AP Calculus AB and BC

Derivative of 1/x^3: Answer, Proof, and Mistakes

The derivative of 1/x^3 with respect to x is -3/x^4. Rewrite 1/x^3 as x^(-3) and apply the power rule: the exponent -3 moves out front and drops by one to -4, giving -3 times x^(-4) = -3/x^4. Since x^4 is positive for every x except 0, this slope is negative everywhere the function is defined.

ddx[1x3]=−3x4\frac{d}{dx}\left[\frac{1}{x^3}\right] = -\frac{3}{x^4}

The proof: rewrite 1/x^3 as a power

Move the cube to the numerator as a negative exponent: 1x3=x−3\frac{1}{x^3} = x^{-3}. Now the power rule ddxxn=n xn−1\frac{d}{dx}x^n = n\,x^{n-1} applies directly with n=−3n = -3.

ddx[x−3]=(−3) x−3−1=−3x−4=−3x4\frac{d}{dx}\left[x^{-3}\right] = (-3)\,x^{-3-1} = -3x^{-4} = -\frac{3}{x^4}

The exponent −3-3 comes down as the coefficient, and the new exponent is −3−1=−4-3 - 1 = -4. That leading −3-3 is where the minus sign in the answer comes from.

The slope is negative wherever the function is defined

For every x≠0x \neq 0, x4>0x^4 > 0, so −3x4-\frac{3}{x^4} is negative. The slope is never zero and never positive, so 1x3\frac{1}{x^3} has no critical points and decreases on each of its two branches.

As ∣x∣|x| grows the slope flattens toward 00; as xx approaches 00 the slope steepens sharply, matching the vertical asymptote at x=0x = 0 where the function is undefined.

Common mistakes

  • Dropping the minus sign and writing 3x4\frac{3}{x^4}. The derivative is negative everywhere, so a positive answer is wrong on sight.
  • Getting the new exponent wrong: from x−3x^{-3} you subtract 11 to reach x−4x^{-4}, not x−2x^{-2}. Subtracting from a negative exponent is the common slip.
  • Leaving the coefficient as 11 or forgetting to bring the exponent down, giving −x−4-x^{-4} instead of −3x−4-3x^{-4}.
  • Treating 1x3\frac{1}{x^3} as a quotient and misapplying the quotient rule; the power-rule route is faster and less error-prone.

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

What is the derivative of 1/x^3?

It is −3x4-\frac{3}{x^4}. Rewrite 1x3\frac{1}{x^3} as x−3x^{-3} and apply the power rule: the exponent −3-3 comes down and drops to −4-4, giving −3x−4=−3x4-3x^{-4} = -\frac{3}{x^4}.

Why is the derivative negative?

The coefficient −3-3 from the power rule is negative, and x4>0x^4 > 0 for every x≠0x \neq 0, so −3x4-\frac{3}{x^4} stays negative. The function decreases on each branch.

What is the second derivative of 1/x^3?

Differentiate −3x−4-3x^{-4} again: (−3)(−4)x−5=12x−5=12x5(-3)(-4)x^{-5} = 12x^{-5} = \frac{12}{x^5}. So the second derivative of 1x3\frac{1}{x^3} is 12x5\frac{12}{x^5}.