AP Calculus AB and BC

Derivative of 1/sqrt(x): Answer, Proof, Mistakes

Q: How do you write 1/sqrt(x) as a power of x?

\frac{1}{\sqrt{x}} = x^{-1/2}. The square root is the exponent \frac{1}{2}, and the denominator flips its sign to -\frac{1}{2}. That form is what lets the power rule apply.

Q: Why is the domain x > 0?

\sqrt{x} needs x \ge 0, and dividing by it rules out x = 0. So \frac{1}{\sqrt{x}} exists only for x > 0, and its derivative -\frac{1}{2x\sqrt{x}} is stated on the same set.

Q: Can you get this derivative with the quotient rule?

Yes. With numerator 1 and denominator \sqrt{x}, the quotient rule simplifies to -\frac{1}{2}x^{-3/2}, the same answer. The power-rule rewrite is shorter and less error prone.

The derivative of 1 over the square root of x is negative 1 divided by 2x times the square root of x, valid for x > 0. Rewriting 1/sqrt(x) as x^(-1/2) and applying the power rule gives -(1/2)x^(-3/2), which is the same as -1/(2x sqrt(x)). The derivative is negative everywhere the function is defined.

\frac{d}{dx}\left[\frac{1}{\sqrt{x}}\right] = -\frac{1}{2x\sqrt{x}}

How to differentiate 1/sqrt(x)

The trick is to stop reading it as a fraction and rewrite it as a single power of $x$ . A square root in the denominator is a negative, fractional exponent.

\frac{1}{\sqrt{x}} = x^{-1/2}

Now the power rule applies directly: multiply by the exponent $-\frac{1}{2}$ and subtract one from it.

\frac{d}{dx}x^{-1/2} = -\frac{1}{2}x^{-3/2}

Convert back to root form. The exponent $-\frac{3}{2}$ means one factor of $x$ and one factor of $\sqrt{x}$ in the denominator.

-\frac{1}{2}x^{-3/2} = -\frac{1}{2x\sqrt{x}}

Mind the domain

$\frac{1}{\sqrt{x}}$ is defined only for $x > 0$ , so the derivative is stated there too. The function has no values for $x \le 0$ , and the graph shoots up toward a vertical asymptote as $x \to 0^{+}$ .

Why the derivative is always negative

As $x$ grows, $\frac{1}{\sqrt{x}}$ shrinks, so the function decreases on its whole domain and the derivative must be negative. The formula confirms it: $x$ and $\sqrt{x}$ are both positive for $x > 0$ , so $-\frac{1}{2x\sqrt{x}}$ is negative.

The reverse operation shows up in Unit 6. Antidifferentiating $x^{-1/2}$ gives $2\sqrt{x} + C$ , the same power rule run backward by adding one to the exponent instead of subtracting.

Where this derivative shows up on the AP exam

Rewriting radicals and reciprocals as powers before differentiating is Topic 2.5 (Applying the Power Rule), on both AB and BC. Unit 2 is worth 10 to 15 percent of the AB exam and 5 to 10 percent of BC.

Expect $\frac{1}{\sqrt{x}}$ inside a chain rule, such as $\frac{d}{dx}\frac{1}{\sqrt{x^{2}+1}}$ , where the same negative-fractional-exponent setup runs one layer deeper.

Common mistakes with the derivative of 1/sqrt(x)

Forgetting the negative sign. The exponent $-\frac{1}{2}$ carries a minus that must survive to the answer.
Using the exponent $+\frac{1}{2}$ . The square root sits in the denominator, so the power is $-\frac{1}{2}$ , not $+\frac{1}{2}$ .
Subtracting one as if the exponent were positive. From $-\frac{1}{2}$ the new exponent is $-\frac{3}{2}$ , more negative, not $\frac{1}{2}$ .
Reaching for the quotient rule and mishandling the constant numerator. It works, but the $x^{-1/2}$ rewrite is faster and avoids sign slips.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

How do you write 1/sqrt(x) as a power of x?

$\frac{1}{\sqrt{x}} = x^{-1/2}$ . The square root is the exponent $\frac{1}{2}$ , and the denominator flips its sign to $-\frac{1}{2}$ . That form is what lets the power rule apply.

Why is the domain x > 0?

$\sqrt{x}$ needs $x \ge 0$ , and dividing by it rules out $x = 0$ . So $\frac{1}{\sqrt{x}}$ exists only for $x > 0$ , and its derivative $-\frac{1}{2x\sqrt{x}}$ is stated on the same set.

Can you get this derivative with the quotient rule?

Yes. With numerator $1$ and denominator $\sqrt{x}$ , the quotient rule simplifies to $-\frac{1}{2}x^{-3/2}$ , the same answer. The power-rule rewrite is shorter and less error prone.