On this screen we’re going to develop “The Quotient Rule,” which we need to find the derivative of the quotient of two differentiable functions, $\dfrac{f}{g}.$ This is a natural next step after we found the Product Rule on the preceding screen. With this new rule, we’ll be able to find the derivative of functions like $\dfrac{e^x}{x}$ and $\dfrac{\sin \theta}{\cos \theta}.$ We of course have Practice Problems below for you to use to become comfortable with this new rule.
As we saw for the Product Rule, let’s emphasize immediately that the naive approach does not work: the derivative of the quotient of two functions is not the quotient of the derivatives: \[\left[\frac{f(x)}{g(x)}\right]’ \ne \frac{f'(x)}{g'(x)} \]
Show/Hide a quick example of the naive approach not working
Consider the quotient function $\dfrac{x^2}{x}.$ Since this function simplifies to $x,$ we know immediately that its derivative is $\dfrac{d(x)}{dx} = 1.$ By contrast, if we compute the derivative naively as simply the derivative of the numerator divided by the derivative of the denominator, we get \begin{align*} [x]’ = \left[\frac{x^2}{x} \right]’ &\overbrace{=}^? \frac{\left[x^2 \right]’}{[x]’} &&\text{[Does naive approach work??]} \\[8px]
&\overbrace{=}^? \frac{2x}{1} \\[8px]
&\overbrace{=}^? 2 \ne 1 \quad \xmark &&\color{red}{\left[\text{No!} \quad \left[\frac{f(x)}{g(x)}\right]’ \ne \frac{f'(x)}{g'(x)} \right]} \end{align*} That is, the naive approach does not work. We need to develop a Quotient Rule that lets us quickly compute the derivative of the quotient of two functions.
[collapse]
After working to develop the Product Rule, we expect a more involved calculation than that of the naive approach. After all, we know from algebra that we can think of division as multiplication by the reciprocal of what’s in the denominator: \[\left[\frac{f(x)}{g(x)}\right]’ = \left[f(x) \cdot \frac{1}{g(x)}\right]’\]
Hence we could think about the quotient instead as a product … but to use this approach we need one more tool, which we’ll see on the next screen.
Deriving the Quotient Rule For now, let’s stick with viewing the function $\dfrac{f}{g}$ as a quotient, and develop a Quotient Rule we can use to easily compute the derivative in terms of $f$, $g$, $f’$ and $g’$. We begin, of course, with the Definition of the Derivative, now applied to the quotient $\dfrac{f}{g}:$ \begin{align*} \left[\frac{f(x)}{g(x)}\right]’ &= \lim_{h \to 0} \frac{\displaystyle{\frac{f(x+h)}{g(x+h)}\, -\, \frac{f(x)}{g(x)}}}{h} \\[8px]
&= \lim_{h \to 0} \frac{\displaystyle{\frac{f(x+h)}{g(x+h)} \cdot \frac{g(x)}{g(x)} \, -\, \frac{f(x)}{g(x)} \cdot \frac{g(x+h)}{g(x+h)}} }{h} \\[8px]
&= \lim_{h \to 0} \frac{\displaystyle{\frac{f(x+h)\,g(x)\, – \,f(x)\,g(x+h)}{g(x)\,g(x+h)}} }{h} \\[8px]
&= \lim_{h \to 0} \frac{1}{h} \cdot \bbox[5px, border: 1px dashed]{\frac{f(x+h)\,g(x)\, – \,f(x)\,g(x+h)}{g(x)\,g(x+h)}} &&(*) \end{align*} Let’s focus briefly on what’s in the dashed box, setting aside the limit and $\dfrac{1}{h}$ for a moment. Similar to what we encountered in our development of the Product Rule, it’s unclear how to proceed given this collection of terms, especially those in the numerator. But let’s remember that we’re aiming to develop a Quotient Rule that involves the terms $f'(x)$ and $g'(x),$ and so we can work to force those to appear. In fact, in our “algebra-based derivation of the Product Rule,” we did just that, using an “algebraic trick” to cleverly introduce the terms we need: we subtract and add the same term $f(x)\,g(x)$ to force the definitions of the derivatives $f'(x)$ and $g'(x)$ to pop up. Let’s do the same thing here: \begin{align*} \bbox[5px, border: 1px dashed]{\frac{f(x+h)\,g(x)\, – \,f(x)\,g(x+h)}{g(x)\,g(x+h)}} &= \frac{f(x+h)\,g(x) \overbrace{- f(x)\,g(x) + f(x)\,g(x)}^{=0, \text{ so we’re good}}\, -\, f(x)\,g(x+h)}{g(x)\,g(x+h)}\,\, \text{[now factor]} \\[8px]
&= \frac{[f(x+h)-f(x)]\,g(x) + f(x)\,[g(x)-g(x+h)]}{g(x)\,g(x+h)} \\[8px]
&= \frac{ \color{#14349c}{[f(x+h)-f(x)]}\, \color{#000000}{g(x)\, -\, f(x)}\, \color{#006d24}{[g(x+h)-g(x)]}}{g(x)\,g(x+h)} \end{align*} The bits in blue and green, the first and last […] terms in the numerator, will turn into $f'(x)$ and $g'(x)$ respectively when we divide by h and take the limit in a minute. Substituting back into the equation (*) where we left off above, \begin{align*} \left[\frac{f(x)}{g(x)}\right]’ &= \lim_{h \to 0} \frac{1}{h} \cdot \bbox[5px, border: 1px dashed]{\frac{f(x+h)\,g(x) – f(x)\,g(x+h)}{g(x)\,g(x+h)}} &&(*) \\[8px]
&= \lim_{h \to 0} \frac{1}{h} \frac{ \color{#14349c}{[f(x+h)-f(x)]}\, \color{#000000}{g(x)\, – \,f(x)}\, \color{#006d24}{[g(x+h)-g(x)]}}{g(x)\,g(x+h)} \\[8px]
&= \lim_{h \to 0} \frac{ \frac{1}{h} \color{#14349c}{[f(x+h)-f(x)]}\, \color{#000000}{g(x)\, – \,\frac{1}{h} f(x)}\, \color{#006d24}{[g(x+h)-g(x)]}}{g(x)\,g(x+h)} \\[12px]
&= \lim_{h \to 0} \frac{ \color{#14349c}{\frac{[f(x+h)-f(x)]}{h}}\, \color{#000000}{g(x)\, -\, f(x)}\, \color{#006d24}{\frac{[g(x+h)-g(x)]}{h}} }{g(x)\,g(x+h)} \\[8px]
&= \frac{\left[ \displaystyle{\lim_{h \to 0}} \color{#14349c}{\frac{[f(x+h)-f(x)]}{h}}\right]\, \color{#000000}{g(x)\, – \, f(x)}\, \left[\displaystyle{\lim_{h \to 0}} \color{#006d24}{\frac{[g(x+h)-g(x)]}{h}}\right] }{\displaystyle{\lim_{h \to 0}} [g(x)\,g(x+h)]} \\[8px]
&= \frac{\color{#14349c}{f'(x)}\,\color{#000000}{g(x) – f(x)}\,\color{#006d24}{g'(x)}}{g(x)\,g(x)} \end{align*} As you saw, in the second-to-last line when we evaluated the limit, the pesky h in the denominator went away as part of the definition of the derivative for $f’$ and $g’,$ because of the algebraic trick we used. We have thus developed the Quotient Rule we were after:
Quotient Rule Consider two differentiable functions $f$ and $g$, and $g(x) \ne 0.$
In prime notation: \[ \left[\frac{f(x)}{g(x)}\right]’ = \frac{f'(x)\,g(x)\, -\, f(x)\,g'(x)}{g(x)^2}\]
And in Leibniz notation: \[\dfrac{d}{dx}\left(\frac{f(x)}{g(x)} \right) = \frac{\left( \dfrac{d}{dx}f(x)\right)g(x)\, -\, f(x) \left(\dfrac{d}{dx}g(x) \right)}{g(x)^2}\]
In words, $\left[\dfrac{f(x)}{g(x)}\right]’$ \[=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \]
This rule is usually harder for students to remember than, say, the Product Rule, so be on guard. Typical mistakes include switching the terms in the numerator, and forgetting to square the denominator.
♫ Low d-high, minus high d-low, all over low-squared ♫
Many students remember the quotient rule by thinking of the numerator as “high,” the demoninator as “low,” the derivative as “d,” and then singing
♫ “low d-high, minus high d-low, all over low-squared” ♫
Every teacher who’s given an exam on this material has seen students silently mouthing that phrase — to good use! — to make sure they use the Quotient Rule correctly.
The Quotient Rule may seem more intimidating to use than the Product Rule was, but we promise that with some practice (see below!) you’ll quickly get it down and have it as part of your working toolkit. Let’s consider a few examples to see how it works.
Example 1: Derivative of $\dfrac{e^x}{x}$
Find the derivative of $\dfrac{e^x}{x}.$
Solution. Here, our “high” function in the numerator is $f(x) = e^x,$ while our “low” function in the denominator is $g(x) = x.$
Find the derivative of the tangent function, $\tan\theta = \dfrac{\sin\theta}{\cos\theta}.$
Solution. Here, our “high” function in the numerator is $f(\theta) = \sin \theta,$ while our “low” function in the denominator is $g(\theta) = \cos \theta$. The Quotient Rule says $\left[\dfrac{f(\theta)}{g(\theta)}\right]’ = \dfrac{f'(\theta)\,g(\theta)\, -\, f(\theta)\,g'(\theta)}{g(\theta)^2},$ so: \begin{align*} [\tan \theta]’ &= \left[\frac{\sin \theta}{\cos \theta}\right]’ \\[8px]
&= \frac{(\sin \theta)’\cdot\cos(\theta)\, -\, \sin \theta\cdot(\cos \theta)’}{\cos(\theta)^2} &&[\text{Recall }(\sin \theta)’ = \cos \theta \text{ and } (\cos \theta)’ = -\sin \theta )]\\[8px]
&= \frac{\cos \theta \cdot\cos \theta\, – \,\sin \theta \cdot(-\sin \theta )}{(\cos \theta)^2} \\[8px]
&= \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} && \left[ \text{Recall } \cos^2 \theta + \sin^2 \theta = 1\right] \\[8px]
&= \frac{1}{\cos^2 \theta } \\[8px]
&= \sec^2 \theta \quad \cmark \end{align*} That is, the derivative of the tangent function is $\dfrac{d (\tan \theta)}{d \theta} = \sec^2 \theta.$
That’s actually a big “trig function derivative” result that we’ll use on occasion, and that appears in our “Trig Function Derivatives” and “Table of Derivatives” screens that’s accessible from the “Key Formulas” menu at the top of every screen:
Derivative of $\tan x$ \[\dfrac{d}{dx}\tan x = \sec^2 x\]
As we wrote above, while the Quotient Rule may initially look awkward to use, with practice it will feel as routine as using something like the Quadratic Formula has probably become for you. So let’s practice!
Practice Problems
Show/Hide Statement of the Quotient Rule
Quotient Rule Consider two differentiable functions $f$ and $g$, and $g(x) \ne 0.$
In prime notation: \[ \left[\frac{f(x)}{g(x)}\right]’ = \frac{f'(x)\,g(x)\, -\, f(x)\,g'(x)}{g(x)^2}\]
And in Leibniz notation: \[\dfrac{d}{dx}\left(\frac{f(x)}{g(x)} \right) = \frac{\left( \dfrac{d}{dx}f(x)\right)g(x)\, -\, f(x) \left(\dfrac{d}{dx}g(x) \right)}{g(x)^2}\]
In words, $\left[\dfrac{f(x)}{g(x)}\right]’$ \[=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \]
[collapse]
Practice Problem #1
If $h(x) = \dfrac{x^2}{e^x}$, then find $h'(x)$ using the Quotient Rule.
\begin{array}{lllll} \text{(A) }x \left( 2-x \right) && \text{(B) }\dfrac{x}{e^x} \left( 2+x \right)
&& \text{(C) }\dfrac{x}{e^x} \left( 2-x \right) && \text{(D) }\dfrac{x}{e^x} \left( x-2 \right)
&& \text{(E) None of the above} \end{array}
In the Quotient Rule, note that the function in the denominator is squared. $h'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$.
In the Quotient Rule, note the negative sign in the numerator. $h'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$
Correct!
In the Quotient Rule, note the numerator is given by: $f'g-fg'$.
The Quotient Rule is $h'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$ where $f(x) = x^2$ and $f'(x) = 2x$. What are $g(x)$ and $g'(x)$?
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #2
If $y= \dfrac{ \sin x }{ \sqrt{x} }$, find $\dfrac{dy}{dx}$ using the Quotient Rule.
\begin{array}{lll} \text{(A) } \dfrac{1}{\sqrt{x}} \left ( \cos x - \dfrac{ \sin x }{2x} \right )
&& \text{(B) } \sqrt{x} \left ( \cos x + \dfrac{ \sin x }{x} \right ) && \text{(C) } \dfrac{1}{\sqrt{x}} \left ( \cos x - \dfrac{ \sin x }{x} \right )\end{array}
\begin{array}{ll} \text{(D) } \dfrac{1}{\sqrt{x}} \left ( \sin x + \dfrac{\cos x}{2x} \right ) && \text{(E) None of the above} \end{array}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #3
Given $f(1)=4$, $f'(1)=-2$, $g(1)= \dfrac{1}{2} $, and $g'(1)= 1$, evaluate $ \left ( \dfrac{ f(x) }{ g(x) } \right )' $ at $x=1$.
\begin{array}{lllll} \text{(A) }-2 && \text{(B) }20 && \text{(C) } 12 && \text{(D) }-20
&& \text{(E) None of the above} \end{array}
The Quotient Rule for two general functions $ \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$.
Close! Be mindful of the order of the functions in the numerator in the result of the Quotient Rule:
$ \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$.
Close! Note, however, the negative sign in the numerator of the result of the Quotient Rule:
$ \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$.
Correct!
The Quotient Rule for two general functions $ \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$.
Here, $f=4$ and $f'=-2$. What is $g$ and $g'$?
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #4
value of $x$
$f(x)$
$f'(x)$
$g(x)$
$g'(x)$
$\left( \dfrac{f}{g} \right )'(x)$
$x=-1$
2
0
1
-5
$\left( \dfrac{f}{g} \right )'(-1)$
$x=0$
5
-2
$ \dfrac{1}{2} $
$g'(0)$
-4
$x=2$
-1
$f'(2)$
3
-8
2
$x=5$
$f(5)$
3
-2
4
$\dfrac{1}{2}$
Given the values for $g$, $g'$, $f'$ and $\left( \dfrac{f}{g} \right)'$ from the table, $f(5)=$ \begin{array}{lllll} \text{(A) } \dfrac{1}{4} && \text{(B) } \dfrac{1}{8} && \text{(C) } 12 && \text{(D ) } -2 && \text{(E) } -8 \end{array}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #5
If $f(3)=2$, $f'(3)=-4$, $g(3)=3$, $g'(3)=- \dfrac{1}{2}$, $h(3)= -1$, and $h'(3)= 2$, find $\left [ \dfrac{ f(x) }{ g(x) h(x) } \right ]'$ at $x=3.$
\begin{array}{lllll} \text{(A) }\dfrac{14}{9} && \text{(B) }-\dfrac{1}{9} && \text{(C) }-\dfrac{11}{18}
&& \text{(D) } \dfrac{23}{9} && \text{(E) None of the above} \end{array}
Since the denominator of this function is a product of two functions, namely $(gh)$, the Product Rule must be used wherever the denominator is differentiated inside the Quotient Rule.
So, $\left[ \dfrac{f}{gh} \right]'=\dfrac{f'(gh)-f(gh)'}{\left[ (gh)^2 \right]} = \dfrac{f'gh-f\left( g'h+gh' \right)}{\left( gh \right)^2}$.
Right!
Since the denominator of this function is a product of two functions, namely $(gh)$, the Product Rule must be used wherever the denominator is differentiated inside the Quotient Rule.
So, $\left[ \dfrac{f}{gh} \right]'=\dfrac{f'(gh)-f(gh)'}{\left[ (gh)^2 \right]} = \dfrac{f'gh-f\left( g'h+gh' \right)}{\left( gh \right)^2}$.
Close! Note that using the Product Rule yields $(fg)'=f'g+fg'$.
Since the denominator of this function is a product of two functions, namely $(gh)$, the Product Rule must be used wherever the denominator is differentiated inside the Quotient Rule.
So, $\left[ \dfrac{f}{gh} \right]'=\dfrac{f'(gh)-f(gh)'}{\left[ (gh)^2 \right]} = \dfrac{f'gh-f\left( g'h+gh' \right)}{\left( gh \right)^2}$.
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #6
If $p(x) = \dfrac{ \cos x }{ x e^x }$, find $p'(x)$. Note: Since there's a product in the denominator, finding this derivative requires a combination of the Quotient and Product Rules. There are several ways to approach this. If you'd like to compare your work to ours: in the solution below we use $p(x) =\dfrac{ f(x) }{ g(x) }$ and will apply the Product Rule to the denominator function $g(x).$ \begin{array}{lll} \text{(A) } - \dfrac{ x \sin x + \cos x }{ x^2 e^x } && \text{(B) } -\dfrac{x \sin x + \cos x - x \cos x }{ \left(x e^x \right)^2 } && \text{(C) } - \dfrac{ \sin x + \cos x }{ xe^x }\end{array} \begin{array}{ll}\text{(D) }- \dfrac{ x \sin x + \cos x + x \cos x }{ x^2 e^x } && \text{(E) None of the above} \end{array}
When applying the Quotient Rule, $ p'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$, note that the Product Rule must be used to differentiate $g(x)=xe^x$.
Then, $g'=e^x+xe^x$.
When applying the Quotient Rule, $ p'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$, recall the Product Rule for $g(x)$ yields $(xe^x)'=e^x+xe^x$.
When applying the Quotient Rule, $ p'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$, recall the Product Rule for $g(x)$ yields $(xe^x)'=e^x+xe^x$.
Right!
Recall the Quotient Rule: $ p'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )' = \dfrac{ f'(x)\,g(x) - f(x)\,g'(x) }{ \left ( g(x) \right )^2 }$, note that the Product Rule must be used to differentiate $g(x)=xe^x$.
So $g'(x)=(xe^x)'=e^x+xe^x$
Show/Hide Solution
Let $f(x)= \cos x$ and $g(x)= xe^x$. Then $f'(x) = – \sin x$ and, using the Product Rule, $g'(x) = e^x + xe^x$. Using the Quotient Rule:
\begin{align*}
p'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )’ &= \dfrac{ f'(x)\,g(x) – f(x)\,g'(x) }{ \left ( g(x) \right )^2 } \\[8px]
&= \dfrac{ \left ( \cos x \right )’ \left ( xe^x \right ) – \left ( \cos x \right ) \left ( xe^x \right )’ }{ \left ( x e^x \right )^2 } \\[8px]
&= \dfrac{ \left (- \sin x \right ) \left ( xe^x \right ) – \left ( \cos x \right ) \left ( e^x + xe^x \right ) }{ \left ( xe^x \right )^2 } \\[8px]
&= \dfrac{ -e^x \left ( x \sin x + \cos x + x \cos x \right ) }{ x^2 \left ( e^x \right )^2 } \\[8px]
&= – \dfrac{ x \sin x + \cos x + x \cos x }{ x^2 e^x } \implies \quad \text{ (D) } \quad \cmark
\end{align*}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #7
If $h(x)=\dfrac{f(x)}{g(x)} $ where f and g are shown in the figure, find $h'(0)$.
\begin{array}{lllll} \text{(A) } - \dfrac{2}{3} && \text{(B) } \dfrac{1}{9} && \text{(C) } 3
&& \text{(D) } 6 && \text{(E) } - \dfrac{1}{15} \end{array}
Show/Hide Solution
Taking the values for $f$ and $g$ at $x=0$ from the graph;
\[ f(0) = 6 \qquad \text{and} \qquad g(0) = -3 \]
Since the curve $y=f(x)$ is horizontal at $x=0$:
\[f'(0)=0\]
To find $g'(0),$ we compute the slope of $g$ at $x=0$ by using the convenient two points $(0,-3)$ and $(3,0)$. (You can use whichever two points on the line segment containing $x=0$ you wish; your slope value will be the same.)
\[g'(0) = \dfrac{ 0 – (-3) }{ 3 – 0 } = \dfrac{3}{3} = 1 \]
Using the Quotient Rule:
\begin{align*}
h'(x) = \left ( \dfrac{f}{g} \right )'(x) &= \dfrac{ f'(x)\,g(x) – f(x)\,g'(x) }{ \left ( g(x) \right )^2 } \\[8px]
h'(0) = \left ( \dfrac{f}{g} \right )'(0) &= \dfrac{ f'(0)\,g(0) – f(0)\,g'(0) }{ \left ( g(0) \right )^2 } \\[8px]
h'(0) &= \dfrac{(0)(-3)-(6)(1)}{ (-3)^2 } \\[8px]
&= – \dfrac{ 6 }{9} = – \dfrac{2}{3} \implies \quad \text{ (A) } \quad \cmark
\end{align*}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #8
If $h(x)= \dfrac{f(x)}{g(x)} $ where $f$ and $g$ are shown in the figure, find $h'(8)$.
\begin{array}{lllll} \text{(A) } \dfrac{5}{16} && \text{(B) } - \dfrac{9}{8} && \text{(C) } \dfrac{4}{5}
&& \text{(D) }- \dfrac{1}{4} && \text{(E) None of the above} \end{array}
Show/Hide Solution
Taking the values for $f$ and $g$ at $x=8$ from the graph;
\[f(8) = 9 \qquad \text{and} \qquad g(8) = -2 \]
To find $f'(8),$ we compute the slope of $f$ at $x=8$ using the convenient points $(2,6)$ and $(10,10)$:
\[f'(8)= \dfrac{ 10 – 6 }{ 10 – 2 } = \dfrac{ 4 }{ 8 } = \dfrac{1}{2} \]
To find $g'(8),$ we compute the slope of $g$ at $x=6$ using the convenient points $(0,0)$ and $(20,-5)$:
\[g'(8) = \dfrac{ -5 – 0 }{ 20 – 0 } = \frac{-5}{20} = – \dfrac{1}{4} \]
Using the Quotient Rule:
\begin{align*}
h'(x) = \left ( \dfrac{ f(x) }{ g(x) } \right )’ &= \dfrac{ f'(x)\,g(x) – f(x)\,g'(x) }{ \left ( g(x) \right )^2 } \\[8px]
h'(8) = \left ( \dfrac{ f }{ g } \right )'(8) &= \dfrac{ f'(8)\,g(8) – f(8)\,g'(8) }{ \left ( g(8) \right )^2 } \\[8px]
&= \dfrac{ \left ( \frac{1}{2} \right ) (-2) – (9) \left (- \frac{1}{4} \right ) }{ \left ( -2 \right )^2 } \\[8px]
&= \dfrac{ -1 + \frac{9}{4} }{4} \\[8px]
&= \dfrac{ – \frac{4}{4} + \frac{9}{4} }{4} = \dfrac{ \frac{5}{4} }{4} \\[8px]
&= \dfrac{5}{16} \implies \quad \text{ (A) } \quad \cmark
\end{align*}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #9
If $f(x) = \dfrac{ x^2 }{1-x^3}$, then find the equation for the line tangent to the curve $y=f(x)$ at $x=-1$.
\begin{array}{lllll} \text{(A) } y=- \dfrac{1}{3}x && \text{(B) } y=x-\dfrac{1}{2} && \text{(C) } y= \dfrac{7}{2}\left(x+3\right)
&& \\[8px] \text{(D) }y=\dfrac{1}{4}\left(1-x\right) && \text{(E) } y= - \dfrac{1}{8}(1+x) \end{array}
Show/Hide Solution
There are two steps to solving this problem: 1. Find the slope of the line tangent to the curve $y=f(x)$ at $x=-1$. 2. Write down the equation for the line in point slope form: $y-y_{0} = m_{ \text{tangent} }(x-x_{0})$.
Step 1: Find the slope $m_{ \text{tangent} } = f'(x)$ at $x=-1$;
\begin{align*}
m_{ \text{tangent} } = f'(x) &= \left ( \dfrac{x^2}{1-x^3} \right )’ \\[8px]
&= \frac{\left( x^2\right)’\left( 1-x^3\right) – x^2 \left(1-x^3 \right)’}{\left( 1-x^3\right)^2} \\[8px]
&= \dfrac{ \left( 2x \right ) \left ( 1-x^3 \right ) – \left (x^2 \right ) \left ( -3x^2 \right ) }{ \left ( 1-x^3 \right )^2 }
\end{align*}
Evaluating at the point of interest, $x=-1$:
\begin{align*}
m_{ \text{tangent} } = f'(-1) &= \dfrac{ (-2) (1+1) – ( 1 )( -3 ) }{ (1+1)^2} \\[8px]
&= \dfrac{ -4 + 3 }{4} \\[8px]
&= – \dfrac{1}{4} \\[8px]
\end{align*} Step 2: Find $y_{0}$ and write the equation for the line in Point Slope form, $y-y_{0} = m_{ \text{tangent} }(x-x_{0})$.
We need the $y$-value on the function’s curve at $x_0 = 1$:
\begin{align*}
y = f(x) &= \dfrac{ x^2 }{1-x^3} \\[8px]
y_{0} = f(-1) &= \dfrac{ 1 }{ 1+ 1 } = \dfrac{1}{2}
\end{align*}
Hence the line passes through (is tangent to) the point $\left(x_0, y_0 \right) = (-1, \dfrac{1}{2}),$ and has slope $m_{ \text{tangent} } = f'(-1) = -\dfrac{1}{4}.$ We can thus write the line’s equation in Point-Slope form as
\begin{align*}
y- \dfrac{1}{2} &= -\dfrac{1}{4}(x + 1) \\[8px]
y &= – \dfrac{1}{4}x – \dfrac{1}{4} + \dfrac{2}{4} \\[8px]
y &= \dfrac{1}{4} \left ( 1 – x \right ) \implies \quad \text{ (D) } \quad \cmark
\end{align*}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
Practice Problem #10
If $f(x)= \dfrac{- \sin x \cos x }{x^2 - 1}$, then find the equation normal to the curve $y=f(x)$ at $x=0$.
\begin{array}{lll} \text{(A) }y=-x && \text{(B) }y=-1 && \text{(C) } y=-(x+1) \end{array}
\begin{array}{ll} \text{(D) }y=x && \text{(E) }y=1-x \end{array}
Show/Hide Solution
There are two steps to solve this problem: 1. Find the slope of the line normal to the curve $y=f(x)$ at $x=0$. 2. Write down the equation for the line in Point Slope form: $y-y_{0}=m_{ \text{normal} }(x-x_{0})$.
Step 1: Find the slope $m_{ \text{normal} }= \dfrac{1}{-f'(x)}$ at $x=0$; Let’s first find $f'(x):$
\begin{align*}
f'(x) &= -\frac{(\sin x \cos x)’\left(x^2 – 1 \right) – (\sin x \cos x)\left(x^2 – 1 \right)’}{\left(x^2 – 1 \right)^2} \\[8px]
&= -\frac{[(\sin x )’\cos x + \sin x (\cos x)’]\left(x^2 – 1 \right) – (\sin x \cos x)(2x)}{\left(x^2 – 1 \right)^2} \\[8px]
&= -\frac{[(\cos x \cos x + \sin x (-\sin x)]\left(x^2 – 1 \right) – 2x\sin x \cos x}{\left(x^2 – 1 \right)^2} \\[8px]
&= – \frac{\left(\cos^2 x – \sin^2 x \right)\left(x^2 – 1 \right) – 2x\sin x \cos x}{\left(x^2 – 1 \right)^2} \\[8px]
\end{align*}
Evaluating at the point of interest, $x=0$:
\begin{align*}
f'(0) & = – \frac{\left(\cos^2 0 – \sin^2 0 \right)\left((0)^2 – 1 \right) – 2(0)\sin 0 \cos 0}{\left[(0)^2 – 1 \right]^2} \\[8px]
&= -\frac{(1-0)(-1) – 0}{1} = 1
\end{align*}
Hence the slope of the normal line at $x=1$ is $m_{ \text{normal} } = – \dfrac{1}{f'(1)}$:
\[m_{ \text{normal} } = – \dfrac{1}{f'(1)} = -1 \] Step 2: Find $y_{0}$ and write the equation for the line in Point Slope form, $y-y_{0} = m_{ \text{normal} }(x-x_{0})$. We need the $y$-value on the function’s curve at $x_0 = 0$:
\begin{align*}
y = f(x) &= – \dfrac{ \sin x \cos x }{x^2 – 1} \\[8px]
y_{0} = f(0) &= – \dfrac{0}{-1} = 0
\end{align*}
Hence the line passes through the point $\left(x_0, y_0 \right) = (0, 0)$ and has slope $m_{ \text{normal} } = – \dfrac{1}{f'(1)} = -1.$ We can then write the line’s equation in Point-Slope form:
\begin{align*}
y-0 &= – 1(x – 0) \\[8px]
y &= -x \implies \quad \text{ (A) } \quad \cmark
\end{align*}
If you’d like to track your learning on our site (for your own use only!), simply log in for free
with a Google, Facebook, or Apple account, or with your dedicated Matheno account. [Don’t have a dedicated Matheno account but would like one? Create your free account in 60 seconds.] Learn more about the benefits of logging in.
It’s all free: our only aim is to support your learning via our community-supported and ad-free site.
[hide solution]
The Upshot
The Quotient Rule as applied to two differentiable functions $f$ and $g$, and $g(x) \ne 0$ is
in prime notation: \[ \left[\frac{f(x)}{g(x)}\right]’ = \frac{f'(x)\,g(x)\, -\, f(x)\,g'(x)}{g(x)^2}\]
and in Leibniz notation: \[\dfrac{d}{dx}\left(\frac{f(x)}{g(x)} \right) = \frac{\left( \dfrac{d}{dx}f(x)\right)g(x)\, -\, f(x) \left(\dfrac{d}{dx}g(x) \right)}{g(x)^2}\]
In words, \[ =\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}}\]
The derivative of $\tan x$ is $\dfrac{d}{dx}\tan x = \sec^2 x.$
In the next section, we’ll develop a final Big Rule that we use to compute most derivatives: the Chain Rule. We’ll see what’s being “chained,” and will provide lots of practice so finding the derivative of complicated functions becomes automatic for you. For now, what do you think about the Quotient Rule? Easier, harder, or the same difficulty as using the Product Rule? Let us know over on the Forum! And if you have any problems that you’re working on and could use some help with, if you post we’ll do our best to assist. : )
If we've helped you learn something, perhaps you'd like to give back a little something and
☕ Buy us a coffeeWe're working to add more, and would appreciate your help to keep going! 😊
Once you log in with your free account, the site will record and then be able to recall for you:
Your answers to multiple choice questions;
Your self-chosen confidence rating for each problem, so you know which to return to before an exam (super useful!);
Your progress, and specifically which topics you have marked as complete for yourself.
Your selections are for your use only, and we do not share your specific data with anyone else. We do use aggregated data to help us see, for instance, where many students are having difficulty, so we know where to focus our efforts.
You will also be able to post any Calculus questions that you have on our Forum, and we'll do our best to answer them!
We believe that free, high-quality educational materials should be available to everyone working to learn well. There is no cost to you for having an account, other than our gentle request that you contribute what you can, if possible, to help us maintain and grow this site.
Please join our community of learners. We'd love to help you learn as well as you can!
We use cookies to provide you the best possible experience on our website. By continuing, you agree to their use.OK
If $h(x)=\dfrac{f(x)}{g(x)} $ where f and g are shown in the figure, find $h'(0)$.
\begin{array}{lllll} \text{(A) } - \dfrac{2}{3} && \text{(B) } \dfrac{1}{9} && \text{(C) } 3
&& \text{(D) } 6 && \text{(E) } - \dfrac{1}{15} \end{array}
What are your thoughts or questions?