Lesson

In this chapter, we use the relation

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\times\frac{\mathrm{d}u}{\mathrm{d}x}$`d``y``d``x`=`d``y``d``u`×`d``u``d``x`

This is the chain rule in the notation of Leibniz. We use the chain rule to differentiate functions that are a function of a function. In particular, given a function $y=f\left(u(x)\right)$`y`=`f`(`u`(`x`)), its derivative is given by the expression above.

It may be that we know how to express a variable $y$`y` as a function of $u$`u`. Thus, we can differentiate $y$`y` with respect to $u$`u` to obtain $\frac{\mathrm{d}y}{\mathrm{d}u}$`d``y``d``u`. We may also have an expression for $u$`u` as a function of $x$`x` and therefore we can find $\frac{\mathrm{d}u}{\mathrm{d}x}$`d``u``d``x`.

Putting these together, we obtain $\frac{\mathrm{d}y}{\mathrm{d}x}$`d``y``d``x` as the product of the two derivatives. That is, we find the rate of change of $y$`y` with respect to $x$`x`.

It can happen that we know how to write the derivatives $\frac{\mathrm{d}y}{\mathrm{d}t}$`d``y``d``t` and $\frac{\mathrm{d}x}{\mathrm{d}t}$`d``x``d``t` but we require $\frac{\mathrm{d}y}{\mathrm{d}x}$`d``y``d``x`. To make this fit into the framework of the chain rule as described above, we need an additional fact:

If we have the rate of change of $y$`y` with respect to $x$`x`, then the rate of change of $x$`x` with respect to $y$`y` is its reciprocal.

In the present example, we have $\frac{\mathrm{d}x}{\mathrm{d}t}$`d``x``d``t` and we need $\frac{\mathrm{d}t}{\mathrm{d}x}$`d``t``d``x` in order to use the chain rule. So, we use the fact that $\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}t}}$`d``t``d``x`=1`d``x``d``t`.

A continuous supply of ink is seeping onto a porous plane surface so that a circular spot forms and grows over time. The area of the spot is increasing at the rate of $0.5\text{cm}^2$0.5cm2 per second. However, the radius of the spot is increasing at a rate that reduces as the spot grows. Find an expression for the time-rate of change of the radius and deduce the rate of change when the radius is $6\text{cm}$6cm.

The first thing to do in problems of this kind is to define some variables so that the problem can be expressed algebraically.

Let $A$`A` be the area of the spot, let $r$`r` be its radius and let $t$`t` be the elapsed time. The goal is to find $\frac{\mathrm{d}r}{\mathrm{d}t}$`d``r``d``t` when $r=6$`r`=6.

We are given that $\frac{\mathrm{d}A}{\mathrm{d}t}=0.5$`d``A``d``t`=0.5, and we can make use of the fact that $A=\pi r^2$`A`=π`r`2 and hence, $\frac{\mathrm{d}A}{\mathrm{d}r}=2\pi r$`d``A``d``r`=2π`r`.

The three derivatives combine to form $\frac{\mathrm{d}A}{\mathrm{d}r}.\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{\mathrm{d}A}{\mathrm{d}t}$`d``A``d``r`.`d``r``d``t`=`d``A``d``t`. That is, $2\pi r\times\frac{\mathrm{d}r}{\mathrm{d}t}=0.5$2π`r`×`d``r``d``t`=0.5. On rearranging, this is $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{0.5}{2\pi r}=\frac{1}{4\pi r}$`d``r``d``t`=0.52π`r`=14π`r`.

Therefore, when $r=6$`r`=6, we have $\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{1}{24\pi}\approx0.01\text{cm}/\text{s}$`d``r``d``t`=124π≈0.01cm/s.

The fuel supply to a certain rocket engine is regulated in such a way that the rocket travels with a constant acceleration of $20\text{m/}s^2$20m/`s`2 for $90$90 seconds after launch. What is the rate of change of velocity with respect to displacement when the velocity reaches $1000\text{km/s}$1000km/s?

Let $a$`a` be the acceleration, $v$`v` the velocity, $s$`s` the displacement, and let $t$`t` be the elapsed time. We are asked to find $\frac{\mathrm{d}v}{\mathrm{d}s}$`d``v``d``s` when $v=1000$`v`=1000.

Acceleration is the time rate of change of velocity. In symbols, $a=\frac{\mathrm{d}v}{\mathrm{d}t}$`a`=`d``v``d``t`. But, according to the chain rule, this is $\frac{\mathrm{d}v}{\mathrm{d}s}.\frac{\mathrm{d}s}{\mathrm{d}t}$`d``v``d``s`.`d``s``d``t`. We recall that velocity is the time rate of change of displacement, $v=\frac{\mathrm{d}s}{\mathrm{d}t}$`v`=`d``s``d``t`. It follows that an alternative characterisation of acceleration is $a=v\frac{\mathrm{d}v}{\mathrm{d}s}$`a`=`v``d``v``d``s`.

Therefore, $a=20=\frac{\mathrm{d}v}{\mathrm{d}s}\times1000$`a`=20=`d``v``d``s`×1000 and so, $\frac{\mathrm{d}v}{\mathrm{d}s}$`d``v``d``s` at $v=1000$`v`=1000 is $\frac{1}{50}\ \text{s}^{-1}$150 s−1. This means that for a brief time the velocity increases by $\frac{1}{50}\ \text{m/s}$150 m/s in the space of one metre.

The volume $V$`V` of oxygen in a scuba diver’s oxygen cylinder is given by $V=\frac{22}{P}$`V`=22`P`, where $P$`P` is the pressure inside the tank.

Find the rate of change of $V$

`V`with respect to $P$`P`.During a dive, the pressure $P$

`P`inside the cylinder increases at $0.5$0.5 units per second. Find the rate of change of the volume of oxygen when $P=2$`P`=2.Let $t$

`t`represent time in seconds.

A spherical hot air balloon, whose volume and radius at time $t$`t` are $V$`V` m^{3} and $r$`r` m respectively, is filled with air at a rate of $4$4 m^{3}/min.

At what rate is the radius of the balloon increasing when the radius is $2$2 m?

A point moves along the curve $y=5x^3$`y`=5`x`3 in such a way that the $x$`x`-coordinate of the point increases by $\frac{1}{5}$15 units per second.

Let $t$`t` be the time at which the point reaches $\left(x,y\right)$(`x`,`y`).

Find the rate at which the $y$`y`-coordinate is changing with respect to time when $x=9$`x`=9.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems