# The chain rule

- Recall \(g(x)^{-1}\)
- Recall that yesterday we were talking about that instead of using the quotient rule to derivate \(g(x)^{-1}\), we could use the chain rule.
- The Chain Rule in general involves derivating a composite of functions.
- In this particular case, we have \(F(x)=(f\circ g)(x)=f(g(x))\), where \(f(x)=x^{-1}\).
- In general, the chain rule is the following.
*Theorem.* Let \(f(x)\) and \(g(x)\) be functions, and let \(F(x)=f(g(x))\). If \(g\) is differentiable at \(x\), and \(f\) is differentiable at \(g(x)\), then \(F(x)\) is differentiable at \(x\), and we have \[
F'(x)=f'(g(x))g'(x).
\]
- In the particular case that \(f(x)=x^{-1}\), we have \(f'(x)=-x^{-2}\).
- Therefore, the chain rule gives \[
(g(x)^{-1})'=f'(g(x))g'(x)=-g(x)^{-2}g'(x),
\] as we've seen using the quotient rule.
- More generally, by setting \(f(x)=x^n\), using \(f'(x)=nx^{n-1}\), we get the following. \[
(g(x)^n)'=f'(g(x))g'(x)=ng(x)^{n-1}g'(x).
\]

# The chain rule 2

- There is an alternative formalism, which is quite popular, although the notation is potentially confusing. Use at your own risk.
- We introduce the new variable \(u\), and then we set \(u=g(x)\).
- The outside function \(f\) is at first defined with variable \(u\): \(f=f(u)\).
- With this notation, \(f(x)\) means the composite function \(f(u=g(x))=f(g(x))\).
- Using this trickery, we can formulate the chain rule as follows. \[
\frac{\mathrm df}{\mathrm dx}=\frac{\mathrm df}{\mathrm du}\frac{\mathrm du}{\mathrm dx}.
\]
- The popularity of this formalism comes from that it looks as if you could simplify with \(\mathrm du\)
- It is important to remember that \(\frac{\mathrm df}{\mathrm du}\) is not an actual quotient, it's just notation for the derivative. Therefore, you can't simplify with \(\mathrm du\).
- This formalism can still come in handy, but please do handle it with care.
- Exercises. 3.4. 2, 4, 6, 8, 12, 14, 24, 26, 30, 34, 40, 42, 44, 50, 25, 60, 62, 72, 76, 78, 80, 82, 86, 96, 99