Tutorial: The chain rule

This tutorial: Part A: Chain rule: Basics

Go to Part B: Chain rule and Leibniz notation

(This topic is also in Section 4.4 in Applied Calculus or Section 11.4 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!]#

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:

last

* #[See the %%orderoperations for a discussion of order of operations][Ve el %%orderoperations para una discusión del orden de operactiones]#.

%%Examples

1. $(3x^2-4)(2x+1)$ 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. $\dfrac{2x-1}{x}$ 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. $(4x-1)(x+2) + x^2$ is calculated by first calculating the product $(4x-1)(x+2)$, 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. $(3x^2-1)^5$ 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

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

$\displaystyle {}= 3\color{indianred}{(2x + 3x^4)}^2 \frac{d}{dx}\color{indianred}{(2x + 3x^4)}$

$\dfrac{d}{dx}\text{(quantity)}^3 = 3\text{(quantity)}^2\dfrac{d}{dx}\text{(quantity)}$

\\ \t \t $\displaystyle {}=3(1+x^2)^2 \cdot 2x$ \\ \t \t $\displaystyle {}=6x(1+x^2)^2$

2. \gap[20] \t $\displaystyle \frac{d}{dx}[u^n] = nu^{n-1} \frac{du}{dx}$ \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]#

\\ \t \t

$\displaystyle {}=(2)(-3)\color{indianred}{(-x+x^3)}^{-4}\frac{d}{dx}\color{indianred}{(-x+x^3)}$

$\dfrac{d}{dx}\text{(quantity)}^{-3} = -3\text{(quantity)}^{-4}\dfrac{d}{dx}\text{(quantity)}$

\\ \t \t $\displaystyle {}=-6(-x+x^3)^{-4}(-1+3x^2)$ \\ \t \t $\displaystyle {}=\frac{-8(-1+3x^2)}{-x+x^3}$

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 $\displaystyle \frac{d}{dx}\sqrt{\color{indianred}{x^2-1}}$ \t

$\displaystyle {}=\frac{1}{2\sqrt{\color{indianred}{x^2-1}}}\frac{d}{dx}\color{indianred}{(x^2-1)}$

$\dfrac{d}{dx}\sqrt{\text{quantity}} = \frac{1}{2\sqrt{\text{quantity}}}\dfrac{d}{dx}(\text{quantity})$

\\ \t \t $\displaystyle {}=\frac{1}{2\sqrt{x^2-1}}\cdot (2x)$ \\ \t \t $\displaystyle {}=\frac{x}{\sqrt{x^2-1}}$

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 $\displaystyle \frac{d}{dx}|\color{indianred}{4x-5}|$ \t

$\displaystyle {}=\frac{|\color{indianred}{4x-5}|}{\color{indianred}{4x-5}}\frac{d}{dx}\color{indianred}{(4x-5)}$

$\dfrac{d}{dx}|\text{quantity}| = \dfrac{|\text{quantity}|}{\text{quantity}}\dfrac{d}{dx}(\text{quantity})$

\\ \t \t $\displaystyle {}=\frac{|4x-5|}{4x-5}\cdot (4)$ \\ \t \t $\displaystyle {}=\frac{4|4x-5|}{4x-5}$

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

$\displaystyle y = \left(\frac{x^2}{x-1} + \sqrt{3x^2-1}\right)^4$? %%A It is in cases like this that we go back to our old friend the CTE (Calculation Thought Experiment), which tells us which rule to use at each step. Following is a quiz on using the CTE for this purpose, just as we did in the %%prevsectut.

Now that you are an expert on taking derivatives of all sorts of things, here is a final quiz to challenge you.

Now try the exercises in Section 4.4 in Applied Calculus or Section 11.4 in Finite Mathematics and Applied Calculus.