Derivative of Roots (Radicals)

Roots differentiate cleanly if you rewrite them as powers. Use the identity $\sqrt[n]{u} = u^{1/n}$ and then apply the power rule together with the chain rule.

📜 Core Formulas

Square Root

For $u = u(x) \ge 0$:

$$ \frac{d}{dx}\big(\sqrt{u}\big) = \frac{d}{dx}\big(u^{1/2}\big) = \frac{1}{2}\,u^{-1/2}\,u'(x) = \frac{u'(x)}{2\sqrt{u(x)}}. $$

General $n$-th Root

For integer $n \ge 2$ and $u = u(x)$ in the domain where $\sqrt[n]{u}$ is defined:

$$ \frac{d}{dx}\big(\sqrt[n]{u}\big) = \frac{d}{dx}\big(u^{1/n}\big) = \frac{1}{n}\,u^{\frac{1}{n}-1}\,u'(x) = \frac{u'(x)}{n\,u(x)^{\frac{n-1}{n}}}. $$

💡 Idea: Rewrite $\sqrt[n]{u}$ as $u^{1/n}$ and apply the power rule $\frac{d}{dx}(u^p) = p\,u^{p-1}\,u'(x)$.
⚠️ Domain: For real functions, ensure the root is defined (e.g., $\sqrt{u}$ requires $u(x)\ge 0$).

🪜 Application in 3 Simple Steps

Rewrite the root as a power: Use $\sqrt{u}=u^{1/2}$ or $\sqrt[n]{u}=u^{1/n}$.
Apply power rule + chain rule: Differentiate $u^{p}$ as $p\,u^{p-1}\,u'(x)$.
Simplify and check the domain: Express negative exponents as fractions and note any domain restrictions.

✏️ Examples for Illustration

Example 1: $\boldsymbol{\sqrt{x}}$

Given: $f(x) = \sqrt{x}$ (for $x \ge 0$)

Rewrite: $f(x) = x^{1/2}$
Differentiate: $f'(x) = \tfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$

Result: $$ f'(x) = \dfrac{1}{2\sqrt{x}}. $$

Example 2: $\boldsymbol{\sqrt{3x^2+1}}$ (chain rule)

Given: $f(x) = \sqrt{3x^2 + 1}$ (always defined for real $x$)

$u(x) = 3x^2 + 1 \Rightarrow u'(x) = 6x$
$f'(x) = \dfrac{u'(x)}{2\sqrt{u(x)}} = \dfrac{6x}{2\sqrt{3x^2+1}} = \dfrac{3x}{\sqrt{3x^2+1}}$

Result: $$ f'(x) = \dfrac{3x}{\sqrt{3x^2+1}}. $$

Example 3: $\boldsymbol{\sqrt[3]{x}}$ (cube root)

Given: $f(x) = \sqrt[3]{x} = x^{1/3}$ (defined for all real $x$)

$f'(x) = \tfrac{1}{3}x^{-2/3} = \dfrac{1}{3\,\sqrt[3]{x^2}}$

Result: $$ f'(x) = \dfrac{1}{3\,\sqrt[3]{x^2}}. $$

Example 4: $\boldsymbol{\dfrac{1}{\sqrt{x}}}$

Given: $f(x) = \dfrac{1}{\sqrt{x}} = x^{-1/2}$ (for $x > 0$)

$f'(x) = -\tfrac{1}{2}x^{-3/2} = -\dfrac{1}{2x^{3/2}} = -\dfrac{1}{2x\sqrt{x}}$

Result: $$ f'(x) = -\dfrac{1}{2x\sqrt{x}}. $$

Example 5: $\boldsymbol{\sqrt{\,\sin x\,}}$ (domain restriction)

Given: $f(x) = \sqrt{\sin x}$ with $\sin x \ge 0$

$u(x) = \sin x \Rightarrow u'(x) = \cos x$
$f'(x) = \dfrac{u'(x)}{2\sqrt{u(x)}} = \dfrac{\cos x}{2\sqrt{\sin x}}$ (where $\sin x > 0$)

Result: $$ f'(x) = \dfrac{\cos x}{2\sqrt{\sin x}}. $$

Example 6: $\boldsymbol{\sqrt[n]{\,x^2+2x\,}}$

Given: $f(x) = \sqrt[n]{x^2+2x} = (x^2+2x)^{1/n}$ (ensure $(x^2+2x)$ is in the domain for the chosen $n$)

$u(x) = x^2+2x \Rightarrow u'(x) = 2x+2$
$f'(x) = \dfrac{1}{n}(x^2+2x)^{\frac{1}{n}-1}\,(2x+2) = \dfrac{2(x+1)}{n\,\big(x^2+2x\big)^{\frac{n-1}{n}}}$

Result: $$ f'(x) = \frac{2(x+1)}{n\,\big(x^2+2x\big)^{\frac{n-1}{n}}}. $$