10G

Differentiating powers with fractional indices

Differentiating square roots, cube roots and other fractional powers, using the chain rule with roots, and where these derivatives are undefined.

Roots as powers

The power rule works for any rational index. Write roots as fractional powers first.

Square root

$$\frac{d}{dx}\sqrt{x} = \frac{d}{dx}\,x^{\frac12} = \frac12 x^{-\frac12} = \frac{1}{2\sqrt{x}}$$

Example 1●

Differentiate $y = 6\sqrt[3]{x}$ and $y = x^{\frac32}$.

$y = 6x^{\frac13}$, so $\dfrac{dy}{dx} = 2x^{-\frac23} = \dfrac{2}{\sqrt[3]{x^2}}$.

$\dfrac{dy}{dx} = \tfrac32 x^{\frac12} = \tfrac32\sqrt{x}$.

Example 2: with the chain rule●●

Differentiate $y = \sqrt{4x + 1}$.

$y = (4x + 1)^{\frac12}$

$\dfrac{dy}{dx} = \tfrac12(4x + 1)^{-\frac12} \times 4 = \dfrac{2}{\sqrt{4x + 1}}$

Where the derivative fails

$y = \sqrt{x}$ is defined at $x = 0$, but $\dfrac{1}{2\sqrt{x}}$ is not. The tangent at the origin is vertical, so the function is not differentiable there.

Tangents to y = √x

Slide a towards 0: the tangent becomes steeper and steeper, heading towards vertical.

Try this

Find $a$ so the tangent has gradient $\tfrac14$.

Flashcards

$rac{d}{dx}sqrt{x}$

$\dfrac{1}{2\sqrt{x}}$

$rac{d}{dx},x^{rac32}$

$\tfrac32\sqrt{x}$

$rac{d}{dx},rac{1}{sqrt{x}}$

$-\tfrac12 x^{-\frac32}$

Practice

$\dfrac{d}{dx}\sqrt{x} = $
For $y = \sqrt{x}$, find $\dfrac{dy}{dx}$ at $x = 9$.
For $y = x^{\frac32}$, find $\dfrac{dy}{dx}$ at $x = 4$.
$\dfrac{d}{dx}\sqrt{4x + 1} = $
For $y = \sqrt{4x + 1}$, find $\dfrac{dy}{dx}$ at $x = 2$.
For $y = 6\sqrt[3]{x}$, find $\dfrac{dy}{dx}$ at $x = 8$.
At $x = 0$, $y = \sqrt{x}$:
The tangent to $y = 4\sqrt{x}$ at $x = 4$ is $y = x + b$. Find $b$.
$\dfrac{d}{dx}\left(\dfrac{1}{\sqrt{x}}\right) = $