(D3) Quotient Rule#

By the end of the lesson you will be able to:

  • compute the derivative of a quotient of functions.

Lecture Videos#

Quotient Rule#

Quotient Rule

\[\begin{split} \dfrac{d}{dx}\left[\dfrac{F(x)}{S(x)}\right]&=\dfrac{\tfrac{d}{dx}\left[F(x)\right]S(x)-F(x)\tfrac{d}{dx}\left[S(x)\right]}{\left(S(x)\right)^{2}}\\ &\\ \left[\dfrac{F(x)}{S(x)}\right]' &= \dfrac{\big[F(x)\big]' \cdot S(x) \;-\; F(x)\cdot \big[S(x)\big]'}{\big(S(x)\big)^2}\\ \end{split}\]

Interpretation

This rule tells us how to differentiate the quotient of two functions. Essentially, if we see a fraction with variable terms on the top and bottom, we need to use quotient rule.

You can implement this rule similar to what we do for Product Rule:

  1. Write 2 copies of the quotient as a product (top times bottom).

  2. In the 1st copy, apply the derivative to the 1st term.

  3. In the 2nd copy, apply the derivative to the 2nd term.

  4. Put a minus sign \(-\) in between the two copies.

  5. Divide by the original denominator squared.

\[ \left[\dfrac{F(x)}{S(x)}\right]' = \color{blue}{F(x)\cdot S(x)}\qquad \color{blue}{F(x)\cdot S(x)} \]

We start by writing 2 copies of the quotient as a product, top times bottom (in that order).

\[ \left[\dfrac{F(x)}{S(x)}\right]' = \color{blue}{\big[}\color{black}F(x)\color{blue}{\big]'} \color{black}\cdot S(x)\qquad F(x)\cdot S(x) \]

In the 1st copy, we apply the derivative to the 1st term.

\[ \left[\dfrac{F(x)}{S(x)}\right]' = \big[F(x)\big]' \cdot S(x) \qquad F(x)\cdot \color{blue}{\big[} \color{black} S(x)\color{blue}{\big]'} \]

In the 2nd copy, we apply the derivative to the 2nd term.

\[ \left[\dfrac{F(x)}{S(x)}\right]' = \big[F(x)\big]' \cdot S(x) \color{blue}{\;-\;} \color{black} F(x)\cdot \big[S(x)\big]' \]

Next we put a minus sign in between the two copies.

\[ \left[\dfrac{F(x)}{S(x)}\right]' = \dfrac{\big[F(x)\big]' \cdot S(x) \;-\; F(x)\cdot \big[S(x)\big]'}{\color{blue}{\big(S(x)\big)^2}} \]

Finally, we divide everything by the original denominator squared. And there we have our quotient rule! (From here, we would then continue differentiating the terms in the brackets.)

Example 1#

Differentiate the function \(y=\dfrac{x^2+x-2}{x^3-6}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two \(x\)-terms … and we do!

\[ y=\dfrac{x^2+x-2}{x^3-6} \]

Notice that the top term \((x^2+x-2)\) contains variable \(x\) and the bottom term \((x^3-6)\) does as well. This shows that we really do have a quotient of two functions of \(x\), so the quotient rule is applicable.

Once we’ve recognized our function as a quotient, the next step is to actually implement the quotient rule:

\[ y'=\dfrac{\color{blue}{\big[}x^2+x-2\color{blue}{\big]'}(x^3-6)\; - \; (x^2+x-2)\color{blue}{\big[}x^3-6\color{blue}{\big]'}}{(x^3-6)^2} \]

Once we’ve applied the quotient rule, we need to continue differentiating each of the factors:

\[\begin{split} y'&=\dfrac{\color{blue}{\big[}x^2+x-2\color{blue}{\big]'}(x^3-6)\; - \; (x^2+x-2)\color{blue}{\big[}x^3-6\color{blue}{\big]'}}{(x^3-6)^2}\\ &\\ y'&=\dfrac{\color{blue}{\big(}2x+1\color{blue}{\big)}(x^3-6)\; - \; (x^2+x-2)\color{blue}{\big(}3x^2\color{blue}{\big)}}{(x^3-6)^2} \\\end{split}\]

This is our derivative, so it is fine to stop here. If we needed to, we could also simplify this result by multiplying out the terms in the numerator.

Example 2#

Differentiate the function \(f(x)=\dfrac{\sec x}{1+\tan x}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two \(x\)-terms … and we do!

\[ f(x)=\dfrac{\sec x}{1+\tan x} \]

Notice that the top term \((\sec x)\) contains variable \(x\) and the bottom term \((1+\tan x)\) does as well. This shows that we really do have a quotient of two functions of \(x\), so the quotient rule is applicable.

Once we’ve recognized our function as a quotient, the next step is to actually implement the quotient rule:

\[ f'(x)=\dfrac{\color{blue}{\big[}\sec x\color{blue}{\big]'}(1+\tan x)\; - \; (\sec x)\color{blue}{\big[}1+\tan x\color{blue}{\big]'}}{(1+\tan x)^2} \]

Once we’ve applied the quotient rule, we need to continue differentiating each of the factors:

\[\begin{split} f'(x)&=\dfrac{\color{blue}{\big[}\sec x\color{blue}{\big]'}(1+\tan x)\; - \; (\sec x)\color{blue}{\big[}1+\tan x\color{blue}{\big]'}}{(1+\tan x)^2}\\ &\\ &=\dfrac{\color{blue}{\big(} \sec x \tan x \color{blue}{\big)} (1+\tan x)\; - \; (\sec x)\color{blue}{\big(}\sec^2 x\color{blue}{\big)}}{(1+\tan x)^2} \\\end{split}\]

This is our derivative, so it is fine to stop here. If we needed to, we could also simplify this result by multiplying out the terms in the numerator.

Example 3#

Calculate the derivative of the function \(g(x)=\dfrac{x^2}{\ln x}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two \(x\)-terms … and we do!

\[ g(x)=\dfrac{x^2}{\ln x} \]

Notice that the top term \((x^2)\) contains variable \(x\) and the bottom term \((\ln x)\) does as well. This shows that we really do have a quotient of two functions of \(x\), so the quotient rule is applicable.

Once we’ve recognized our function as a quotient, the next step is to actually implement the quotient rule:

\[ g'(x)=\dfrac{\color{blue}{\big[}x^2 \color{blue}{\big]'}(\ln x)\; - \; (x^2 )\color{blue}{\big[} \ln x \color{blue}{\big]'}}{(\ln x)^2} \]

Once we’ve applied the quotient rule, we need to continue differentiating each of the factors:

\[\begin{split} g'(x) & =\dfrac{\color{blue}{\big[}x^2 \color{blue}{\big]'}(\ln x)\; - \; (x^2 )\color{blue}{\big[} \ln x \color{blue}{\big]'}}{(\ln x)^2} \\ &\\ &=\dfrac{\color{blue}{\big(} 2x \color{blue}{\big)} (\ln x)\; - \; (x^2 )\color{blue}{\big(}\tfrac{1}{x} \color{blue}{\big)}}{(\ln x)^2} \\\end{split}\]

This is our derivative, so it is fine to stop here. If we needed to, we could also simplify this result by multiplying out the terms in the numerator.

Example 4#

Calculate the derivative of the function \(v(t)=\dfrac{e^t}{1+t^2}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two variable-terms … and we do!

\[ v(t)=\dfrac{e^t}{1+t^2} \]

Notice that the top term \((e^t)\) contains variable \(t\) and the bottom term \((1+t^2)\) does as well. This shows that we really do have a quotient of two functions of \(t\), so the quotient rule is applicable.

Once we’ve recognized our function as a quotient, the next step is to actually implement the quotient rule:

\[ v'(t)=\dfrac{\color{blue}{\big[}e^t \color{blue}{\big]'}(1+t^2)\; - \; (e^t )\color{blue}{\big[} 1+t^2 \color{blue}{\big]'}}{(1+t^2)^2} \]

Once we’ve applied the quotient rule, we need to continue differentiating each of the factors:

\[\begin{split} v'(t) & =\dfrac{\color{blue}{\big[}e^t \color{blue}{\big]'}(1+t^2)\; - \; (e^t )\color{blue}{\big[} 1+t^2 \color{blue}{\big]'}}{(1+t^2)^2} \\ &\\ &=\dfrac{\color{blue}{\big(} e^t \color{blue}{\big)} (1+t^2)\; - \; (e^t )\color{blue}{\big(}2t \color{blue}{\big)}}{(\1+t^2)^2} \\\end{split}\]

This is our derivative, so it is fine to stop here. If we needed to, we could also simplify this result by multiplying out the terms in the numerator. We don’t normally do anything with the denominator though.

Is Quotient Rule Necessary?#

Example 5#

Calculate the derivative of the function \(g(x)=-\dfrac{5}{x^4}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two variable-terms … Do we???

\[ g(x)=-\dfrac{5}{x^4} \]

Notice that the top term \((5)\) does not contain a variable.

While we do have division, the top function is not dependent on variable \(x\) and so the quotient rule is not really necessary. (We could still use it and get the correct answer, but there is an easier way.)

Instead of the quotient rule, we go back to one of our usual strategies when we have a fraction with a variable in the denominator: rewrite using negative exponents.

\[ g(x)=-\dfrac{5}{x^4} = -5 \cdot \dfrac{1}{x^4} = -5 \cdot x^{-4} \]

In this form, we have a constant-multiple of a power function, so we can just use the power rule (and constant-multiple rule) to differentiate.

Using the constant-multiple rule and power rule, we get:

\[\begin{split} g'(x) & =\color{blue}{\big[} -5 \cdot x^{-4} \color{blue}{\big]'} \\ & \\ & = -5\cdot \color{blue}{\big[} x^{-4} \color{blue}{\big]'} \qquad \text{Constant-Multiple Rule}\\ & \\ & = -5\cdot \color{blue}{\big(} (-4) x^{-5} \color{blue}{\big)} \qquad \text{Power Rule}\\ & \\ & = 20 x^{-5} \\ \end{split}\]

And with practice, you probably do these power rule calculations in your head and do:

\[ g(x)= -5 \cdot x^{-4} \xrightarrow{\quad\text{Differentiate}\quad} g'(x)= 20x^{-5} \]

Example 6#

Calculate the derivative of the function \(F(x)=\dfrac{3x^2+2\sqrt{x}}{x}\).

Click through the tabs to see the steps of our solution.

  • We start by first identifying which derivative rules we might need to use.

The first step, is to identify that we actually have a quotient containing two variable-terms … Do we???

\[ F(x)=\dfrac{3x^2+2\sqrt{x}}{x} \]

Notice that the top term \((3x^2+2\sqrt{x})\) contains variable \(x\) and the bottom term \((x)\) does as well. This shows that we really do have a quotient of two functions of \(x\), so the quotient rule is applicable.

However, it is actually easier to do the division first to do for this function. (Normally this is the case, when all terms in the function are powers of \(x\) and we have a lone term in the denominator.)

Doing the division in this function, will allow us to simplify down to the point where quotient rule is not necessary.

\[\begin{split} F(x)=\dfrac{3x^2+2\sqrt{x}}{x} & = \dfrac{3x^2}{x}+\dfrac{2\sqrt{x}}{x} \\ & \\ & = \dfrac{3x^2}{x}+\dfrac{2x^{1/2}}{x} \\ & \\ & = 3x+ 2x^{-1/2} \\ \end{split}\]

Remember: when we divide powers of \(x\), we subtract the exponents.

Now we can just use our basic derivative rules to differentiate:

\[\begin{split} F'(x) & =\color{blue}{\big[} 3x+ 2x^{-1/2} \color{blue}{\big]'} \\ & \\ & = \color{blue}{\big[} 3x \color{blue}{\big]'} + \color{blue}{\big[} 2x^{-1/2} \color{blue}{\big]'}\\ & \\ & = \color{blue}{\big(} 3 \color{blue}{\big)}+ 2 \color{blue}{\big[} x^{-1/2} \color{blue}{\big]'}\\ & \\ & = 3 + 2 \color{blue}{\big(} \left(\tfrac{-1}{2}\right) x^{-3/2} \color{blue}{\big)}\\ & \\ & = 3 - x^{-3/2} \\ \end{split}\]

And with practice, you probably do these calculations in your head and get:

\[ F(x)= 3x+ 2x^{-1/2} \xrightarrow{\quad\text{Differentiate}\quad} F'(x)= 3- \tfrac{1}{2}x^{-3/2} \]