Elementary Derivative Rules, calculus lesson

Going back to the limit definition for every derivative works, but it is slow and easy to get wrong. A handful of rules, each proved once from the definition, let you differentiate every polynomial in a single line. The only care needed is to know exactly where each rule applies.

Key terms

Constant rule

Definition If $f(x) = c$ for every real $x$ , where $c$ is a constant, then $f'(x) = 0$ for every $x$ . In Leibniz notation,

\frac{d}{dx}(c) = 0.

Proof. The difference quotient is $\frac{c - c}{h} = 0$ for every $h \neq 0$ , so its limit is $0$ .

In plain words A horizontal line has slope zero, and a quantity that never changes has rate of change zero.

Example Both of these derivatives are zero,

\frac{d}{dx}(7) = 0, \qquad \frac{d}{dx}\left(\pi^2\right) = 0.

The number $\pi^2$ looks complicated but it does not depend on $x$ .

Power rule and its real-domain restrictions

Definition The formula

\frac{d}{dx}\left(x^r\right) = r x^{r - 1}

holds in the following cases.

If $r = n$ is a positive integer, it holds for every real $x$ , reading $x^0$ as $1$ when $n = 1$ .
If $r = n$ is a negative integer, it holds for every $x \neq 0$ .
If $r = \frac{p}{q}$ is a rational number in lowest terms with $q > 0$ odd, then $x^r = \left(\sqrt[q]{x}\right)^p$ is defined for negative $x$ too, and the formula holds for every $x \neq 0$ . If moreover $r > 1$ , it also holds at $x = 0$ , where both sides equal $0$ .
For every real $r$ , with $x^r$ defined through exponentials and logarithms when $r$ is irrational, it holds for every $x > 0$ .

For a positive integer $n$ the proof uses the binomial theorem, which gives

(x + h)^n - x^n = n x^{n - 1} h + (\text{terms with } h^2 \text{ or higher}).

Dividing by $h$ and letting $h \to 0$ leaves $n x^{n - 1}$ . The other cases are proved later in the course, and we use them now.

In plain words Bring the exponent down in front and lower it by one. Then check that the input is allowed. The derivative can have a smaller domain than the function, especially at $x = 0$ and for negative $x$ .

Example The rule gives $\frac{d}{dx}\left(x^5\right) = 5x^4$ everywhere, and $\frac{d}{dx}\left(x^{-2}\right) = -2x^{-3}$ for $x \neq 0$ . For $x^{2/3}$ , defined for all real $x$ , the derivative is $\frac{2}{3}x^{-1/3}$ for $x \neq 0$ , and at $x = -8$ this gives $\frac{2}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{3}$ . At $x = 0$ the formula breaks down, which matches the cusp of $x^{2/3}$ at the origin. Finally $x^{1/2} = \sqrt{x}$ is defined on $[0, \infty)$ , but its derivative $\frac{1}{2}x^{-1/2}$ exists only for $x > 0$ .

Constant multiple rule

Definition If $f$ is differentiable at $x$ and $c$ is a constant, then $c f$ is differentiable at $x$ and

\frac{d}{dx}\left(c f(x)\right) = c f'(x).

Proof. As $h \to 0$ ,

\frac{c f(x + h) - c f(x)}{h} = c \cdot \frac{f(x + h) - f(x)}{h} \to c f'(x).

In plain words Stretching a graph vertically by a factor $c$ multiplies every slope by $c$ .

Example For a cubic term the rule gives $\frac{d}{dx}\left(4x^3\right) = 4 \cdot 3x^2 = 12x^2$ . For $x > 0$ , writing $\frac{5}{\sqrt{x}} = 5x^{-1/2}$ gives

\frac{d}{dx}\left(\frac{5}{\sqrt{x}}\right) = 5 \cdot \left(-\frac{1}{2}\right) x^{-3/2} = -\frac{5}{2}x^{-3/2}.

Sum and difference rules

Definition If $f$ and $g$ are both differentiable at $x$ , then so are $f + g$ and $f - g$ , and

\frac{d}{dx}\left(f(x) \pm g(x)\right) = f'(x) \pm g'(x).

Proof. The difference quotient of $f + g$ is the sum of the difference quotients of $f$ and $g$ , and the limit of a sum is the sum of the limits. The difference works the same way. By induction the rule extends to any finite sum.

In plain words Differentiate term by term. The rate of change of a total is the total of the rates.

Example Term by term,

\frac{d}{dx}\left(x^4 + \sqrt{x}\right) = 4x^3 + \frac{1}{2\sqrt{x}} \quad \text{for } x > 0,

and

\frac{d}{dx}\left(x^3 - x^{-1}\right) = 3x^2 + x^{-2} \quad \text{for } x \neq 0.

Derivatives of polynomials

Definition Every polynomial

p(x) = a_n x^n + a_{n - 1} x^{n - 1} + \dots + a_1 x + a_0

is differentiable at every real $x$ , and

p'(x) = n a_n x^{n - 1} + (n - 1) a_{n - 1} x^{n - 2} + \dots + a_1.

This follows from the power, constant multiple, sum and constant rules. If $a_n \neq 0$ and $n \ge 1$ , so that $p$ has degree $n$ , then $p'$ has degree $n - 1$ .

In plain words Handle each term on its own. The constant term disappears.

Example Let $p(x) = 4x^3 - 5x^2 + 7x - 2$ . Then $p'(x) = 12x^2 - 10x + 7$ . At $x = 1$ we get $p(1) = 4$ and $p'(1) = 9$ , so the tangent line there is $y = 4 + 9(x - 1)$ , or $y = 9x - 5$ .

Derivative of the absolute value away from zero

Definition The absolute value is differentiable at every $x \neq 0$ , with

\frac{d}{dx} \lvert x \rvert = \frac{x}{\lvert x \rvert} = \begin{cases} 1, & x > 0, \\ -1, & x < 0, \end{cases}

and it is not differentiable at $x = 0$ . More generally, if $f$ is differentiable at $a$ and $f(a) \neq 0$ , then $\lvert f \rvert$ is differentiable at $a$ with

\lvert f \rvert'(a) = \frac{f(a)}{\lvert f(a) \rvert} f'(a).

The reason is that $f$ is continuous at $a$ , so it keeps the sign of $f(a)$ on some open interval around $a$ , and there $\lvert f \rvert$ equals $f$ or $-f$ . Where $f(a) = 0$ the rule says nothing. Then $\lvert f \rvert$ may fail to be differentiable, as $\lvert x \rvert$ does at $0$ , or it may be differentiable, as $\lvert x^3 \rvert$ is at $0$ with derivative $0$ .

In plain words Away from the points where the inside is zero, the absolute value just leaves the function alone or flips its sign, and the derivative follows along.

Example Take $g(x) = \lvert x^2 - 4 \rvert$ at $x = 1$ . There $x^2 - 4 = -3 < 0$ , so near $x = 1$ we have $g(x) = 4 - x^2$ and $g'(1) = -2 \cdot 1 = -2$ . At $x = 2$ , where $x^2 - 4 = 0$ , the rule does not apply, and in fact $g$ has a corner there.

Check yourself

Differentiate $f(x) = 3x^4 - \frac{2}{x} + 6$ and state where your answer is valid.

Show the answer

Rewrite the function as $f(x) = 3x^4 - 2x^{-1} + 6$ . Then $f'(x) = 12x^3 + 2x^{-2}$ , valid for every $x \neq 0$ .

Find the equation of the tangent line to $y = x^3 - 3x + 1$ at $x = 2$ .

Show the answer

The derivative is $y' = 3x^2 - 3$ , so the slope at $x = 2$ is $9$ . The point is $(2, 3)$ , and the tangent line is $y = 3 + 9(x - 2)$ , or $y = 9x - 15$ .

Where is $f(x) = \sqrt[3]{x}$ differentiable, and what is $f'(-1)$ ?

Show the answer

The power rule gives $f'(x) = \frac{1}{3}x^{-2/3}$ for every $x \neq 0$ . At $0$ the graph has a vertical tangent and no derivative. Since $(-1)^{-2/3} = 1$ , the value is $f'(-1) = \frac{1}{3}$ .

Elementary Derivative Rules

Theory, part 43

Practice, part 44

Key terms

Constant rule

Power rule and its real-domain restrictions

Constant multiple rule

Sum and difference rules

Derivatives of polynomials

Derivative of the absolute value away from zero

Check yourself

How this lesson connects

Builds on

Used later in