When two quantities are linked by a formula, their rates of change are linked too — the chain rule is the bridge between them.
9 min read4 worked examples
Connecting two rates via the chain rule
Many real situations involve two quantities that change over time and are related by a formula — the volume and radius of an inflating balloon, say. If a rate involving one quantity is known, the chain rule connects it to a rate involving the other, without needing an explicit formula for either quantity in terms of time.
Here $\dfrac{dV}{dr}$ comes from differentiating the formula connecting $V$ and $r$ (e.g. $V=\dfrac{4}{3}\pi r^3$ for a sphere), while $\dfrac{dr}{dt}$ or $\dfrac{dV}{dt}$ is given by the context of the problem. The chain rule links whichever rate is known to whichever rate is wanted.
The general strategy
Identify the two quantities linked by a formula (e.g. $V$ and $r$), write down that connecting formula, differentiate it to get $\dfrac{dV}{dr}$, then multiply or divide by the given rate using the chain rule to reach the rate that's wanted. Time, $t$, is almost always the variable both original rates are measured against.
Setting up a differential equation from a description
Some problems describe how a rate of change behaves in words, and the first task is to translate that description into an equation involving a derivative — a differential equation — rather than to calculate a numerical rate directly.
For example, "the rate of decrease of the radius $r$ is inversely proportional to the square of the radius" translates piece by piece: "rate of decrease of $r$" is $-\dfrac{dr}{dt}$ (negative, since $r$ is decreasing); "inversely proportional to $r^2$" means it equals $\dfrac{k}{r^2}$ for some constant $k$. Combining:
Constructing a differential equation like $\dfrac{dr}{dt}=-\dfrac{k}{r^2}$ from a worded description is the skill covered here. Actually solving such an equation — finding $r$ explicitly as a function of $t$ — requires separating the variables and integrating, which is covered later in the Integration topic's "Differential Equations" lesson.
Worked example 1
Inflating balloon — rate of change of volume
A spherical balloon is inflated so that its radius increases at a constant rate of $0.5\text{ cm/s}$. Find the rate of increase of its volume when the radius is $10\text{ cm}$. (Volume of a sphere: $V=\dfrac{4}{3}\pi r^3$.)
MethodConnect dV/dr and dr/dt via the chain rule
1
$\dfrac{dV}{dr}=4\pi r^2$, from differentiating $V=\dfrac{4}{3}\pi r^3$.
Differentiate the connecting formula with respect to r.
At $r=10$: $\dfrac{dV}{dt}=2\pi(10)^2=200\pi\approx628\text{ cm}^3\text{/s}$.
Substitute the given radius.
▪ solved
Worked example 2
Expanding puddle — working backwards from area
Oil spilled on a flat surface spreads out in a circle. The area of the circle is increasing at a constant rate of $8\text{ cm}^2\text{/s}$. Find the rate of increase of the radius when the radius is $4\text{ cm}$.
MethodConnect dA/dr and dA/dt, then rearrange for dr/dt
1
$A=\pi r^2$, so $\dfrac{dA}{dr}=2\pi r$.
Differentiate the connecting formula for a circle's area.
2
Given: $\dfrac{dA}{dt}=8$. By the chain rule, $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}$, so $8=2\pi r\times\dfrac{dr}{dt}$.
Set up the chain rule connection with the known rate on the left.
At $r=4$: $\dfrac{dr}{dt}=\dfrac{4}{4\pi}=\dfrac{1}{\pi}\approx0.318\text{ cm/s}$.
Substitute the given radius.
▪ solved
Look deeper
Even though the area grows at a constant rate, the radius does not — $\dfrac{dr}{dt}=\dfrac{4}{\pi r}$ shrinks as $r$ grows. This makes geometric sense: a fixed increase in area spreads over an ever-larger circumference as the puddle grows, so each additional bit of area buys a smaller increase in radius.
Worked example 3
Setting up a differential equation from a description
A tank is draining so that the rate of decrease of the volume of water, $V$, is proportional to the square root of the volume remaining. Write down a differential equation for $\dfrac{dV}{dt}$.
MethodTranslate the description phrase by phrase
1
"Rate of decrease of $V$" means $-\dfrac{dV}{dt}$.
A decreasing quantity has a negative dV/dt, so "rate of decrease" is its negative.
2
"Proportional to the square root of $V$" means it equals $k\sqrt{V}$ for some constant $k>0$.
Direct proportionality to √V translates to a constant multiple of √V.
3
Combining: $-\dfrac{dV}{dt}=k\sqrt{V}$.
Set the two translated phrases equal.
4
$\dfrac{dV}{dt}=-k\sqrt{V}$.
Rearrange to the standard form, with dV/dt as the subject.
▪ solved
Not solved here
This equation isn't solved for $V$ in terms of $t$ in this lesson — that requires separating variables and integrating, which is covered in the Integration topic.
Worked example 4
Connected rates with two chained formulas · Challenge
A cone-shaped pile of sand has height always equal to twice its base radius ($h=2r$), and its volume increases at a constant rate of $30\text{ cm}^3\text{/s}$. Find the rate of increase of the height when $h=12\text{ cm}$. (Volume of a cone: $V=\dfrac{1}{3}\pi r^2h$.)
MethodEliminate r to connect V and h directly, then chain to dh/dt
1
Since $h=2r$, $r=\dfrac{h}{2}$. Substitute into $V=\dfrac{1}{3}\pi r^2h$: $V=\dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2h=\dfrac{\pi h^3}{12}$.
Rewrite V purely in terms of h, using the given relationship between h and r.
Given $\dfrac{dV}{dt}=30$. By the chain rule, $30=\dfrac{\pi h^2}{4}\times\dfrac{dh}{dt}$.
Connect the known rate to the wanted rate via dV/dh.
4
$\dfrac{dh}{dt}=\dfrac{120}{\pi h^2}$.
Rearrange to isolate dh/dt.
5
At $h=12$: $\dfrac{dh}{dt}=\dfrac{120}{\pi(144)}=\dfrac{120}{144\pi}=\dfrac{5}{6\pi}\approx0.265\text{ cm/s}$.
Substitute the given height and simplify.
▪ solved
Look deeper
Because $r$ and $h$ were linked by a fixed ratio, $V$ could be rewritten purely in terms of $h$ before differentiating — turning a three-variable problem ($V$, $r$, $h$) into the same two-variable chain-rule pattern as Examples 1 and 2. Spotting a hidden constraint like $h=2r$ is often the key step in harder connected-rates problems.
Common mistake
Forgetting the negative sign when a quantity is decreasing — "rate of decrease of $V$" is $-\dfrac{dV}{dt}$, not $\dfrac{dV}{dt}$ itself, so setting up the equation without that sign gives $\dfrac{dV}{dt}$ the wrong sign throughout. Also common: multiplying the two rates in the wrong order, or forgetting to substitute a given numerical value into the correct variable (e.g. substituting into $r$ when the derivative $\dfrac{dV}{dr}$ was left in terms of $r$, not $t$).
Check yourself
A cube's side length $x$ increases at a constant rate of $2\text{ mm/s}$. Find the rate of increase of the cube's volume when $x=5\text{ mm}$. (Hint: connect dV/dx and dx/dt via the chain rule, using $V=x^3$.)
Method
1
$\dfrac{dV}{dx}=3x^2$, from differentiating $V=x^3$.