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

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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:

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.

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 $g(x) = x.$

The Quotient Rule says $\left[\dfrac{f(x)}{g(x)}\right]’ = \dfrac{f'(x)\,g(x) – f(x)\,g'(x)}{g(x)^2},$ so:

\begin{align*}

\left[ \frac{e^x}{x}\right]’ &= \frac{(e^x)’\cdot x\, -\, e^x \cdot (x)’}{x^2} \\[8px]
&= \frac{e^x \cdot x\, -\, e^x \cdot 1}{x^2} \\[8px]
&= \frac{x e^x\, -\, e^x}{x^2} \quad \cmark

\end{align*}

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:

\[\dfrac{d}{dx}\tan x = \sec^2 x\]

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!

Show/Hide Statement of the 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, $\left[\dfrac{f(x)}{g(x)}\right]’$

\[=\dfrac{{[{\small \text{(derivative of the numerator) } \times \text{ (the denominator)}]}\\ \quad – \, [{\small \text{ (the numerator) } \times \text{ (derivative of the denominator)}}]}}{{\small \text{all divided by [the denominator, squared]}}} \]

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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}

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}

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}

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}

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}

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}

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}

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}

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}

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}

- 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. : )

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