The sum, product and quotient rules show us how to differentiate a great many different functions from the reals to the reals. However some functions, such as $f(x)=\sin 2x$ are a composition of functions, and these rules don’t tell us what the derivative of $\sin 2x$ is. There is, however, a theorem called the chain rule that tells us how to differentiate these functions. Here we present the proof. In class we won’t prove this assertion but we will make one attempt to explain why it takes the form it does. In general only practise can make you proficient in the use of the chain rule. See http://euclid.ucc.ie/pages/staff/wills/teaching/ms2001/exercise3.pdf or any other textbook (such as a LC text book) with exercises.

Proposition 4.1.8 (Chain Rule)

Let $f, g:\mathbb{R}\rightarrow\mathbb{R}$ be functions, and let $F$ denote the composition $F=g\circ f$ (that is $F(x)=g(f(x))$ for each $x\in\mathbb{R}$). If $a\in\mathbb{R}$ such that $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $F$ is differentiable at $a$ with $F'(a)=g'(f(a))f'(a)$

Proof: Define a function $G:\mathbb{R}\rightarrow\mathbb{R}$ by $G(k):=\left\{\begin{array}{cc}\frac{g(f(a)+k)-g(f(a))}{k}-g'(f(a)) & \text{ if }k\neq 0\\ 0 & \text{ if }k=0\end{array}\right.$

Then $\lim_{k\rightarrow 0}G(k)=0$, since we assumed that $g$ is differentiable at $f(a)$. Hence $G$ is continuous at $0$. Rearranging the equation above we see that $g(f(a)+k)-g(f(a))=k[G(k)+g'(f(a))]\,\,,\,\text{for all }k\in\mathbb{R}.$ (*)

Now for any $h\neq 0$, $F(a+h)-F(a)=g(f(a+h)\overbrace{-f(a)+f(a)}^{=0})-g(f(a))$ $=g(r(h)+f(a))-g(f(a))$

where we define $r(h)=f(a+h)-f(a)$ for all $h\in\mathbb{R}$. The right hand side of this last equation is of the form of the left hand side of (*), with $k=r(h)$, and so $F(a+h)-F(a)=r(h)[G(r(h))+g'(f(a))].$

But $f$ is differentiable at $a$, hence continuous there as well, thus $\lim_{h\rightarrow 0}\frac{r(h)}{h}=f'(a)$ and $\lim_{h\rightarrow 0} r(h)=r(0)=0.$

That is, $r$ is a function on $\mathbb{R}$ that is continuous at $0$ and satisfies $r(0)=0$, and $G$ is continuous at $0$ and satisfies $G(0)=0$. Hence $x\mapsto (G\circ r)(x)$ is continuous at $0$, with $\lim_{h\rightarrow 0} (G\circ r)(h)=(G\circ r)(0)=0$ by Proposition 3.3.5. So now $\lim_{h\rightarrow 0}\frac{F(a+h)-F(a)}{h}=\lim_{h\rightarrow 0}\frac{r(h)}{h}\times\lim_{h\rightarrow 0}[G(r(h))+g'(f(a))]$ $=f'(a)\times [0+g'(f(a))]=g'(f(a))f'(a)$

as required $\bullet$

# The Chain Rule… by Rule

Suppose $f(x)$ is some function with derivative $f'(x)$. :The following is a nice, rough and ready definition of the chain rule:

Let $F(x)=g(f(x))$ where $g(x)$ is a function with derivative $g'(x)$. Then $F(x)$ has derivatative $F'(x)=g'(f(x))f'(x)$.

Translating this into some common examples (note that you are expected to know the derivatives of the ‘outside’ functions):

Logarithms:

Suppose $F(x)=\log(f(x))$. Now $\log x$ has derivative $1/x$, hence: $F'(x)=\frac{1}{f(x)}\times f'(x)$.

Inverse Trigonometric

Suppose $F(x)=\sin^{-1}(f(x))$, then: $F'(x)=\frac{1}{\sqrt{1-[f(x)]^2}}\times f'(x)$.

Suppose that $F(x)=\tan^{-1}(f(x))$, then: $F'(x)=\frac{1}{1+[f(x)]^2}\times f'(x)$.

Powers

Suppose $F(x)=[f(x)]^n$, then $F'(x)=n[f(x)]^{n-1}\times f'(x)$.

Trignometric

Suppose $F(x)=\sin (f(x))$, then $F'(x)=\cos(f(x))\times f'(x)$.

Suppose $F(x)=\cos (f(x))$, then $F'(x)=-\sin (f(x))\times f'(x)$.

Suppose $F(x)=\tan (f(x))$, then $F'(x)=\sec^2(f(x))\times f'(x)$.

Exponential

Suppose that $F(x)=e^{f(x)}$, then $F'(x)=e^{f(x)}\times f'(x)$.