Derivative of x^{1/3}. In this article, we will find the derivative of the cube root of x by applying the power rule of derivatives. We will also use the limit definition to evaluate the derivative of the cube root of x.

**What is the derivative of cube root of x?**

**First Method of Finding Derivative of Cube Root of x: **At first, we will calculate the derivative of cube root x by the power rule of derivatives. See the below steps.

Step 1: We rewrite the cube root of x using the rule of indices.

\[\sqrt[3]{x}=x^{1/3}\]

Step 2: Now apply the following power rule of derivatives:

\[\frac{d}{dx}(x^n)=nx^{n-1}.\]

Therefore, we get that

$\frac{d}{dx}(\sqrt[3]{x})$ $=\frac{d}{dx}(x^{1/3})$ $=\frac{1}{3}x^{1/3-1}$ $\cdots (*)$

Step 3: Finally, simplify the above expression. Thus from $(*)$ we have

$\therefore \dfrac{d}{dx}(\sqrt[3]{x})=\dfrac{1}{3}x^{-2/3}$

$=\dfrac{1}{3} \times \dfrac{1}{x^{2/3}}$

$=\dfrac{1}{3x^{2/3}}$

So the derivative of the cube root of $x$ by power rule is $\frac{1}{3x^{2/3}}$, that is,

\[\dfrac{d}{dx}(\sqrt[3]{x})=\dfrac{1}{3x^{2/3}}.\]

**Also Read:**

Derivative of sin 3x : The derivative of sin 3x is 3cos 3x. |

Derivative of log : The derivative of log_{e} 3x_{e} 3x is 1/x. |

Derivative of e: The derivative of e^{sin x}^{sin x }is cos x e^{sin x}. |

**Second Method of Finding Derivative of Cube Root of x: **In the second method on how to calculate the derivative of the cube root of x, we will use the substitution method. Let $y=\sqrt[3]{x}.$ Taking cubes on both sides, we get that

$y^3=x$

Differentiating both sides with respect to $x$, we obtain that

$3y^2 \dfrac{dy}{dx}=\dfrac{dx}{dx}=1$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{3y^2}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{3(\sqrt[3]{x})^2}$ $[\because y=\sqrt[3]{x}]$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{3(x^{1/3})^2}$

$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{3x^{2/3}}$

In other words, $\frac{d}{dx}(\sqrt[3]{x})$ $=\frac{1}{3x^{2/3}}$. So the derivative of third root of $x$ is $\frac{1}{3x^{2/3}}$.

**Also Read: Derivative of Square Root of x**

Now, we will find the derivative of cube root of $x$ by the first principle.

**Derivative of cube root of x from first principle**

Let $f(x)=\sqrt{x}.$ We need to find the derivative of $f(x).$ From first principle of derivatives, we have

$\frac{d}{dx}(f(x)) = \lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$

$=\lim\limits_{h \to 0} \dfrac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}$ $\cdots (i)$

We will multiply both numerator and denominator by $\sqrt[3]{x+h}^2 + \sqrt[3]{x+h}\sqrt[3]{x}$ $+\sqrt[3]{x}^2$. Doing so and applying the formula of a^{3}-b^{3} = (a-b)(a^{2}-ab+b^{2}), the numerator becomes

$(\sqrt[3]{x+h}-\sqrt[3]{x})$ $(\sqrt[3]{x+h}^2 + \sqrt[3]{x+h}\sqrt[3]{x}$ $+\sqrt[3]{x}^2)$

$=\sqrt[3]{x+h}^3- \sqrt[3]{x}^3$

$=x+h-x=h$

Myltiplying the numerator and the denominator of $(i)$ by the above quantity, we get that

$\dfrac{d}{dx}(f(x))$

$=\lim\limits_{h \to 0} {\small \dfrac{h}{h(\sqrt[3]{x+h}^2 + \sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2)} }$

$=\lim\limits_{h \to 0} {\small\dfrac{1}{\sqrt[3]{x+h}^2 + \sqrt[3]{x+h}\sqrt[3]{x}+\sqrt[3]{x}^2} }$

$=\dfrac{1}{\sqrt[3]{x+0}^2 + \sqrt[3]{x+0}\sqrt[3]{x}+\sqrt[3]{x}^2}$

$=\dfrac{1}{3\sqrt[3]{x}^2}$

$=\dfrac{1}{3(x^{1/3})^2}$

$=\dfrac{1}{3x^{2/3}}$

Thus, the derivative of the cube root of x by first principle is 1/3x^{2/3}

**Derivative of Cube Root x by Logarithmic Differentiation**

Now, we will find the derivative of the cube root of x with the help of the logarithmic derivative. Note that cube root x can be written as x^{1/3}. Write

y= x^{1/3}

We will take natural logarithms on both sides (i.e, logarithm with base e, denoted by ln). Thus we get that

ln y = 1/3 ln x

Differentiating with respect to x, we have

$\frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \cdot \frac{1}{x}$

⇒ $\frac{dy}{dx} =\frac{y}{3x}$ $= \frac{\sqrt[3]{x}}{3x}$ $=\frac{1}{3x^{2/3}}$.

Hence, by the logarithmic differentiation, the derivative of cube root x is $\frac{1}{3x^{2/3}}$.

**Application of Derivative of cube root of x**

From above we get the derivative of root $x$ which is given below.

$\dfrac{d}{dx}(\sqrt{x})$ $=\dfrac{1}{3x^{2/3}}$ $\cdots (I)$

This will help us to find the derivatives of many functions involving cube roots by the chain rule of derivatives. At first, we will calculate the derivative of the cube root of x+2.

**Example 1:** Find $\dfrac{d}{dx}(\sqrt[3]{x+2})$

Let $z=x+2$

∴ $\dfrac{d}{dx}(\sqrt[3]{x+2})$ $=\dfrac{d}{dx}(\sqrt[3]{z})$

$=\dfrac{d}{dz}(\sqrt[3]{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=\dfrac{1}{3z^{2/3}} \cdot \dfrac{d}{dx}(x+2)$ by Equation (I)

$=\dfrac{1}{3z^{2/3}} \cdot 1$

$=\dfrac{1}{3(x+2)^{2/3}}$ $[\because z=x+2]$

Next, we will find the derivative of the cube root of e^{x}, that is, the derivative of e^{x/3}.

**Example 2:** Find $\dfrac{d}{dx}(\sqrt[3]{e^x})$

Let $z=e^x$

∴ $\dfrac{d}{dx}(\sqrt[3]{e^x})$ $=\dfrac{d}{dx}(\sqrt[3]{z})$

$=\dfrac{d}{dz}(\sqrt[3]{z}) \cdot \dfrac{dz}{dx}$ (by the chain rule)

$=\dfrac{1}{3z^{2/3}} \cdot \dfrac{d}{dx}(e^x)$ by Equation (I)

$=\dfrac{1}{3z^{2/3}} \cdot e^x$

$=\dfrac{1}{3e^{2x/3}}$ $[\because z=e^x]$

## FAQs on Derivative of Cube Root of x

**Q1: What is the derivative of cube root of x?**

Ans: The derivative of cube root of x is 1/3x^{2/3}.

**Q2: How a cube root of x defined?**

Ans: A cube root of x is a number b such that b^{3} =x. Symbolically, a cube root of x can be written as $\sqrt[3]{x}$.