Tutorial: The product and quotient rules
(This topic is also in Section 4.3 in Applied Calculus or Section 11.3 in Finite Mathematics and Applied Calculus)) #[I don't like this new tutorial. Take me back to the old tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo!]#
How not to differentiate products and quotients
In the #[tutorial on derivatives of powers, sums, and constant multiples][tutorial sobre derivadas de potencias, sumas, y múltiples constantes]# #[we saw that the derivative of a sum (or difference) is the sum (or difference) of the derivatives, so it is natural to ask whether the same is true for products and quotients: Is the derivative of a product (or quotient) the product (or quotient) of the derivatives?][vimos que la derivada de una suma (o resta) es la suma (o resta) de las derivadas, por lo que es natural pregunar si lo mismo es válido para los productos y los cocientes: ¿És la derivada de un producto (o cociente) el producto (o cociente) de las derivadas?]#
#[The following warmup quiz puts this hypothesis to the test][La siguiente prueba de calentamiento pone esta hipótesis a prueba.]#
%%Q So how do we calculate the derivatives of products and quotients?%%A That is exactly the purpose of this tutorial.
The product and quotient rules
Product rule
If $f$ and $g$ are differentiable functions, then so is their product $fg$, and

$\displaystyle \frac{d}{dx}[f(x)\,g(x)] = \color{blue}{f'(x)}\,g(x) + f(x)\,\color{blue}{g'(x)}$
 The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.
%%Examples
1. \t $\displaystyle \frac{d}{dx}[ (3x^3 + 7)(4x^2  x + 4) ]$ \t $\displaystyle {}=$ #[Deriv of first][Deriv del primero]# × #[Second][Segundo]# + #[First][Primero]# × #[Deriv of second][Deriv del segundo]#
\\ \t \t $\displaystyle {}=\color{blue}{(9x^2)}(4x^2  x + 4) + (3x^3 + 7)\color{blue}{(8x  1)}$
\\ 2. \t $\displaystyle \frac{d}{dx}\left[ x(x^2 + 4)\right]$ \t !2! $\displaystyle {}=$ #[Deriv of first][Deriv del primero]# × #[Second][Segundo]# + #[First][Primero]# × #[Deriv of second][Deriv del segundo]#
\\ \t \t $\displaystyle {}=\color{blue}{\frac{x}{x}}(x^2 + 4) + x\color{blue}{(2x)} \qquad $ \t $\displaystyle \frac{d}{dx}x = \frac{x}{x}$
(#[See the][Ve el]# .) \\ \t \t !2! $\displaystyle {}=x\left(x+\frac{4}{x} + 2x\right) = x\left(3x+\frac{4}{x}\right) $
%%Note In some circumstances, it is helpful to simplify the answer; you should use your best judgment in each situation to decide how much to simplify. Following are some for you to try. You need not simplify the answers for these.
(#[See the][Ve el]# .) \\ \t \t !2! $\displaystyle {}=x\left(x+\frac{4}{x} + 2x\right) = x\left(3x+\frac{4}{x}\right) $
Quotient rule
If $f$ and $g$ are differentiable functions, then so is their quotient $f/g$ whenever the denominator is nonzero, and

$\displaystyle \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{\color{blue}{f'(x)}\,g(x)  f(x)\,\color{blue}{g'(x)}}{g(x)^2}$.
 The derivative of a quotient is the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.
%%Examples
1. \t $\displaystyle \frac{d}{dx}\left[\frac{3x^3 + 7}{4x^2  x + 4}\right]$ \t #[$\displaystyle {}=\frac{\color{blue}{\text{Deriv of top }} \times \text{ Bottom }  \text{ Top }\times \color{blue}{\text{ Deriv of bottom }}}{\text{Bottom}^2}$][$\displaystyle {}=\frac{\color{blue}{\text{Deriv del Numer }} \times \text{ Denom }  \text{ Numer }\times \color{blue}{\text{ Deriv del denom }}}{\text{Denom}^2}$]#
\\ \t \t $\displaystyle {}= \frac{(\color{blue}{9x^2})(4x^2x+4) \ \  \ \ (3x^3 + 7)(\color{blue}{8x1})}{(4x^2  x + 4)^2}$
\\ 2. \t $\displaystyle \frac{d}{dx}\left[\frac{x}{x^2 + 4}\right]$ \t !2! #[$\displaystyle {}=\frac{\color{blue}{\text{Deriv of top }} \times \text{ Bottom }  \text{ Top }\times \color{blue}{\text{ Deriv of bottom }}}{\text{Bottom}^2}$][$\displaystyle {}=\frac{\color{blue}{\text{Deriv del Numer }} \times \text{ Denom }  \text{ Numer }\times \color{blue}{\text{ Deriv del denom }}}{\text{Denom}^2}$]#
\\ \t \t $\displaystyle {}=\frac{\color{blue}{\frac{x}{x}}(x^2 + 4)  x\color{blue}{(2x)}}{(x^2+4)^2} \qquad $ \t #[This looks ugly so let's simplify:][Esto se ve feo, así que simplifiquemos:]#
\\ \t \t !2! $\displaystyle {}=\frac{x\left(x+\frac{4}{x}  2x\right)}{(x^2+4)^2}$
\\ \t \t $\displaystyle = \frac{x\left(x+\frac{4}{x}\right)}{(x^2+4)^2}$
\\ \t \t $\displaystyle = \frac{x\left(4x^2\right)}{x(x^2+4)^2}$ \t #[A lot nicer!][¡Mucho mejor!]#
Some for you
%%A You can find a proof of the product rule in Section 4.3 in Applied Calculus or Section 11.3 in Finite Mathematics and Applied Calculus) #[Press here for a proof of the quotient rule.][Clic aquí para una prueba de la regla del cociente.]#
Calculation thought experiment
%%Q This is all very well if what you're given is an obvious product or quotient. But we all know that instructors are fond of "inbetween" expressions, such as $\displaystyle (3x + 1)\frac{x^2 + 4}{x^2  3}$,
%%A To deal with things like that—or any mathematical expression whatsoever, we use the following little secret described in Section 4.3 in Applied Calculus or Section 11.3 in Finite Mathematics and Applied Calculus), called the Calculation Thought Experiment:
Calculation thought experiment (CTE)
The calculation thought experiment is a technique to determine whether to treat an algebraic expression as a product, quotient, sum, difference, or power:

Given an expression, consider the operations you would use in computing its value following the standard order of operations.* If the last operation is multiplication, treat the expression as a product; if the last operation is division, treat the expression as a quotient, and so on.
%%Examples
1. $(3x^24)(2x+1)$ is calculated by first calculating the expressions in parentheses and then multiplying. Since the last step is multiplication, we treat the expression as a product.
2. $\dfrac{2x1}{x}$ is calculated by first calculating the numerator and denominator separately, and then dividing one by the other. Since the last step is division, we treat the expression as a quotient.
3. $(4x1)(x+2) + x^2$ is calculated by first calculating the product $(4x1)(x+2)$, then calculating $x^2$, and finally adding the two answers. Since the last step is addition, we treat the expression as a sum.
4. $(3x^21)^5$ is calculated by first calculating the expression in parentheses, and then raising the answer to the fifth power. Since the last step is raising to a power, we treat the expression as a power. Derivatives of powers of functions other than $x$ are in the %%nexttut. Some for you
2. $\dfrac{2x1}{x}$ is calculated by first calculating the numerator and denominator separately, and then dividing one by the other. Since the last step is division, we treat the expression as a quotient.
3. $(4x1)(x+2) + x^2$ is calculated by first calculating the product $(4x1)(x+2)$, then calculating $x^2$, and finally adding the two answers. Since the last step is addition, we treat the expression as a sum.
4. $(3x^21)^5$ is calculated by first calculating the expression in parentheses, and then raising the answer to the fifth power. Since the last step is raising to a power, we treat the expression as a power. Derivatives of powers of functions other than $x$ are in the %%nexttut. Some for you
Using the calculation thought experiment (CTE) to differentiate a function
If the CTE says, for instance, that the function is written a sum of two smaller expressions, then apply the rule for sums as a first step. This will leave you having to differentiate the two smaller expressions, and you can then use the CTE on these, and so on...
%%Example
Let us use the CTE to find the derivative of

$\displaystyle f(x) = (3x + 1)\frac{x^2 + 4}{x^2 + x}$.

$f(x) = (3x + 1) \times \frac{x^2 + 4}{x^2 + x}$.
$\displaystyle f'(x)$ \t ${}={}$ \t
Remember that the expressions "$\frac{d}{dx}$" are shorthand for "the derivative of ..." In other words, we haven't finished the job yet; the line (I) above tells us what we still need to do: take two derivatives. (If we wanted, we could take a coffee break at this point and come back to it later to do the work, as the instructions as to what to do next are encoded in what we did above.) To finish the calculation, we must compute the pink and bluecolored derivatives oneatatime and plug them in to the expression (I):
#[The derivative of the first (pink) expression is easy:][La derivada de la primera expresión (rosita) es fácil]#:
$\displaystyle \frac{d}{dx}(3x+1)$
$\displaystyle {}\times \ \frac{x^2 + 4}{x^2 + x}$ \t $\ \ + \ \ $ \t $\displaystyle (3x+1) \ \times \ $ \t $\displaystyle \frac{d}{dx}\left[ \frac{x^2 + 4}{x^2 + x}\right]$
\t ... (I)
\\ \t \t #[Deriv of first][Deriv del primero]# $ \ \times \ $ #[Second][Segundo]# \t $\ \ + \ \ $ \t !r! #[First][Primero]# $ \ \times \ $\t #[Deriv of second][Deriv del segundo]#
$\displaystyle \frac{d}{dx}(3x + 1)$
$\displaystyle {} = 3$
$\displaystyle \frac{d}{dx}\left[\frac{x^2 + 4}{x^2 + x}\right]$
$\displaystyle {}=\frac{(2x)(x^2+x)  (x^2+4)(2x+1)}{(x^2 + x)^2} = \frac{x^2  8x  4}{(x^2 + x)^2}$

$\displaystyle f'(x) = 3\frac{x^2 + 4}{x^2 + x} + (3x + 1)\left[\frac{x^2  8x  4}{(x^2 + x)^2}\right]$
Now try the exercises in Section 4.3 in Applied Calculus or Section 11.3 in Finite Mathematics and Applied Calculus).
Copyright © 2019 Stefan Waner and Steven R. Costenoble