When two quantities are linked by an equation, their rates of change are linked too. Differentiating the equation reveals how.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Why the Chain Rule appears
Differentiating a term like V = s³ with respect to time gives 3s² · ds/dt. The inner derivative is the rate you are tracking.
The method
Write the relationship, differentiate both sides with respect to time, then substitute the known values and solve.
Substitute last
Numbers that change must stay as variables until after differentiating. Substituting early freezes a moving quantity into a constant.
Signs carry meaning
A negative rate means the quantity is shrinking. A draining tank has a negative dV/dt.
Draw it first
Label what changes and what stays fixed. The fixed lengths are constants; the moving ones are functions of time.
One thing changes, so another must
When two quantities are related by an equation, their rates of change are related by the derivative of that equation. Differentiating the relationship with respect to time links the two rates.
The procedure
Draw and label; write the equation relating the quantities; differentiate both sides with respect to time; substitute the known values last. Substituting before differentiating is the classic mistake — it freezes the variable you needed to vary.
The chain rule does the work
Differentiating x² with respect to t gives 2x·dx/dt, not 2x. Every variable that depends on time picks up its own rate factor. Forgetting one is what makes related rates feel harder than it is.
Check the sign and the units
A shrinking quantity has a negative rate. If a problem says the radius is decreasing, dr/dt must be entered negative. Checking that the answer's units and sign match the situation catches most errors.
Step 2: Try It Yourself
Tap and try it out.
- Point(2, 8)
- Slope of the tangent12
Step 3: Watch an Example
One step at a time.
Watch Ines Track a Growing Cube
A cube’s side grows at 2 cm/s. Ines needs the rate of volume change when the side is 5 cm.
- Step 1
The relationship is V = s³, where both V and s change with time.
Step 4: Your Turn
Practice makes it stick.
The Cube
Problem 1 of 2
dV/dt = 3s² · ds/dt. With s = 2 and ds/dt = 1, what is dV/dt?
The Circle
Problem 2 of 2
A circle’s area is A = πr². dA/dt = 2πr · dr/dt. With r = 3 and dr/dt = 1, what is dA/dt divided by π?
Linked Rates
1 of 8
dV/dt = 3s² · ds/dt. With s = 4 and ds/dt = 2, what is dV/dt?
2 of 8
dA/dt = 2πr · dr/dt. With r = 5 and dr/dt = 2, what is dA/dt divided by π?
3 of 8
A tank drains. Is dV/dt positive or negative? 1 positive, 2 negative.
4 of 8
A square’s area is A = s². dA/dt = 2s · ds/dt. With s = 6 and ds/dt = 3, what is dA/dt?
5 of 8
Should a changing quantity be substituted before differentiating? 1 for yes, 0 for no.
6 of 8
dV/dt = 3s² · ds/dt. With s = 1 and ds/dt = 5, what is dV/dt?
7 of 8
Put the related rates method in order.
- 1Write an equation relating the quantities.
- 2Differentiate both sides with respect to time.
- 3Substitute the known values and solve.
- 4Draw the situation and label what changes.
8 of 8
dA/dt = 2s · ds/dt. With s = 10 and ds/dt = 1, what is dA/dt?
Step 5: Quick Check
Show what you know.
Question 1 of 2
dV/dt = 3s² · ds/dt. With s = 3 and ds/dt = 2, what is dV/dt?
Question 2 of 2
Why must substitution wait until after differentiating?
What You Learned
- Differentiating a relationship with respect to time links the rates.
- The Chain Rule supplies the rate factor on every changing variable.
- Substitute known values only after differentiating.