10I

The quotient rule

Differentiating a quotient with (u/v)′ = (u′v − uv′)/v², simplifying the result, and choosing between the quotient rule and rewriting.

Quotient rule

$$\text{If } y = \frac{u}{v}, \text{ then } \frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$

Example 1●

Differentiate $y = \dfrac{x}{x + 1}$.

$u = x$, $v = x + 1$, $u' = 1$, $v' = 1$.

$\dfrac{dy}{dx} = \dfrac{(x + 1) - x}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}$

Example 2●●

Differentiate $y = \dfrac{x^2}{2x - 1}$ and find where the gradient is zero.

$\dfrac{dy}{dx} = \dfrac{2x(2x - 1) - 2x^2}{(2x - 1)^2} = \dfrac{2x^2 - 2x}{(2x - 1)^2} = \dfrac{2x(x - 1)}{(2x - 1)^2}$

Zero when the numerator is zero: $x = 0$ or $x = 1$.

Flashcards

Quotient rule

$\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$

Gradient zero

Set the numerator equal to zero.

Practice

If $y = \dfrac{u}{v}$, then $\dfrac{dy}{dx} = $
For $y = \dfrac{x}{x + 1}$, find $\dfrac{dy}{dx}$ at $x = 1$.
For $y = \dfrac{2x + 3}{x - 1}$, find $\dfrac{dy}{dx}$ at $x = 2$.
$\dfrac{d}{dx}\left(\dfrac{x^2}{2x - 1}\right) = $
Find both $x$-values where $y = \dfrac{x^2}{2x - 1}$ has gradient zero.
For $y = \dfrac{x^2 + 1}{x}$, find $\dfrac{dy}{dx}$ at $x = 1$.
Why is the gradient of $y = \dfrac{x}{x + 1}$ never negative?
For $y = \dfrac{1}{x^2 + 1}$, find $\dfrac{dy}{dx}$ at $x = 1$.
The tangent to $y = \dfrac{x}{x + 1}$ at $x = 1$ is $y = \tfrac14 x + b$. Find $b$.