms cancel exactly when $p=q$, not in general. For instance $P=x^{2}$, $Q=x$ give $P'Q-PQ'=2x\cdot x-x^{2}\cdot 1=x^{2}$, of the full degree $p+q-1=2$; while $P=x^{2}+1$, $Q=x^{2}$ give $P'Q-PQ'=-2x$, of degree $1<3$.

Worked example 1 — linearity plus the power rule with negative and fractional exponents

Differentiate $f(x)=5x^{4}-\dfrac{3}{x^{2}}+7\sqrt{x}-9$.

Step 1 — rewrite every term as a power of $x$. $f(x)=5x^{4}-3x^{-2}+7x^{1/2}-9.$ The natural domain is $x>0$ (the $\sqrt x$ needs $x\ge0$, the $x^{-2}$ needs $x\ne0$).

Step 2 — differentiate term by term (linearity), each term by the power rule.

$$\begin{aligned} \frac{d}{dx}\big(5x^4\big)&=5\cdot 4x^{3}=20x^{3},\\ \frac{d}{dx}\big(-3x^{-2}\big)&=-3\cdot(-2)x^{-3}=6x^{-3},\\ \frac{d}{dx}\big(7x^{1/2}\big)&=7\cdot\tfrac12 x^{-1/2}=\tfrac72 x^{-1/2},\\ \frac{d}{dx}(-9)&=0. \end{aligned}$$

Step 3 — assemble and tidy. $f'(x)=20x^{3}+\frac{6}{x^{3}}+\frac{7}{2\sqrt{x}},\qquad x>0.$ Note the sign: differentiating $-3x^{-2}$ produced a $+$, because $-3\times(-2)=+6$.

Worked example 2 — the quotient rule, in full

Differentiate $F(x)=\dfrac{2x^{3}-x}{x^{2}+1}$.

Step 1 — name the pieces. $u=2x^{3}-x$, $v=x^{2}+1$, so $u'=6x^{2}-1$ and $v'=2x$. Since $x^2+1\ge1>0$ for all real $x$, the hypothesis $v\ne0$ holds everywhere.

Step 2 — apply the rule. $F'(x)=\frac{u'v-uv'}{v^{2}}=\frac{(6x^{2}-1)(x^{2}+1)-(2x^{3}-x)(2x)}{(x^{2}+1)^{2}}.$

Step 3 — expand the numerator carefully.

$$\begin{aligned} (6x^{2}-1)(x^{2}+1)&=6x^{4}+6x^{2}-x^{2}-1=6x^{4}+5x^{2}-1,\\ (2x^{3}-x)(2x)&=4x^{4}-2x^{2},\\ \text{numerator}&=\big(6x^{4}+5x^{2}-1\big)-\big(4x^{4}-2x^{2}\big)=2x^{4}+7x^{2}-1. \end{aligned}$$

The minus sign multiplies the whole second product — that is where most sign errors happen.

Step 4 — state the answer and sanity-check it. $F'(x)=\frac{2x^{4}+7x^{2}-1}{(x^{2}+1)^{2}}\quad\text{for all }x\in\mathbb{R}.$ Check at $x=1$: $F'(1)=\frac{2+7-1}{4}=2$. A numerical slope check gives $\frac{F(1.001)-F(0.999)}{0.002}=1.9999992\ldots$, which rounds to $2.00000$ and so agrees with the computed value $F'(1)=2$.

Simplify before differentiating

The quotient rule is never wrong, but it is often the slowest route. Rewrite first whenever the denominator is a single power of $x$, or when the fraction cancels.

Two warnings. First, cancelling changes the domain: $\frac{x^2-9}{x-3}$ is undefined at $x=3$, so its derivative is $1$ on $x \ne 3$ and undefined at $x=3$. Second, "simplify first" is a tactic, not a rule — when the denominator has two or more terms, the quotient rule is usually the right tool.

Common mistakes

Fast self-checks

• Degree check. Differentiating a polynomial of degree $n\ge1$ must lower the degree by exactly one. (For $n=0$ the derivative is the zero polynomial, whose degree is left undefined, so the check does not apply.)
• Numeric check. Compute $\frac{f(a+0.001)-f(a-0.001)}{0.002}$ and compare with $f'(a)$; they should agree to about three decimals.
• Zero check. For $f'=0$ from a quotient, only the numerator can vanish — solutions must still lie in the domain of $f$.