AP Calculus AB and BC

Derivative of the Cube Root of x: Answer, Proof, Mistakes

The derivative of the cube root of x is 1 divided by the quantity 3 times the cube root of x squared. In prime notation, if f(x) = x^(1/3) then f'(x) = (1/3)x^(-2/3), which equals 1/(3 times the cube root of x^2). Rewrite the root as the power x^(1/3), then apply the power rule.

ddx[x3]=13x23\frac{d}{dx}\left[\sqrt[3]{x}\right] = \frac{1}{3\sqrt[3]{x^2}}

How to differentiate the cube root of x

Rewrite the radical as a fractional power, x3=x13\sqrt[3]{x} = x^{\frac{1}{3}}, then apply the power rule ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1} with n=13n = \frac{1}{3}.

ddxx13=13x13−1=13x−23\frac{d}{dx}x^{\frac{1}{3}} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}}

Convert the negative fractional exponent back to radical form to state the answer.

13x−23=13x23=13x23\frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3x^{\frac{2}{3}}} = \frac{1}{3\sqrt[3]{x^2}}

What the derivative says about the graph

The cube root is defined for every real xx, including x=0x = 0. Its derivative 13x23\frac{1}{3\sqrt[3]{x^2}} is not, because the denominator is 00 at x=0x = 0.

That gap is a vertical tangent: the curve passes through the origin but its slope shoots to infinity there. This is a standard example of a function that is continuous at a point yet not differentiable at it. The power rule itself is Unit 2 material, Topic 2.5.

Common mistakes with the derivative of the cube root of x

  • Subtracting 11 from the exponent wrong. From 13\frac{1}{3} the new exponent is 13−1=−23\frac{1}{3} - 1 = -\frac{2}{3}, not −13-\frac{1}{3} or 23\frac{2}{3}.
  • Dropping the coefficient 13\frac{1}{3} and writing x−23x^{-\frac{2}{3}} alone.
  • Claiming the derivative exists at x=0x = 0. The function does, but the derivative does not; there is a vertical tangent.

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 x3\sqrt[3]{x}?

It is 13x−23\frac{1}{3}x^{-\frac{2}{3}}, which is the same as 13x23\frac{1}{3\sqrt[3]{x^2}}.

Why rewrite the cube root as x13x^{\frac{1}{3}}?

The power rule applies to any power of xx, including fractional ones. Writing x3=x13\sqrt[3]{x} = x^{\frac{1}{3}} lets you differentiate in one step.

Is the cube root differentiable at x=0x = 0?

No. The function is defined and continuous there, but the derivative 13x23\frac{1}{3\sqrt[3]{x^2}} blows up, so the graph has a vertical tangent at the origin.