chain rule

Related Rates: Understanding How Changing Quantities Influence One Another

August 17, 2026 by Splendid Leave a Comment

One of the most fascinating ideas in calculus is that many quantities in the real world do not change independently.

When one quantity changes, another quantity often changes with it.

As a balloon expands, its radius changes and its volume changes.

As a ladder slides down a wall, its height changes and its distance from the wall changes.

As a company’s production increases, inventory, costs, and revenue change simultaneously.

Calculus provides a powerful technique for analyzing these interconnected changes.

This technique is called related rates.

What are related rates?

Related rates are problems involving two or more quantities that change over time and are connected by a mathematical relationship.

Instead of asking:

What is the value of a quantity?

Related rates ask:

How quickly is one quantity changing compared with another?

In mathematical language, we study relationships between derivatives.

The fundamental idea

Suppose two variables are connected by an equation:

$x^2+y^2=25$

Both $x$ and $y$ change over time.

Therefore:

$x=x(t)$

and

$y=y(t)$

Differentiate both sides with respect to time:

$\frac{d}{dt}(x^2+y^2)=\frac{d}{dt}(25)$

Applying the chain rule:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

This equation relates the rates of change of $x$ and $y$ .

Why is the chain rule essential?

Without the chain rule, related rates would not exist.

Suppose:

$y=x^2$

and $x$ changes over time.

Then:

$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$

Applying the derivative:

$\frac{dy}{dt}=2x\frac{dx}{dt}$

Notice what happened.

The derivative with respect to $x$ became a derivative with respect to time.

The chain rule created a bridge between two changing quantities.

Example 1: An expanding circle

Suppose the radius of a circle increases at a rate of:

$\frac{dr}{dt}=3\text{ cm/min}$

The area of a circle is:

$A=\pi r^2$

Differentiate with respect to time:

$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$

Suppose:

$r=5\text{ cm}$

Substitute the known values:

$\frac{dA}{dt}=2\pi(5)(3)$

Therefore:

$\boxed{\frac{dA}{dt}=30\pi\text{ cm}^2/\text{min}}$

The area increases at a rate of $30\pi$ square centimeters per minute.

Interpreting the result

Notice that the radius changes at a constant rate.

However, the area does not.

As the circle becomes larger, the same increase in radius produces larger increases in area.

This illustrates an important principle:

A constant rate in one variable does not necessarily produce a constant rate in another.

Example 2: A sliding ladder

A ladder 10 meters long leans against a wall.

The distance from the wall is:

$x$

The height on the wall is:

$y$

The relationship is:

$x^2+y^2=100$

Suppose:

$\frac{dx}{dt}=2\text{ m/s}$

and:

$x=6\text{ m}$

Find:

$\frac{dy}{dt}$

Differentiate:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

First, find $y$ :

$6^2+y^2=100$

Therefore:

$y=8$

Substitute:

$2(6)(2)+2(8)\frac{dy}{dt}=0$

Simplify:

$24+16\frac{dy}{dt}=0$

Therefore:

$\boxed{\frac{dy}{dt}=-1.5\text{ m/s}}$

The negative sign indicates that the top of the ladder is moving downward.

A business example

Suppose total revenue is:

$R=P\times Q$

where:

$P$ is price.
$Q$ is quantity sold.

Suppose both price and quantity change over time.

Differentiate:

$\frac{dR}{dt}=P\frac{dQ}{dt}+Q\frac{dP}{dt}$

Revenue growth now depends on two separate rates.

Businesses frequently analyze these relationships when forecasting sales.

A manufacturing example

Suppose the volume of a cylindrical storage tank is:

$V=\pi r^2h$

If the radius and height both change over time, then:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2\frac{dh}{dt}$

Manufacturing engineers use related rates to analyze production systems, storage capacities, and material flow.

A practical problem-solving strategy

Whenever you encounter a related-rates problem:

Step 1: Identify all changing variables.

Step 2: Write the equation connecting them.

Step 3: Differentiate with respect to time.

Step 4: Substitute the known values.

Step 5: Solve for the unknown rate.

Common mistakes

Forgetting the chain rule

Incorrect:

$\frac{d}{dt}(r^2)=2r$

Correct:

$\frac{d}{dt}(r^2)=2r\frac{dr}{dt}$

Substituting values too early

Differentiate first.

Substitute numerical values later.

Otherwise, important variables may disappear.

The deeper philosophical idea

Related rates reveal something profound about the world.

Very few systems exist in isolation.

Everything is connected.

A changing radius changes an area.

A changing position changes a velocity.

A changing price changes revenue.

Calculus allows us to quantify these relationships.

Related rates versus implicit differentiation

Concept	Purpose
Implicit differentiation	Find a derivative
Related rates	Find a relationship between rates of change

Related rates often depend on implicit differentiation.

The two topics are closely connected.

Conclusion

Related rates extend the idea of derivatives beyond individual functions.

Instead of studying how a single quantity changes, they study how multiple quantities influence one another.

Perhaps the simplest way to remember the idea is this:

Ordinary derivatives describe change.

Related rates describe how one change causes another.

This insight makes related rates one of the most powerful applications of differential calculus.

Integration by Substitution: Reversing the Chain Rule

August 14, 2026 by Splendid Leave a Comment

After learning the chain rule in differentiation, many students discover a surprising fact:

The chain rule not only helps us differentiate complicated functions, but it also teaches us how to integrate them.

This idea leads to one of the most powerful techniques in calculus:

Integration by substitution.

At first, substitution appears to be another formula that must be memorized.

In reality, it is simply the process of reversing the chain rule.

Understanding this connection makes substitution feel natural rather than mechanical.

Revisiting the chain rule

Suppose we have the function:

$y=(x^2+1)^5$

The chain rule tells us:

$\frac{dy}{dx}=5(x^2+1)^4(2x)$

Notice what happened.

The derivative contains two parts:

The derivative of the outer function:

$5(x^2+1)^4$

The derivative of the inner function:

$2x$

The chain rule combines them.

Running the chain rule backward

Now suppose we want to evaluate:

$\int5(x^2+1)^4(2x),dx$

Instead of asking:

How do I integrate this?

Ask a different question:

Which function would produce this expression if I differentiated it?

We already know the answer.

Differentiating produces:

$5(x^2+1)^4(2x)$

Therefore:

$\int5(x^2+1)^4(2x),dx=(x^2+1)^5+C$

Integration by substitution is simply the reverse of the chain rule.

Why do we need substitution?

Simple functions are easy to integrate.

For example:

$\int x^3,dx$

can be evaluated directly.

But consider:

$\int x(x^2+1)^4,dx$

The ordinary power rule doesn’t work.

The expression contains one function nested inside another.

This is exactly the situation where substitution becomes useful.

The basic idea

Suppose we have:

$\int f(g(x))g'(x),dx$

We define a new variable:

$u=g(x)$

Differentiating gives:

$\frac{du}{dx}=g'(x)$

Therefore:

$du=g'(x),dx$

The integral becomes:

$\int f(u),du$

After integrating, we replace $u$ with the original expression.

This process is called u-substitution.

Example 1: A simple substitution

Evaluate:

$\int2x(x^2+1)^3,dx$

Step 1: Choose a substitution

Let:

$u=x^2+1$

Step 2: Differentiate

Differentiate both sides:

$du=2x,dx$

Step 3: Rewrite the integral

Substitute:

$u=x^2+1$

and

$du=2x,dx$

The integral becomes:

$\int u^3,du$

Step 4: Integrate

Using the power rule:

$\int u^3,du=\frac{u^4}{4}+C$

Step 5: Replace the variable

Substitute the original expression:

$=\frac{(x^2+1)^4}{4}+C$

Why does this work?

Substitution works because it temporarily hides the complicated expression.

Instead of integrating:

$2x(x^2+1)^3$

we integrate:

$u^3$

We simplify the problem and then translate the answer back into the original language.

Example 2: An exponential function

Evaluate:

$\int e^{3x},dx$

Choose:

$u=3x$

Differentiate:

$du=3,dx$

Therefore:

$dx=\frac{du}{3}$

Substitute:

$\int e^{3x},dx=\int e^u\frac{du}{3}$

Move the constant outside:

$=\frac13\int e^u,du$

Integrate:

$=\frac13e^u+C$

Replace $u$ :

$=\frac13e^{3x}+C$

Example 3: A trigonometric function

Evaluate:

$\int\sin(5x),dx$

Choose:

$u=5x$

Differentiate:

$du=5,dx$

Therefore:

$dx=\frac{du}{5}$

Rewrite:

$\int\sin(5x),dx=\frac15\int\sin(u),du$

Integrate:

$=-\frac15\cos(u)+C$

Replace $u$ :

$=-\frac15\cos(5x)+C$

A geometric interpretation

Imagine a complicated machine.

The machine receives:

$x$

The first stage transforms it into:

$x^2+1$

The second stage raises it to a power.

Differentiation follows the entire journey through the machine.

Integration by substitution temporarily enters the machine and focuses on one stage at a time.

Instead of analyzing every component simultaneously, we isolate the important transformation.

A business example

Suppose a company’s customer acquisition rate is:

$r(t)=2t(t^2+100)^2$

The total number of acquired customers is:

$\int2t(t^2+100)^2,dt$

Choose:

$u=t^2+100$

Then:

$du=2t,dt$

The integral becomes:

$\int u^2,du$

Integrating:

$=\frac{u^3}{3}+C$

Returning to the original variable:

$=\frac{(t^2+100)^3}{3}+C$

Substitution allows us to reconstruct the accumulated number of customers.

How to recognize when substitution should be used

Look for these patterns:

A function inside another function.
An expression raised to a power.
Exponential functions such as:

$e^{g(x)}$

Trigonometric functions such as:

$\sin(g(x))$

A derivative that almost appears elsewhere in the integral.

A simple checklist

When solving an integral:

Step 1: Find the inner function.

Step 2: Let the inner function become $u$ .

Step 3: Differentiate to obtain $du$ .

Step 4: Rewrite the integral.

Step 5: Integrate.

Step 6: Replace $u$ with the original expression.

The connection between differentiation and integration

The relationship can be summarized as follows:

Chain rule:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

Substitution:

$\int f'(g(x))g'(x),dx=f(g(x))+C$

The two formulas are mirror images of one another.

Conclusion

Integration by substitution is not a new idea.

It is simply the chain rule running in reverse.

Differentiation breaks a complicated function into smaller pieces.

Integration reconstructs those pieces and combines them into an accumulated quantity.

Perhaps the easiest way to remember substitution is this:

The chain rule takes a composite function apart.

Substitution puts the composite function back together.

Understanding this relationship transforms substitution from a memorized technique into an intuitive mathematical process.

Before Understanding the Quotient Rule, Understand the Chain Rule

August 9, 2026 by Splendid Leave a Comment

The quotient rule is often taught as a formula to memorize:

$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$

But there is a problem with learning it this way.

The formula tells us what the derivative is, but not necessarily why it has this particular structure.

Why is there a subtraction?

Why is the denominator squared?

Why does g'(x) appear?

And why does a changing denominator behave differently from a changing numerator?

To answer these questions naturally, we should actually take one step backward.

Before understanding the quotient rule, we need to understand the chain rule.

1. The Chain Rule: Understanding Layers of Change

Imagine a function built in two stages.

First, x changes.

That change affects an intermediate quantity, say u.

Then the change in u affects another quantity, say y.

We can write this as

$x\longrightarrow u\longrightarrow y$

or mathematically,

$u=g(x)$

and

$y=f(u)$

Therefore,

$y=f(g(x))$

This is a function inside another function.

2. The Intuitive Meaning of the Chain Rule

Suppose a small change in x produces a change in u.

Then u itself produces a change in y.

So the total rate at which y changes with respect to x should depend on two rates:

$\frac{du}{dx}$

and

$\frac{dy}{du}$

The remarkable thing is that these rates multiply:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

This is the chain rule.

3. Why Do the Rates Multiply?

Think about units.

Suppose

x is measured in seconds,

u is measured in metres,

and y is measured in litres.

Then

$\frac{du}{dx}=\frac{\text{metres}}{\text{second}}$

and

$\frac{dy}{du}=\frac{\text{litres}}{\text{metre}}$

Multiplying them gives

$\frac{\text{litres}}{\text{metre}}\times\frac{\text{metres}}{\text{second}}=\frac{\text{litres}}{\text{second}}$

The intermediate unit cancels.

That is exactly what we want:

$\frac{dy}{dx}$

So the chain rule can be understood as passing change through a sequence of layers.

4. A Simple Example of the Chain Rule

Consider

$y=(x^2)^3$

There are two layers.

The inner function is

$u=x^2$

and the outer function is

$y=u^3$

Differentiate the outer layer:

$\frac{dy}{du}=3u^2$

Differentiate the inner layer:

$\frac{du}{dx}=2x$

Now multiply the two rates:

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

Therefore,

$\frac{dy}{dx}=3u^2(2x)$

Replace u with x^2:

$\frac{dy}{dx}=3(x^2)^2(2x)$

and hence

$\frac{dy}{dx}=6x^5$

The important idea is not merely the calculation.

It is this:

When change passes through several layers, each layer contributes its own rate of change, and the rates multiply.

This idea will soon become crucial for the quotient rule.

5. Now Consider the Reciprocal

A quotient contains division.

But division can always be rewritten as multiplication by a reciprocal:

$\frac{f(x)}{g(x)}=f(x)\frac{1}{g(x)}$

And the reciprocal can be written as a negative power:

$\frac{1}{g(x)}=g(x)^{-1}$

Therefore,

$\frac{f(x)}{g(x)}=f(x)g(x)^{-1}$

This is the key transformation.

Instead of thinking:

“How do I differentiate a quotient?”

we can think:

“How do I differentiate a product involving a reciprocal?”

That is a much easier question.

But first, we need to understand how the reciprocal changes.

6. Why Does the Reciprocal Move in the Opposite Direction?

Consider

$y=\frac{1}{x}$

As x increases, 1/x decreases.

For example,

$\frac{1}{2}=0.5$

while

$\frac{1}{4}=0.25$

So the reciprocal has a negative rate of change.

Using the power rule,

$\frac{1}{x}=x^{-1}$

therefore,

$\frac{d}{dx}x^{-1}=-x^{-2}$

$\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{1}{x^2}$

The negative sign makes intuitive sense:

When the denominator of a reciprocal gets larger, the reciprocal gets smaller.

But now comes the crucial question.

What if the denominator itself is not simply x, but some function g(x)?

This is exactly where the chain rule enters.

7. A Changing Denominator Creates a Function Inside a Function

Consider

$y=\frac{1}{g(x)}$

Rewrite it:

$y=[g(x)]^{-1}$

Look carefully at its structure.

The outer function is

$u^{-1}$

while the inner function is

$u=g(x)$

So we have:

$x\longrightarrow g(x)\longrightarrow [g(x)]^{-1}$

This is a textbook example of the chain rule.

The reciprocal is the outer layer.

The changing denominator is the inner layer.

8. Apply the Chain Rule

First differentiate the outer function with respect to g:

$\frac{d}{dg}g^{-1}=-g^{-2}$

Then differentiate the inner function:

$\frac{d}{dx}g(x)=g'(x)$

The chain rule tells us to multiply these two effects:

$\frac{d}{dx}[g(x)]^{-1}=-g(x)^{-2}g'(x)$

Since

$g(x)^{-2}=\frac{1}{[g(x)]^2}$

we obtain

$\boxed{\frac{d}{dx}\left(\frac{1}{g(x)}\right)=-\frac{g'(x)}{[g(x)]^2}}$

This is an extremely important result.

And notice what happened.

We did not memorize it.

We built it from two familiar ideas:

$\boxed{\text{Power Rule}+\text{Chain Rule}=\text{Derivative of a Reciprocal}}$

9. Now the Quotient Rule Is Almost Here

Return to

$y=\frac{f(x)}{g(x)}$

Rewrite division as multiplication:

$y=f(x)\frac{1}{g(x)}$

$y=f(x)[g(x)]^{-1}$

Now we have a product.

So we use the product rule:

u’v+uv’

Therefore,

$y'=f'(x)[g(x)]^{-1}+f(x)\frac{d}{dx}[g(x)]^{-1}$

But we have just learned that

$\frac{d}{dx}[g(x)]^{-1}=-\frac{g'(x)}{[g(x)]^2}$

Substitute it:

$y'=\frac{f'(x)}{g(x)}-\frac{f(x)g'(x)}{[g(x)]^2}$

Put both terms over the same denominator:

$y'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$

And therefore,

$\boxed{\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}}$

The quotient rule has emerged naturally.

10. Why the Minus Sign Is Now Completely Understandable

The famous minus sign in

$f'g-fg'$

is no longer mysterious.

It originates from the derivative of the reciprocal:

$\frac{d}{dx}\left(\frac{1}{g}\right)=-\frac{g'}{g^2}$

Why is that derivative negative?

Because increasing the denominator decreases the reciprocal.

So the quotient rule is telling us something intuitive:

The numerator’s change pushes the ratio in one direction, while the denominator’s change pushes it in the opposite direction.

The subtraction is therefore not an arbitrary algebraic feature.

It represents an actual opposing effect.

11. Why Does the Denominator Become Squared?

The square also has a natural origin.

We started with the reciprocal:

$\frac{1}{g}=g^{-1}$

The power rule changes the exponent from $-1$ to $-2$ :

$\frac{d}{dg}g^{-1}=-g^{-2}$

And

$g^{-2}=\frac{1}{g^2}$

Therefore the denominator becomes squared.

So the two features that students often memorize separately actually have simple explanations:

$\boxed{\text{Minus sign}\longleftarrow\text{reciprocal decreases}}$

and

$\boxed{\text{Squared denominator}\longleftarrow\text{power }-1\text{ becomes }-2}$

12. The Entire Chain of Ideas

The quotient rule can now be reconstructed from a small collection of ideas.

Start with the quotient:

$\frac{f}{g}$

Rewrite division as multiplication:

$\frac{f}{g}=f\cdot g^{-1}$

Apply the product rule:

f’g^{-1}+f(g^{-1})’

Use the power rule:

$\frac{d}{dg}g^{-1}=-g^{-2}$

Use the chain rule because $g$ depends on x:

$\frac{d}{dx}g(x)^{-1}=-g(x)^{-2}g'(x)$

Then simplify:

$\boxed{\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}}$

So there is a beautiful dependency:

$\boxed{\text{Power Rule}\longrightarrow\text{Reciprocal Rule}}$ $\boxed{\text{Chain Rule}+\text{Power Rule}\longrightarrow\text{Changing Reciprocal}}$ $\boxed{\text{Product Rule}+\text{Changing Reciprocal}\longrightarrow\text{Quotient Rule}}$

13. The Deeper Lesson

This is an excellent example of how calculus rules are connected rather than isolated.

The quotient rule does not really need to be thought of as an independent formula.

It can be constructed.

A quotient is multiplication by a reciprocal.

A reciprocal is a negative power.

A negative power is handled by the power rule.

When the base of that power is itself a function, the chain rule is required.

And because the reciprocal is multiplied by the numerator, the product rule finishes the job.

In other words:

$\boxed{\text{Quotient Rule}=\text{Product Rule}+\text{Chain Rule}+\text{Power Rule}}$

with the reciprocal sitting at the center of the construction.

14. The Most Useful Mental Model

Instead of memorizing

$\frac{f'g-fg'}{g^2}$

try remembering the sequence:

Division → Reciprocal → Negative Power → Chain Rule → Product Rule

Or even more intuitively:

A quotient is a product with a reciprocal attached. The reciprocal decreases when its denominator increases, and the chain rule tells us how the changing denominator carries its own rate of change into that reciprocal.

Once this picture is understood, the quotient rule becomes much less like a formula handed down by calculus and much more like something we could derive whenever we need it.

Conclusion

The quotient rule is often introduced as a rule that must be memorized.

But there is a much more satisfying way to see it.

First understand the chain rule as the mathematics of change passing through layers.

Then recognize that

$\frac{1}{g(x)}=[g(x)]^{-1}$

is a layered function: the denominator g(x) changes first, and then the reciprocal acts on it.

The chain rule therefore gives

$\boxed{\frac{d}{dx}\left(\frac{1}{g(x)}\right)=-\frac{g'(x)}{[g(x)]^2}}$

Finally, rewrite

$\frac{f(x)}{g(x)}=f(x)[g(x)]^{-1}$

and apply the product rule.

The result is

$\boxed{\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}}$

The formula is no longer something mysterious to memorize.

It is the logical consequence of a few fundamental ideas:

$\boxed{\text{Power Rule}\rightarrow\text{Chain Rule}\rightarrow\text{Reciprocal}\rightarrow\text{Product Rule}\rightarrow\text{Quotient Rule}}$

And that is one of the beautiful things about calculus: the rules are not a collection of unrelated tricks; they grow naturally out of one another.

chain rule

What are related rates?

The fundamental idea

Why is the chain rule essential?

Example 1: An expanding circle

Interpreting the result

Example 2: A sliding ladder

A business example

A manufacturing example

A practical problem-solving strategy

Common mistakes

Forgetting the chain rule

Substituting values too early

The deeper philosophical idea

Related rates versus implicit differentiation

Conclusion

Revisiting the chain rule

Running the chain rule backward

Why do we need substitution?

The basic idea

Example 1: A simple substitution

Step 1: Choose a substitution

Step 2: Differentiate

Step 3: Rewrite the integral

Step 4: Integrate

Step 5: Replace the variable

Why does this work?

Example 2: An exponential function

Example 3: A trigonometric function

A geometric interpretation

A business example

How to recognize when substitution should be used

A simple checklist

The connection between differentiation and integration

Conclusion

1. The Chain Rule: Understanding Layers of Change

2. The Intuitive Meaning of the Chain Rule

3. Why Do the Rates Multiply?

4. A Simple Example of the Chain Rule

5. Now Consider the Reciprocal

6. Why Does the Reciprocal Move in the Opposite Direction?

7. A Changing Denominator Creates a Function Inside a Function

8. Apply the Chain Rule

9. Now the Quotient Rule Is Almost Here

10. Why the Minus Sign Is Now Completely Understandable

11. Why Does the Denominator Become Squared?

12. The Entire Chain of Ideas

13. The Deeper Lesson

14. The Most Useful Mental Model

Conclusion

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Calculus 1A: Differentiation

Calculus 1B: Integration

Calculus 1C: Coordinate Systems & Infinite Series