Derivative of 1/x: Answer, Proof, and Mistakes

The proof: rewrite 1/x as a power

Start by writing the reciprocal as a power. Since $\frac{1}{x} = x^{-1}$, this is an ordinary power-rule derivative, not a separate rule to memorize.

The power rule says $\frac{d}{dx}\left[x^n\right] = n\,x^{n-1}$. With $n = -1$, the exponent $-1$ comes down as the coefficient out front, and the new exponent is $-1 - 1 = -2$. That leading $-1$ is the source of the minus sign in the answer.

The quotient rule gives the same result. Treating $\frac{1}{x}$ as a quotient with constant numerator $1$:

The derivative of the numerator is $0$, so only the $-1\cdot 1$ term survives, and the $x^2$ in the denominator comes from squaring the bottom.

The slope is negative everywhere, so 1/x is decreasing

For every $x \neq 0$, $x^2 > 0$, so $-\frac{1}{x^2}$ is negative. The slope of $\frac{1}{x}$ is never zero and never positive, which means it has no critical points and no turning points.

So $\frac{1}{x}$ is decreasing on each of its two branches, $(-\infty, 0)$ and $(0, \infty)$, taken separately. Avoid saying it is decreasing for all $x$: the domain is split by the vertical asymptote at $x = 0$, where the function is undefined, and the two branches are not one connected piece.

Every entry in the last column is negative. The slope grows steeper (more negative) as $x$ approaches $0$ and flattens toward $0$ as $|x|$ gets large.

Where 1/x shows up on the AP exam

On the AP Calculus AB and BC exams, $x^{-1}$ is a direct application of the power rule from Unit 2, Topic 2.5 (Applying the Power Rule). Unit 2 carries about 10-15% of the AB exam and 5-10% of the BC exam.

It shows up more often inside composites. Whenever you differentiate $\frac{1}{g(x)}$, the chain rule from Unit 3, Topic 3.1 (The Chain Rule) turns the outer $\frac{1}{u}$ into $-\frac{1}{u^2}$ and multiplies by $g'(x)$. Differentiating $\frac{1}{g(x)}$ is also a special case of the quotient rule (Topic 2.9), with a constant numerator, so you get the same $-\frac{1}{g(x)^2}$ factor either way.

Common mistakes

• Dropping the minus sign and writing $\frac{1}{x^2}$ instead of $-\frac{1}{x^2}$. The derivative is negative everywhere, so a positive answer is wrong on sight.
• Mixing up the two directions: $\frac{d}{dx}\left[\ln x\right] = \frac{1}{x}$, while $\frac{d}{dx}\left[\frac{1}{x}\right] = -\frac{1}{x^2}$. The reciprocal $\frac{1}{x}$ being the derivative of $\ln x$ does not make it the derivative of $\frac{1}{x}$.
• Botching the new exponent: from $x^{-1}$ you subtract $1$ to reach $x^{-2}$, not $x^{0}$ or $x^{-1}$. Subtracting from a negative exponent is where the slip happens.
• Assuming the derivative comes from squaring the function. The $x^2$ in the denominator comes from the power rule (or from squaring the denominator in the quotient rule), not from squaring $\frac{1}{x}$.
• Skipping the chain rule on composites: $\frac{d}{dx}\left[\frac{1}{3x+1}\right]$ is not $-\frac{1}{(3x+1)^2}$, because you still multiply by the inner derivative $3$.

Two chain-rule examples

Example 1. Differentiate $\frac{1}{3x+1}$. Let $u = 3x + 1$, so the outer function is $\frac{1}{u}$ with derivative $-\frac{1}{u^2}$, and the inner derivative is $u' = 3$.

Example 2. Differentiate $\frac{1}{x^2+1}$. Here $u = x^2 + 1$ and $u' = 2x$, so multiply the outer $-\frac{1}{u^2}$ by $2x$.

In both problems the outer step is the same $-\frac{1}{u^2}$ you get from $\frac{1}{x}$; the chain rule just appends the inner derivative.