Chain Rule

Welcome, students 👋 In calculus, one of the biggest ideas is that functions can be built from other functions. The Chain Rule helps us find the derivative of a function inside another function, which is called a composite function. This is very important in IB Mathematics Analysis and Approaches SL because many real-life models are not simple straight-line formulas. They often involve layers, such as temperature depending on time and time depending on distance, or cost depending on the number of items and the number of items depending on production rate.

By the end of this lesson, you should be able to:

explain what the Chain Rule means and why it is needed,
apply the Chain Rule to differentiate composite functions,
connect the Chain Rule to other calculus ideas such as product rule, rate of change, optimisation, and kinematics,
interpret results in context using clear mathematical reasoning.

Think of the Chain Rule as a way to follow the “chain” of dependence between quantities 🔗. If one quantity depends on another, and that second quantity depends on a third, then the total rate of change depends on both links in the chain.

What is a composite function?

A composite function is a function made by putting one function inside another. If $y=f(u)$ and $u=g(x)$, then $y=f(g(x))$. The function $g(x)$ is the inner function, and $f(u)$ is the outer function.

For example, let $y=(3x+1)^5$. This is a composite function because the inside part is $3x+1$ and the outside operation is raising to the power $5$. We can rewrite it as $y=u^5$ where $u=3x+1$.

This “inside-outside” structure appears all the time in IB calculus. You may see expressions like $\sin(2x)$, $\sqrt{5x-4}$, $e^{x^2}$, or $\ln(1+x^3)$. Each one contains a function inside another function.

The key challenge is that differentiating a composite function is not done by simply differentiating each part separately. For example, the derivative of $(3x+1)^5$ is not just $5(3x+1)^4$. That misses the fact that the inside expression also changes with $x$.

The Chain Rule idea

The Chain Rule says that if $y=f(u)$ and $u=g(x)$, then

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

In words, differentiate the outside function with respect to the inside variable, then multiply by the derivative of the inside function with respect to $x$.

Another common way to write it is

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

This formula is one of the most important tools in calculus because it tells us how changes pass through layers of functions.

A simple way to remember it is: differentiate the outside, keep the inside, then multiply by the derivative of the inside ✅

Let’s use the earlier example:

$$y=(3x+1)^5.$$

Set $u=3x+1$, so $y=u^5$.

Then

$$\frac{dy}{du}=5u^4$$

and

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

$$\frac{dy}{dx}=5u^4\cdot 3=15(3x+1)^4.$$

Notice that the derivative is larger than just the derivative of the outside part, because the inside function is also changing.

Why the Chain Rule works

The Chain Rule works because of how small changes move through linked quantities. Suppose a small change in $x$ causes a change in $u$, and that causes a change in $y$. Then the overall change in $y$ depends on both steps.

In IB terms, this is a rate of change idea. If $y$ depends on $u$ and $u$ depends on $x$, then the gradient of $y$ with respect to $x$ is the product of the gradients along the chain.

This idea is very useful in modelling. For example, if the area of a circle depends on radius, and radius depends on time, then the area also changes with time. To find $\frac{dA}{dt}$, you use the Chain Rule.

$$A=\pi r^2$$

and $r=r(t)$, then

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

This tells you how quickly the area changes at any moment, not just how it depends on radius.

Example 1: Power function with an inner linear expression

Differentiate

$$y=(2x-7)^6.$$

First identify the inner function $u=2x-7$ and the outer function $y=u^6$.

Then

$$\frac{dy}{du}=6u^5$$

and

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

Therefore,

$$\frac{dy}{dx}=6(2x-7)^5\cdot 2=12(2x-7)^5.$$

A common mistake is forgetting the factor $2$ from the derivative of the inside expression.

Example 2: Trigonometric function

Differentiate

$$y=\sin(4x).$$

Let $u=4x$, so $y=\sin(u)$.

Then

$$\frac{dy}{du}=\cos(u)$$

and

$$\frac{du}{dx}=4.$$

$$\frac{dy}{dx}=4\cos(4x).$$

This is a classic Chain Rule result in IB Maths 📘

Example 3: Exponential function

Differentiate

$$y=e^{x^2}.$$

Let $u=x^2$, so $y=e^u$.

Then

$$\frac{dy}{du}=e^u$$

and

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

$$\frac{dy}{dx}=2xe^{x^2}.$$

This example shows why the Chain Rule is essential. The derivative of $e^x$ is $e^x$, but because the exponent is $x^2$, the extra factor $2x$ must appear.

Applying the Chain Rule in IB reasoning

In IB Mathematics Analysis and Approaches SL, you may be asked not only to differentiate but also to explain your steps clearly. A strong answer usually includes:

identifying the inner and outer functions,
writing the derivative carefully,
simplifying the final result,
interpreting the meaning if the question is in context.

For example, if a quantity $h$ is given by

$$h=(5t+2)^3,$$

you can write

$$\frac{dh}{dt}=3(5t+2)^2\cdot 5=15(5t+2)^2.$$

If the question asks about the rate of change at $t=1$, then substitute after differentiating:

$$\frac{dh}{dt}\Bigg|_{t=1}=15(5(1)+2)^2=15\cdot 49=735.$$

That means the quantity is increasing at a rate of $735$ units per time unit at $t=1$.

This is the kind of reasoning that connects differentiation to real applications. In optimisation, for example, you may need the derivative to find where a quantity is largest or smallest. In kinematics, you may need the derivative of displacement to find velocity, and the derivative of velocity to find acceleration. If those formulas involve composite expressions, the Chain Rule becomes necessary.

Common forms you should recognize

students, it is helpful to spot common patterns quickly:

$\big(f(x)\big)^n$ gives a power outside a function inside.
$\sin(f(x))$, $\cos(f(x))$, and $\tan(f(x))$ are trig composites.
$e^{f(x)}$ and $a^{f(x)}$ are exponential composites.
$\ln(f(x))$ is a logarithmic composite.
$\sqrt{f(x)}$ can be written as $\big(f(x)\big)^{1/2}$.

For example,

$$\frac{d}{dx}\big(\sqrt{3x+1}\big)=\frac{d}{dx}(3x+1)^{1/2}.$$

Using the Chain Rule,

$$\frac{d}{dx}(3x+1)^{1/2}=\frac{1}{2}(3x+1)^{-1/2}\cdot 3=\frac{3}{2\sqrt{3x+1}}.$$

Recognizing the hidden structure makes differentiation much faster and more accurate.

Conclusion

The Chain Rule is a central calculus technique because it lets you differentiate composite functions, which are everywhere in mathematics and in the real world. Its main idea is simple: differentiate the outside function, then multiply by the derivative of the inside function. This gives a correct rate of change when one quantity depends on another that depends on another.

In IB Mathematics Analysis and Approaches SL, the Chain Rule connects to many parts of calculus, including graphs, modelling, optimisation, and kinematics. If you can identify the inner and outer functions clearly, and apply the rule step by step, you will be well prepared for a wide range of exam questions 🎯

Study Notes

A composite function has one function inside another, such as $f(g(x))$.
The Chain Rule is

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

A useful memory phrase is: differentiate the outside, keep the inside, then multiply by the derivative of the inside.
Common examples include $\sin(4x)$, $e^{x^2}$, $\ln(1+x^3)$, and $(2x-7)^6$.
Always identify the inner function and outer function before differentiating.
Do not forget the derivative of the inside expression.
In applications, the Chain Rule helps find rates of change in geometry, optimisation, and kinematics.
If a question asks for a rate at a specific value, differentiate first, then substitute the value.
The Chain Rule is a fundamental part of calculus because it connects derivatives across linked variables.