The chain rule is used when one function is placed inside another function.
Such functions are called composite functions.
1. Composite function idea
If:
\[
y=f(g(x))
\]
then \(g(x)\) is the inside function and \(f(u)\) is the outside function.
\[
u=g(x),\qquad y=f(u)
\]
2. Main chain rule
The chain rule says:
\[
\boxed{
\frac{d}{dx}f(g(x))=f'(g(x))g'(x)
}
\]
In Leibniz notation:
\[
\boxed{
\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}
}
\]
3. Worked example
Differentiate:
\[
y=\sin(x^2)
\]
The outside function is:
\[
f(u)=\sin u
\]
The inside function is:
\[
g(x)=x^2
\]
Differentiate the outside:
\[
f'(u)=\cos u
\]
Differentiate the inside:
\[
g'(x)=2x
\]
Apply the chain rule:
\[
\frac{dy}{dx}=f'(g(x))g'(x)
\]
\[
\frac{dy}{dx}=\cos(x^2)\cdot2x
\]
Final answer:
\[
\boxed{\frac{dy}{dx}=2x\cos(x^2)}
\]
4. Nested layers
Some expressions have more than one inside layer. For example:
\[
y=\ln\left(\sqrt{x^2+4}\right)
\]
This can be viewed as layers:
\[
\ln(\square)
\rightarrow
\sqrt{\square}
\rightarrow
x^2+4
\]
Each layer contributes one derivative factor.
5. Power as an outside layer
A power such as:
\[
y=(x^2+1)^5
\]
is also a chain-rule problem.
Let:
\[
u=x^2+1
\]
Then:
\[
y=u^5
\]
Differentiate:
\[
\frac{dy}{dx}=5u^4\frac{du}{dx}
\]
Since:
\[
\frac{du}{dx}=2x
\]
the derivative is:
\[
\boxed{
\frac{dy}{dx}=5(x^2+1)^4(2x)
}
\]
6. Common outside derivatives
| Outer function |
Outer derivative |
Chain-rule form |
| \(\sin u\) |
\(\cos u\) |
\(\dfrac{d}{dx}\sin(g(x))=\cos(g(x))g'(x)\) |
| \(\cos u\) |
\(-\sin u\) |
\(\dfrac{d}{dx}\cos(g(x))=-\sin(g(x))g'(x)\) |
| \(e^u\) |
\(e^u\) |
\(\dfrac{d}{dx}e^{g(x)}=e^{g(x)}g'(x)\) |
| \(\ln u\) |
\(\dfrac{1}{u}\) |
\(\dfrac{d}{dx}\ln(g(x))=\dfrac{g'(x)}{g(x)}\) |
| \(\sqrt{u}\) |
\(\dfrac{1}{2\sqrt{u}}\) |
\(\dfrac{d}{dx}\sqrt{g(x)}=\dfrac{g'(x)}{2\sqrt{g(x)}}\) |
| \(u^n\) |
\(nu^{n-1}\) |
\(\dfrac{d}{dx}[g(x)]^n=n[g(x)]^{n-1}g'(x)\) |
7. Composition tree method
A useful way to solve chain-rule problems is to draw a composition tree.
For:
\[
y=e^{\sin(x^2)}
\]
the layers are:
\[
e^{\square}
\rightarrow
\sin(\square)
\rightarrow
x^2
\rightarrow
x
\]
The derivative is the product of the derivative contribution from each layer:
\[
\frac{dy}{dx}
=
e^{\sin(x^2)}
\cdot
\cos(x^2)
\cdot
2x
\]
8. Common mistakes
- Only differentiating the outside and forgetting the inside derivative.
- Only differentiating the inside and forgetting the outside derivative.
- Substituting \(u\) incorrectly after differentiating.
- Forgetting that powers like \((x^2+1)^5\) also require the chain rule.
- Missing a negative sign, especially in derivatives involving cosine.
9. Good practice strategy
Use this four-step process:
- Circle the inside expression.
- Name it \(u\).
- Differentiate the outside with respect to \(u\).
- Multiply by \(\dfrac{du}{dx}\).
In symbols:
\[
y=F(u),\qquad u=g(x)
\]
\[
\frac{dy}{dx}=F'(u)u'
\]
\[
\boxed{
\frac{dy}{dx}=F'(g(x))g'(x)
}
\]
