The Quotient Rule (College Board AP® Calculus BC): Revision Note

Notice that the numerator is the same as the product rule but with a subtraction instead (provided you write the $u^{\text{'}}v$ term first!).

Product rule: If $y=u·v$, then $y^{\text{'}}=u^{\text{'}}v+u{v}^{\text{'}}$.

The functions $f$ and $g$ are differentiable. The table shown gives values of the functions and their first derivatives at $x=2$.

Let $h$ be the function defined by $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$. Find $h\text{'}\left(2\right)$. Show the work that leads to your answer.

Answer:

Apply the quotient rule to the function $h$

$h\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)·g\left(x\right)-f\left(x\right)·g\text{'}\left(x\right)}{{\left(g\left(x\right)\right)}^2}$

Substitute $x=2$

$h\text{'}\left(2\right)=\frac{f\text{'}\left(2\right)·g\left(2\right)-f\left(2\right)·g\text{'}\left(2\right)}{{\left(g\left(2\right)\right)}^2}$

Use the values in the table

$h\text{'}\left(2\right)=\frac{\left(-1\right)·5-3·4}{{\left(5\right)}^2}$

$h\text{'}\left(2\right)=-\frac{17}{25}$

Find the derivative of the following functions.

(a) $f\left(x\right)=\frac{4x+3}{2x+5}$

(b) $g\left(x\right)=\frac{\sin 3x}{e^{4x}}$

Answer:

(a)

Assign $u$ and $v$ to each function

$u=4x+3$

$v=2x+5$

Find the derivatives of $u$ and $v$

$u^{\text{'}}=4$

$v^{\text{'}}=2$

Apply the quotient rule, $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

$y^{\text{'}}=\frac{4\left(2x+5\right)-\left(4x+3\right)·2}{{\left(2x+5\right)}^2}$

Simplify

$y^{\text{'}}=\frac{\left(8x+20\right)-\left(8x+6\right)}{{\left(2x+5\right)}^2}$

$f^{\text{'}}\left(x\right)=\frac{14}{{\left(2x+5\right)}^2}$

(b)

Method 1 - Quotient rule

Assign $u$ and $v$ to each function

$u=\sin 3x$

$v=e^{4x}$

Find the derivatives of $u$ and $v$

$u^{\text{'}}=3\cos 3x$

$v^{\text{'}}=4e^{4x}$

Apply the quotient rule, $y^{\text{'}}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$

$y^{\text{'}}=\frac{3\cos 3x·e^{4x}-\sin 3x·4e^{4x}}{{\left(e^{4x}\right)}^2}$

In the numerator, the $e^{4x}$ term can be factored out

$y^{\text{'}}=\frac{e^{4x}\left(3\cos 3x-4\sin 3x\right)}{{\left(e^{4x}\right)}^2}$

Simplify the powers of $e^{4x}$

$g^{\text{'}}\left(x\right)=\frac{3\cos 3x-4\sin 3x}{e^{4x}}$

Method 2 - Product rule

Rewrite as a product

$g\left(x\right)=\sin 3x·e^{-4x}$

Assign $u$ and $v$ to each function

$u=\sin 3x$

$v=e^{-4x}$

Find the derivatives of $u$ and $v$

$u^{\text{'}}=3\cos 3x$

$v^{\text{'}}=-4e^{-4x}$

Apply the product rule, $y^{\text{'}}=u^{\text{'}}v+u{v}^{\text{'}}$

$y^{\text{'}}=3\cos 3x·e^{-4x}+\sin 3x·\left(-4e^{-4x}\right)$

Simplify

$g^{\text{'}}\left(x\right)=\left(3\cos 3x-4\sin 3x\right)e^{-4x}$

In part (b) of the worked example above, both methods give equivalent answers. In your exam, you are not usually required to simplify your answers, so you would not need to cancel the common term in the fraction in method 1 or factor the common term in method 2.

However, method 1 is definitely safer, as you are less likely to make a sign error. The only time a student should use method 2 is if they forget the formula for the quotient rule.