Tutorial: The chain rule
Adaptive 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)
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 $\displaystyle {}= \frac{8}{3} \pi$ cm^{3}/#[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 $\displaystyle \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$
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
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
$\dfrac{dx}{dy} = \dfrac{1}{\Bigl(\dfrac{dy}{dx}\Bigr)}$, provided that $\dfrac{dy}{dx} \neq 0$.
\t
Notice again how $\dfrac{dy}{dx}$ behaves like a fraction.
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, provided that $\dfrac{dx}{dt} \neq 0$.
\t
The terms $dt$ appear to cancel.
%%Examples
1. %%If $y = 3x + 6$, %%then $\dfrac{dy}{dx} = 3$. %%Therefore,

$\dfrac{dx}{dy} = \dfrac{1}{\Bigl(\dfrac{dy}{dx}\Bigr)} = \dfrac{1}{3}$.

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{2t}{2} = t$.

$\dfrac{dy}{dx} = \dfrac{4x}{2} = \dfrac{x4}{2}$.
Now try the exercises in Section 4.4 in Applied Calculus or Section 11.4 in Finite Mathematics and Applied Calculus.
Copyright © 2020 Stefan Waner and Steven R. Costenoble