Use the quotient rule when the function is written as a fraction:

\[ y=\frac{u}{v} \]

Examples:

• \(\dfrac{x^2+1}{x-3}\)
• \(\dfrac{\sin x}{x}\)
• \(\dfrac{e^x}{x^2+1}\)

If \[ y=\frac{u}{v}, \] then

A common memory aid is:

Notice the order: bottom times derivative of top, minus top times derivative of bottom.

Differentiate

\[ y=\frac{x^2+1}{x-3} \]

Let

\[ u=x^2+1,\qquad v=x-3 \]

Then

\[ \frac{du}{dx}=2x,\qquad \frac{dv}{dx}=1 \]

Using the quotient rule:

\[ \frac{dy}{dx} =\frac{(x-3)(2x)-(x^2+1)(1)}{(x-3)^2} \]

\[ =\frac{2x^2-6x-x^2-1}{(x-3)^2} \]

Differentiate

\[ y=\frac{\sin x}{x} \]

Let

\[ u=\sin x,\qquad v=x \]

Then

\[ \frac{du}{dx}=\cos x,\qquad \frac{dv}{dx}=1 \]

So

\[ \frac{dy}{dx}=\frac{x\cos x-\sin x}{x^2} \]

Differentiate

\[ y=\frac{e^x}{x^2+1} \]

Let

\[ u=e^x,\qquad v=x^2+1 \]

Then

\[ \frac{du}{dx}=e^x,\qquad \frac{dv}{dx}=2x \]

So

\[ \frac{dy}{dx} =\frac{(x^2+1)e^x-e^x(2x)}{(x^2+1)^2} \]

Factorising:

Sometimes it is easier to rewrite the fraction first.

Example:

\[ y=\frac{x^2+3x}{x}=x+3 \]

It is easier to simplify first, then differentiate: \[ \frac{dy}{dx}=1 \]

Sometimes one part of the fraction also needs the chain rule.

Example:

\[ y=\frac{1}{(2x+1)^3} \]

This can be done by quotient rule, but it is usually easier to rewrite first: \[ y=(2x+1)^{-3} \] and then use the chain rule.

• Forgetting to square the denominator.
• Getting the order wrong in the numerator.
• Forgetting the minus sign.
• Not simplifying the final answer when possible.