# 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