The Power Rule and Basic Derivative Rules

Finding derivatives with the limit definition works, but it’s slow. Luckily, patterns appear quickly, and they become rules that let you differentiate in one line. The power rule, together with the rules for constants, sums, and constant multiples, lets you differentiate every polynomial, plus anything with roots or powers of $x$ in the denominator.

Key ideas

The constant rule

The graph of $y = c$ is a horizontal line, so its slope is always $0$:

$$\frac{d}{dx}[c] = 0$$

The power rule

For any real number $n$:

$$\frac{d}{dx}\big[x^n\big] = nx^{n - 1}$$

Bring the exponent down in front, then subtract $1$ from the exponent. For example, $\dfrac{d}{dx}\big[x^5\big] = 5x^4$ and $\dfrac{d}{dx}[x] = 1$.

Why it works: the definition gives $\dfrac{d}{dx}\big[x^2\big] = 2x$ and $\dfrac{d}{dx}\big[x^3\big] = 3x^2$. In $(x + h)^n = x^n + nx^{n-1}h + (\text{terms with } h^2 \text{ or higher})$, only the $nx^{n-1}h$ term survives after dividing by $h$ and letting $h \to 0$. That argument works for positive whole numbers $n$; the rule is true for every real $n$, but proving it needs tools from later topics.

Rewrite roots and fractions as powers first

The power rule works for negative and fractional exponents too, but only once the function is written as a power of $x$:

Original	Rewritten	Derivative
$\sqrt{x}$	$x^{1/2}$	$\tfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$
$\dfrac{1}{x}$	$x^{-1}$	$-x^{-2} = -\dfrac{1}{x^2}$
$\dfrac{1}{x^3}$	$x^{-3}$	$-3x^{-4} = -\dfrac{3}{x^4}$
$\sqrt[3]{x^2}$	$x^{2/3}$	$\tfrac{2}{3}x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}$

Constant multiple, sum, and difference rules

Constants come along for the ride, and you can differentiate term by term:

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

There is no such rule for products or quotients. The derivative of a product is not the product of the derivatives. For those, either expand or divide first, or use the product rule and quotient rule.

A function and its derivative

The derivative is a function in its own right. Where $f$ is rising, $f'$ is positive; where $f$ is falling, $f'$ is negative; and where $f$ has a horizontal tangent, $f'$ is $0$.

$f(x) = x^3 - 3x$ (top) and its derivative $f'(x) = 3x^2 - 3$ (bottom). The horizontal tangents of $f$ line up with the zeros of $f'$.

Worked examples

Example 1: A polynomial

Differentiate $y = 4x^5 - 3x^2 + 7x - 9$.

Solution. Go term by term:

$$\frac{dy}{dx} = 4(5x^4) - 3(2x) + 7(1) - 0 = 20x^4 - 6x + 7$$

Example 2: Roots and fractions

Find $f'(x)$ for $f(x) = 6\sqrt{x} - \dfrac{2}{x^3} + \dfrac{5}{x}$.

Solution. Rewrite every term as a power of $x$:

$$f(x) = 6x^{1/2} - 2x^{-3} + 5x^{-1}$$

Now use the power rule. Watch the negative exponents: $-3 - 1 = -4$ and $-1 - 1 = -2$.

$$\begin{aligned} f'(x) &= 6 \cdot \tfrac{1}{2}x^{-1/2} - 2(-3)x^{-4} + 5(-1)x^{-2} \\ &= 3x^{-1/2} + 6x^{-4} - 5x^{-2} \\ &= \frac{3}{\sqrt{x}} + \frac{6}{x^4} - \frac{5}{x^2} \end{aligned}$$

Example 3: Divide first

Find $g'(x)$ for $g(x) = \dfrac{2x^2 - 5x + 3}{x}$.

Solution. The denominator is a single term, so split the fraction:

$$g(x) = \frac{2x^2}{x} - \frac{5x}{x} + \frac{3}{x} = 2x - 5 + 3x^{-1}$$ $$g'(x) = 2 - 3x^{-2} = 2 - \frac{3}{x^2}$$

Example 4: Horizontal tangents and a tangent line

Let $f(x) = x^3 - 3x$ (the function in the figure).

• (a) Find the points where the tangent line is horizontal.
• (b) Find the tangent line at $x = 2$.

Solution. $f'(x) = 3x^2 - 3$.

(a) Horizontal means slope $0$:

$$3x^2 - 3 = 0 \quad\Rightarrow\quad x^2 = 1 \quad\Rightarrow\quad x = \pm 1$$

$f(1) = 1 - 3 = -2$ and $f(-1) = -1 + 3 = 2$. The points are $(1, -2)$ and $(-1, 2)$.

(b) $f(2) = 8 - 6 = 2$ and $f'(2) = 12 - 3 = 9$:

$$y - 2 = 9(x - 2) \quad\text{or}\quad y = 9x - 16$$

Common mistakes

Thinking the derivative of a constant is the constant. $\dfrac{d}{dx}[7] = 0$, and so is $\dfrac{d}{dx}\big[\pi^2\big]$ or $\dfrac{d}{dx}\big[e^3\big]$. A constant doesn’t change.

Getting negative exponents wrong. Subtracting $1$ makes a negative exponent more negative: $\dfrac{d}{dx}\big[x^{-3}\big] = -3x^{-4}$, not $-3x^{-2}$.

Using the power rule before rewriting. $\dfrac{1}{x^3}$ is not $x^3$ with something on top. Rewrite it as $x^{-3}$ first. Similarly, $\dfrac{5}{2x^4} = \tfrac{5}{2}x^{-4}$ (the $2$ stays in the denominator as a constant).

Differentiating a product factor by factor. $\dfrac{d}{dx}\big[(x^2 + 3)(2x - 1)\big]$ is not $(2x)(2)$. Expand first, or use the product rule.

Using the power rule on an exponential. The power rule is for $x^n$ (variable base, constant exponent). It does not apply to $2^x$ or $e^x$, which have a variable exponent. See derivatives of trig, exponential and log functions.

Practice

1. (Warm-up) Differentiate each.

• (a) $x^7$
• (b) $10$
• (c) $x^{-2}$

Solution

(a) $7x^6$

(b) $0$

(c) $-2x^{-3} = -\dfrac{2}{x^3}$

2. (Warm-up) Find $\dfrac{dy}{dx}$ for $y = 3x^4 - 2x + 8$.

Solution

$$\frac{dy}{dx} = 12x^3 - 2$$

3. (Warm-up) Find $f'(x)$ for $f(x) = \sqrt[3]{x}$.

Solution

$f(x) = x^{1/3}$, so

$$f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}}$$

4. (Core) Find $f'(x)$ for $f(x) = \dfrac{4}{x^2} + x\sqrt{x}$.

Solution

Rewrite: $f(x) = 4x^{-2} + x^{3/2}$ (since $x \cdot x^{1/2} = x^{3/2}$).

$$f'(x) = -8x^{-3} + \frac{3}{2}x^{1/2} = -\frac{8}{x^3} + \frac{3}{2}\sqrt{x}$$

5. (Core) Find $g'(t)$ for $g(t) = \dfrac{t^3 - 2t + 5}{t^2}$.

Solution

Divide each term by $t^2$: $g(t) = t - 2t^{-1} + 5t^{-2}$.

$$g'(t) = 1 + 2t^{-2} - 10t^{-3} = 1 + \frac{2}{t^2} - \frac{10}{t^3}$$

6. (Core) Find the equation of the tangent line to $y = 2\sqrt{x} + \dfrac{1}{x}$ at $x = 4$.

Solution

Point: $y(4) = 2(2) + \dfrac{1}{4} = \dfrac{17}{4}$.

Slope: $y = 2x^{1/2} + x^{-1}$, so $y' = x^{-1/2} - x^{-2}$. At $x = 4$: $y' = \dfrac{1}{2} - \dfrac{1}{16} = \dfrac{7}{16}$.

$$y - \frac{17}{4} = \frac{7}{16}(x - 4) \quad\text{or}\quad y = \frac{7}{16}x + \frac{5}{2}$$

7. (Core) Find all points on the graph of $f(x) = x^3 - 6x^2 + 9x + 1$ where the tangent line is horizontal.

Solution

$$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)$$

$f'(x) = 0$ at $x = 1$ and $x = 3$. Then $f(1) = 1 - 6 + 9 + 1 = 5$ and $f(3) = 27 - 54 + 27 + 1 = 1$.

The points are $(1, 5)$ and $(3, 1)$.

8. (Challenge) Find $a$ and $b$ so that the line $y = 5x - 3$ is tangent to the graph of $f(x) = ax^2 + b$ at $x = 1$.

Solution

Tangent at $x = 1$ means the slopes match and the graphs share the point.

Slope: $f'(x) = 2ax$, so $f'(1) = 2a = 5$, giving $a = \dfrac{5}{2}$.

Point: the line passes through $(1, 5 - 3) = (1, 2)$, so $f(1) = a + b = 2$, giving $b = 2 - \dfrac{5}{2} = -\dfrac{1}{2}$.

9. (Challenge) Show that every tangent line to $y = \dfrac{1}{x}$ (for $x \gt 0$) forms a triangle with the coordinate axes that has area $2$.

Solution

Take the point $\left(a, \dfrac{1}{a}\right)$ with $a \gt 0$. Since $y' = -x^{-2}$, the slope there is $-\dfrac{1}{a^2}$:

$$y - \frac{1}{a} = -\frac{1}{a^2}(x - a) \quad\Rightarrow\quad y = -\frac{x}{a^2} + \frac{2}{a}$$

The $y$-intercept is $\dfrac{2}{a}$. Setting $y = 0$ gives $x = 2a$, the $x$-intercept. The triangle’s area is

$$\frac{1}{2}(2a)\left(\frac{2}{a}\right) = 2$$

The $a$ cancels, so the area is $2$ for every tangent line.