Tutorial: The chain rule

Game version

This tutorial: 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)

Adaptive game tutorial
This adaptive game tutorial will help you master this topic in a way that adapts to your ability as you practice the questions.
If this is your first time studying the material, you may find yourself requiring more help, and as a result your game scores could end up lower, or you could even "die." Do not be discouraged! Just press "New game" to play the game as many times as you need in order to require less help and improve your scores.
You can also try the non-game version (link on the side), which does not score you, is not randomized or adaptive and gives you all the answers, but will not give you additional practice at the level you might need.
To complete this game you must answer all the questions correctly; you will be allowed to have a number of tries for some questions.
The questions are randomized: Expect to see lots of differences differences every time you load the page.
Press "Scores" at any time to show the scores you currently have.
The game is automatically saved. Relaunching the game on the same computer with the same browser will bring up the saved version.
Press "New game" to discard the saved game and start a new game.
Caution: Clicking on the little pictures that appear from time to time on the left can have unexpected consequences! You might want to experiment with them; this is a game after all!

Review of Leibniz d notation

If $y = f(x)$, we have used the notation $f'(x)$ or $y'(x)$ for the derivative of $y$ at $x$ and also, in %%partAtut and the %%powerruletut, we also used "differential notation" $\dfrac{d}{dx}[f(x)]$. There is another notation related to differential notation: Recall from the %%averratetut that we can write the average rate of change of $y$ over the interval $\Delta x$ as

Average rate of change = $\displaystyle \frac{\Delta y}{\Delta x}$. \t \gap[40] $\displaystyle \frac{\text{Change in }y}{\text{Change in }x}$

As we use smaller and smaller values for $\Delta x$, we approach the instantaneous rate of change, or derivative, for which we also have the notation $\dfrac{dy}{dx}$, due to Leibniz:

Instantaneous rate of change = $\displaystyle \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}$.

That is, $\dfrac{dy}{dx}$ is just another notation for $y'(x)$. Do not think of $\dfrac{dy}{dx}$ as an actual quotient of two numbers: Remember that we use an actual quotient $\dfrac{\Delta y}{\Delta x}$ only to approximate the value of $\dfrac{dy}{dx}$.

%%Q If $\dfrac{dy}{dx}$ shouldn't be thought of as a quotient, why is it written like that?
%%A Originally, Lebniz himself defined it to be an actual quotient; not of real numbers, but of infinitely small numbers: $dy$ and $dx$ were infinitely small changes in $x$ and $y$. However, the theory of infinitely small numbers was not mathematically rigorous (although it has been made rigorous more recently) and was replaced by the notion of a limits as we now use.

Chain rule in Leibniz d notation

Chain rule (Leibniz notation)

If $y$ is a differentiable function of $u$ and, in turn, $u$ is a differentiable function of $x$, then $y$ is a differentiable function of $x$, and:

$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$ \gap[30] \t Notice how the "quantities" $du$ appear to cancel.

%%Examples

1. %%If $y = u^3$ %%and $u = 2x + 3x^4$, %%then

$\displaystyle \frac{dy}{dx}$ \t $\displaystyle {}= \frac{dy}{du} \frac{du}{dx}$ \gap[30] \t $y$ is a function of $u$ and $u$ is a function of $x$. \\ \t $\displaystyle {}= 3u^2(2+12x^3)$ \\ \t $\displaystyle {}= 3(2x + 3x^4)^2(2+12x^3)$ \gap[30] \t Substitute for $u$ to express the answer in terms of $x$.

The derivative of a quantity cubed is 3 times the (original) quantity squared, times the derivative of the quantity.

2. %%If $A = \pi r^2$ %%and $r = 3t+1$, %%then

$\displaystyle \frac{dA}{dt}$ \t $\displaystyle {}= \frac{dA}{dr} \frac{dr}{dt}$ \gap[30] \t $A$ is a function of $r$ and $r$ is a function of $t$. \\ \t $\displaystyle {}= 2\pi r(3) = 6\pi r$ \\ \t $\displaystyle {}= 6\pi (3t+1)$ \gap[30] \t Substitute for $r$ to express the answer in terms of $t$.

If the radius of a disc is changing with time $t$ according to $r = 3t+1$, then how fast is the area changing? Answer: $6\pi (3t+1)$.

3. A spherical balloon is being inflated in such a way that its radius after $t$ seconds is $r = t^{1/3}$ cm. How fast is the volume increasing?

We are looking for the rate of change of the volume $V$, $\dfrac{dV}{dt}$, given that its radius is $r = t^{1/3}$. We also know that the volume of a sphere is given by $V = \dfrac{4}{3}\pi r^3$, giving us $V$ as a function of $r$. Thus,

$\displaystyle \frac{dV}{dt}$ \t $\displaystyle {}= \frac{dV}{dr} \frac{dr}{dt}$ \gap[30] \t $V$ is a function of $r$ and $r$ is a function of $t$. \\ \t $\displaystyle {}= 4\pi r^2\left(\frac{2}{3t^{2/3}}\right)$ \\ \t $\displaystyle {}= 4\pi t^{2/3}\left(\frac{2}{3t^{2/3}}\right)$ \gap[30] \t Substitute for $r$ to express the answer in terms of $t$. \\ \t $\displaystyle {}= \frac{8}{3} \pi$ cm³/#[second][segundo]#. \t Simplify.

Manipulating derivatives in differential notation

Although we, and most likely your calculus instructors, have told you repeatedly that $\dfrac{dy}{dx}$ is not a fraction (the fact that Leibniz himself regarded it as a fraction notwithstanding!) the chain rule

$\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$ suggests that it certainly behaves as though it is a fraction; even more so when we look at some consequences of the chain rule!

Manipulating derivatives in differential notation

Suppose $y$ is a function of $x$. Then, thinking of $x$ as a function of $y$ (as, for instance, when we can solve for $x$), we have

$\dfrac{dx}{dy} = \dfrac{1}{\Bigl(\dfrac{dy}{dx}\Bigr)}$, provided that $\dfrac{dy}{dx} \neq 0$.

Notice again how $\dfrac{dy}{dx}$ behaves like a fraction.

Suppose $y$ and $x$ are both functions of $t$. Then, thinking of $y$ as a function of $x$ (as, for example, when we can solve for $t$ as a function of $x$, and hence obtain $y$ as a function of $x$), we have

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, provided that $\dfrac{dx}{dt} \neq 0$.

The terms $dt$ appear to cancel.

%%Examples

1. %%If $y = -3x + 6$, %%then $\dfrac{dy}{dx} = -3$. %%Therefore,

$\dfrac{dx}{dy} = \dfrac{1}{\Bigl(\dfrac{dy}{dx}\Bigr)} = -\dfrac{1}{3}$. 2. %%If $x = 4-2t$ and $y = t^2$, %%then

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{-2} = -t$. As we are thinking of $y$ as a function of $x$ rather than $t$, we would like to have the answer in terms of $x$ also. Solving the first given equation $x = 4-2t$ for $t$ gives $t = \dfrac{4-x}{2}$, so

$\dfrac{dy}{dx} = -\dfrac{4-x}{2} = \dfrac{x-4}{2}$.

Some for you

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