Tutorial: The chain rule
This tutorial: Part A: Chain rule: Basics
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Warmup: The calculation throught experiment
#[Let us start with a quick review and quiz on using the "Calculation Thought Experiment (CTE)" discussed in the %%prevsectut:][Empecemos con un repaso y concurso rápido sobre el uso del "Experimento mental de cálculo (EMC)" descrito en el %%prevsectut:]#
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. Thus, when calculating its derviative, we use the product rule.
2. $7(2x+1)$ is calculated by first calculating the expression in parentheses and then multiplying by $7$, so it too is a product, and in particular a constant multiple as we are mutliplying by the constant $7$. We can therefore calculate its derivative using the constant multiple rule more easily than using the product rule.
3. $\dfrac{2x1}{x}$ 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, and so, in calculating its derivative, we use the product rule.
4. $(4x1)(x+2) + x^2$ 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, and thus we use the rule for sums when taking its derivative.
5. $(3x^21)^5$ 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. Below we see how to take derivatives of powers of functions other than $x$. Some for you
3. $\dfrac{2x1}{x}$ 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, and so, in calculating its derivative, we use the product rule.
4. $(4x1)(x+2) + x^2$ 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, and thus we use the rule for sums when taking its derivative.
5. $(3x^21)^5$ 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. Below we see how to take derivatives of powers of functions other than $x$. Some for you
Introducing the chain rule
%%Q So what, exactly, is the chain rule all about?%%A Here is an example: We know that the derivative of $x^3$ is $3x^2$. What then would you say is the derivative of something more complicated raised to the third power, like for instance $(2x + 3x^4)^3$? %%Q Duh! That's just $3(2x + 3x^4)^2$.
%%A No it isn't; the power rule applies only to powers of $x$, and not powers of more complicated things. In this example, you could find the derivative by first expanding
$(2x+3x^4)^3 = (2x+3x^4)(2x+3x^4)(2x+3x^4) = 8x^3+36x^6+54x^9+27x^12$
and then taking the derivative using the power rule on each term.
%%Q OK then; what about $(2x + 3x^4)^{100}$? Are we really expected to expand that just to find its derivative?%%A Luckily for us, no; and it is the chain rule that tells us how to quickly and easily calculate the derivatives of expressions like that.In what follows, think of, say, $(2x + 3x^4)^{100}$ as $u^{100}$, where $u$ is the "inside:" $u = 2x+3x^4$.
Chain rule
If $u$ is a differentiable function of $x$, and $f$ is a differentiable function of $u$, then $f$ is a differentiable function of $x$, and:
$\displaystyle \frac{d}{dx}[f(u)] = f'(u) \frac{du}{dx}$
Chain rule in words:
The derivative of $f$ of a quantity is the derivative of $f$, evaluated at that (original) quantity, times the derivative of the quantity.
%%Examples
\\ \t \t $\displaystyle {}=3(1+x^2)^2 \cdot 2x$
\\ \t \t $\displaystyle {}=6x(1+x^2)^2$
1. \gap[20] \t $\displaystyle \frac{d}{dx}[u^3] = 3u^2 \frac{du}{dx}$ \gap[30] #[Chain rule with][Relga de la cadena con]# $f(x) = x^3$
The derivative of a quantity cubed is 3 times the (original) quantity squared, times the derivative of the quantity.
Eg.
\t $\displaystyle \frac{d}{dx}[\color{indianred}{(2x + 3x^4)}^3]$ \t $\displaystyle {}= 3\color{indianred}{(2x + 3x^4)}^2 \frac{d}{dx}\color{indianred}{(2x + 3x^4)}$
$\dfrac{d}{dx}\text{(quantity)}^3 = 3\text{(quantity)}^2\dfrac{d}{dx}\text{(quantity)}$
2. \gap[20] \t $\displaystyle \frac{d}{dx}[u^n] = nu^{n1} \frac{du}{dx}$ \gap[30] Chain rule with $f(x) = x^n$
Generalized power rule: The derivative of a quanity raised to the power n is n times the (original) quantity raised to the n − 1, times the derivative of the quantity.
Eg.
\t $\displaystyle \frac{d}{dx}\left[\frac{2}{\color{indianred}{(x+x^3)}^3}\right]$ \t $\displaystyle {}= \frac{d}{dx}\left[2\color{indianred}{(x+x^3)}^{3}\right]$
#[Converted to %%powerform][Convertido en %%powerform]#
$\displaystyle {}=(2)(3)\color{indianred}{(x+x^3)}^{4}\frac{d}{dx}\color{indianred}{(x+x^3)}$
$\dfrac{d}{dx}\text{(quantity)}^{3} = 3\text{(quantity)}^{4}\dfrac{d}{dx}\text{(quantity)}$
3. \gap[20] \t $\dfrac{d}{dx}\sqrt{u} = \dfrac{1}{2\sqrt{u}}\dfrac{du}{dx}$ \gap[30] Chain rule with $f(x) = \sqrt{x}$
The derivative of the square root of a quantity is 1 over twice the square root of the (original) quantity, times the derivative of the quantity.
Eg.
\t $\displaystyle \frac{d}{dx}\sqrt{\color{indianred}{x^21}}$ \t $\displaystyle {}=\frac{1}{2\sqrt{\color{indianred}{x^21}}}\frac{d}{dx}\color{indianred}{(x^21)}$
$\dfrac{d}{dx}\sqrt{\text{quantity}} = \frac{1}{2\sqrt{\text{quantity}}}\dfrac{d}{dx}(\text{quantity})$
4. \gap[20] \t $\dfrac{d}{dx}u = \dfrac{u}{u}\dfrac{du}{dx}$ \gap[30] Chain rule with $f(x) = x$
The derivative of the absolute value of a quantity is the absolute value of the (original) quantity over the quantity, times the derivative of the quantity.
Eg.
\t $\displaystyle \frac{d}{dx}\color{indianred}{4x5}$ \t $\displaystyle {}=\frac{\color{indianred}{4x5}}{\color{indianred}{4x5}}\frac{d}{dx}\color{indianred}{(4x5)}$
$\dfrac{d}{dx}\text{quantity} = \dfrac{\text{quantity}}{\text{quantity}}\dfrac{d}{dx}(\text{quantity})$
Combining the rules for differentiation
%%Q How do we deal with more complicated expressions that are combinations of products or quotients but also appear to require the chain rule, like

$\displaystyle y = \left(\frac{x^2}{x1} + \sqrt{3x^21}\right)^4$?
Now try the exercises in Section 4.4 in Applied Calculus or Section 11.4 in Finite Mathematics and Applied Calculus.
Copyright © 2019 Stefan Waner and Steven R. Costenoble