Let us consider the function $y=\sqrt{x^2+1}$. Notice that this is a composite function $y=\sqrt{u}$ and $u=x^2+1$. In general, a composite function can be written as $y=f(u)$ where $u$ is a function of $x$, $u=g(x)$. While we know how to differentiate $y=\sqrt{u}$ (i.e. finding $\frac{dy}{du}$) and $u=x^2+1$ (i.e. finding $\frac{du}{dx}$), we do not know how to differentiate $y=\sqrt{x^2+1}$ (i.e finding $\frac{dy}{dx}$). In this lecture, we would like to devise a way to differentiate a composite function. This is actually very important because the differentiable functions we stumble onto most of time are composite functions.

Let $y=f(u)$ and $u=g(x)$ and assume that both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist. Now,
\begin{align*}
\frac{\Delta y}{\Delta x}&=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}\\
&=\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\frac{g(\Delta x+x)-g(x)}{\Delta x}.
\end{align*}
Hence,
\begin{align*}
\frac{dy}{dx}&=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\\
&=\lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\ (\Delta u\to 0\ \mbox{as}\ \Delta x\to 0)\\
&=\frac{dy}{du}\cdot\frac{du}{dx}
\end{align*}
or
\begin{align*}
\frac{dy}{dx}&=\lim_{\Delta u\to 0}\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{g(\Delta x+x)-g(x)}{\Delta x}\\
&=f'(u)g'(x).
\end{align*}

Theorem. [The Chain Rule]
Let $y=f(u)$ and $u=g(x)$. If both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist, then $\frac{dy}{dx}$ exists and
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=f'(u)g'(x).
\end{align*}

Remark. The derivation of the chain rule shown above is not rigorously correct. The reason is that $\Delta u$ may become $0$. There is a more rigorous proof of the chain rule but we will not discuss that here.

Remark. Students commonly feel a difficulty with applying the chain rule when they learn it for the first time. The difficulty usually is not about understanding the chain rule itself but identifying the function $u=g(x)$. The candidate for $u$ is usually the function inside parentheses (or brackets) or the innermost function.

Example. We are now ready to find $\frac{dy}{dx}$ when $y=\sqrt{x^2+1}$. In this case, we don’t see parentheses or brackets but the innermost function is $x^2+1$. Let $u=x^2+1$. Then $y=\sqrt{u}$. Now,
\begin{align*}
\frac{dy}{du}&=\frac{1}{2\sqrt{u}}\\
&=\frac{1}{2\sqrt{x^2+1}},\\
\frac{du}{dx}&=2x.
\end{align*}
so, we have by the chain rule
$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{x}{\sqrt{x^2+1}}.$$

Example. Differentiate $y=(x^3-1)^{100}$.

Solution. The function inside parentheses is $x^3-1$. So, it is our candidate. Let $u=x^3-1$. Then $y=u^{100}.$
By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=100u^{99}\cdot(3x^2)\\
&=300x^2(x^3-1)^{99}.
\end{align*}

Example. Find the derivative of each function.

1. $y=\sin 4x$.

Solution. The innermost function is $4x$. Let $u=4x$. Then $y=\sin u$. By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=\cos u\cdot4\\
&=4\cos 4x.
\end{align*}

2. $y=\sqrt{\sin x}$.

Solution. The innermost function is $\sin x$. Let $u=\sin x$. Then $y=\sqrt{u}$. By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=\frac{1}{2\sqrt{u}}\cdot\cos x\\
&=\frac{\cos x}{2\sqrt{\sin x}}.
\end{align*}

Update: For those who are interested, the rigorous proof of the Chain Rule can be found here.

First we begin with the Mean Value Theorem for Definite Integrals.

Theorem. If $f$ is continuous on $[a,b]$, then at some point $c\in [a,b]$,
$$f(c)(b-a)=\int_a^b f(x)dx$$
or
\begin{equation}
\label{eq:mvt}
f(c)=\frac{1}{b-a}\int_a^b f(x)dx.
\end{equation}
Notice that the RHS of \eqref{eq:mvt} is the average of $f(x)$ on $[a,b]$.

Example. Find the average value of $f(x)=4-x$ on $[0,3]$ and $c\in[0,3]$ at which $f$ actually takes on this value.

Solution. The region under $y=4-x$ on $[0,3]$ is a trapezoid and its area is $\frac{15}{2}$. So,
\begin{align*}
av(f)&=\frac{1}{3-0}\int_0^3(4-x)dx\\
&=\frac{5}{2}.
\end{align*}
Let $f(c)=\frac{5}{2}$. Then $4-c=\frac{5}{2}$ and $c=\frac{3}{2}$.

Example. Show that if $f$ is continuous on $[a,b]$ ($a\ne b$) and if $\int_a^b f(x)=0$ then $f(x)=0$ at least once in $[a,b]$.

Solution. By the MVT, there exists $c\in[a,b]$ such that
$$f(c)=\frac{1}{b-a}\int_a^b f(x)dx=0.$$

Suppose that $f(t)$ is an integrable function on a finite intervale $I$. Let $a\in I$. Then
$$F(x):=\int_a^x f(t)dt$$
defines a fuction on the interval $I$. Let $f(x)$ be a continuous function on $[a,b]$. Then

Claim 1: $F(x)=\int_a^x f(t)dt$ is continuous on $[a,b]$. (For a proof click here. For those who are not familiar with $\epsilon$-$\delta$ argument, an alternative proof is available here.)

Claim 2: $F(x)$ is differentiable on $(a,b)$ and $F'(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$. (For a proof click here.)

The claims 1 and 2 constitute:

The Fundamental Theorem of Calculus (Part I)

If $f$ is continuous on $[a,b]$, then $F(x)=\int_a^xf(t)dt$ is continuous on $[a,b]$ and is differentiable on $(a,b)$. Moreover, $F'(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$.

The Fundamental Theorem of Calculus (Part II) relates the definitel integral and an antiderivative of a function $f(x)$.

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$, then
$$\int_a^b f(x)dx=F(b)-F(a).$$(For a proof click here.)

Note: The usual notation for $F(b)-F(a)$ is $F(x)|_a^b$ or $[F(x)]_a^b$.

Example. Find using the Fundamental Theorem.

1. $\frac{d}{dx}\int_a^x\cos tdt$
2. $\frac{d}{dx}\int_0^x\frac{1}{1+t^2}dt$
3. $\frac{d}{dx}\int_x^5 3t\sin tdt$
4. $\frac{d}{dx}\int_1^{x^2}\cos tdt$

Solution.

1. $\frac{d}{dx}\int_a^x\cos tdt=\cos x$
2. $\frac{d}{dx}\int_0^x\frac{1}{1+t^2}dt=\frac{1}{1+x^2}$
3. $\frac{d}{dx}\int_x^5 3t\sin tdt=-\frac{d}{dx}\int_5^x 3t\sin tdt=-3x\sin x$
4. Let $u=x^2$. Then by the chain rule, \begin{align*}\frac{d}{dx}\int_1^{x^2}\cos tdt&=\frac{d}{du}\int_1^u\cos tdt\frac{du}{dx}\\&=2x\cos x^2\end{align*}

Example. Find a function $y=f(x)$ on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ with $\frac{dy}{dx}=\tan x$ and $f(3)=5$.

Solution. $y=\int_3^x\tan t dt+5=\ln\frac{\cos 3}{\cos x}+5$.

Example.

1. $\int_0^{\pi}\cos x dx=[\sin x]_0^{\pi}=\sin\pi-\sin 0=0$. 
   

   The graph of y=cos(x) on [0,pi]

2. \begin{align*}\int_1^4\left(\frac{3}{2}\sqrt{x}-\frac{4}{x^2}\right)dx&=\frac{3}{2}\int_1^4 x^{\frac{1}{2}}dx-4\int_1^4x^{-2}dx\\&=[x^{\frac{3}{2}}]_1^4+4\left[\frac{1}{x}\right]_1^4\\&=4\end{align*}

Example. Find the area bounded by the $x$-axis and the parabola $y=6-x-x^2$.

Solution. Note that when you solve area problems it is extremely important to sketch the graph of the given function in order to correctly set up your integral. The reason is that some part of the function may not be nonnegative. The quadratic equation $6-x-x^2=0$ has two solutions $x=-3,2$, so these are the $x$-intercepts of the function $y=6-x-x^2$.

Since the graph is above the $x$-axis on $[-3,2]$, the area is obtained by the integral $$\int_{-3}^2(6-x-x^2)dx=\frac{125}{6}$$

Example. Find the area of the region bounded by the $x$-axis and $f(x)=x^3-x^2-2x$.

Solution.  The function has three $x$-intercepts $x=-1, 0, 2$. Since $f(x)$ has an odd degree and a positive leading coefficient, it’s graph is

The graph is below the $x$-axis on $[0,2]$. Hence the area is $$\int_{-1}^0(x^3-x^2-2x)dx-\int_0^2(x^3-x^2-2x)dx=\frac{37}{12}$$

If $f: [a,b]\longrightarrow\mathbb{R}$ is a continuous function, the limits from the left-end point, midpoint, and right-end point methods in Areas under Curves exist and they are identical. The limit is denoted by $\int_a^b f(x)dx$ and called the definite integral of $f(x)$ on the closed interval $[a,b]$. It is not necessarily that $f(x)\geq 0$ for all $x\in [a,b]$, but if $f(x)\geq 0$ for all $x\in [a,b]$, then $\int_a^b f(x)dx$ is the area under the curve $y=f(x)$ on the interval $[a,b]$. It can be proven that the definite integral $\int_a^b f(x)dx$ does not depend on the choice of partition of $[a,b]$ or the choice of point in each subinterval to evaluate $f(x)$. More specifically, let $P: a=x_0<x_1<x_2<\cdots<x_n=b$ be an arbitrary partition (or a subdivision) of $[a,b]$ with $\Delta x_k=x_k-x_{k-1}$, $k=1,2,\cdots,n$. Also let $x_k’$ be any point in the subinterval $[x_{k-1},x_k]$, $k=1,2,\cdots,n$. Then
$$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^nf(x_k’)\Delta x_k.$$

The definite integral $\int_a^b f(x)dx$ satisfies the following properties.

Theorem. 1. $\int_a^b f(x)dx=-\int_b^a f(x)dx$.

2. $\int_a^a f(x)dx=0$.

3. $\int_a^b cf(x)dx=c\int_a^b f(x)dx$, where $c$ is a constant.

4. $\int_a^b (f(x)+g(x))dx=\int_a^b f(x)dx+\int_a^b g(x)dx$.

The properties 3 and 4 tell us that the definite integral $\int_a^b f(x)dx$ is linear.

5. $\int_a^c f(x) dx+\int_c^b f(x)dx=\int_a^b f(x)dx$.

6. Let $m$ and $M$ be the minimum and the Maximum values of $f(x)$ on $[a,b]$. Then
$$m(b-a)\leq\int_a^b f(x)dx\leq M(b-a).$$

7. If $f(x)\leq g(x)$ on $[a,b]$, then
$$\int_a^b f(x)dx\leq\int_a^b g(x)dx.$$
As a special case, if $f(x)\geq 0$, then $\int_a^b f(x)dx\geq 0$.

Example. Suppose that
$$\int_{-1}^1 f(x)dx=5,\ \int_1^4 f(x)dx=-2,\ \int_{-1}^1 h(x)dx=7.$$
Find

1. $\int_4^1 f(x)dx$.

Solution.
$$\int_4^1 f(x)dx=-\int_1^4 f(x)dx=2.$$

2. $\int_{-1}^1(2f(x)+3h(x))dx$.

Solution. \begin{align*}
\int_{-1}^1(2f(x)+3h(x))dx&=2\int_{-1}^1f(x)dx+3\int_{-1}^1h(x)dx\\
&=31.
\end{align*}

3. $\int_{-1}^4 f(x)dx$.

Solution.
\begin{align*}
\int_{-1}^4 f(x)dx&=\int_{-1}^1 f(x)dx+\int_1^4 f(x)dx\\
&=3.
\end{align*}

Example. Show that the value of $\int_0^1\sqrt{1+\cos x}dx$ is less than $\frac{3}{2}$.

Solution. Since $-1\leq\cos x\leq 1$ and $\sqrt{x}$ is an increasing function on $[0,\infty)$, $\sqrt{1+\cos x}\leq \sqrt{2}$. Hence,
\begin{align*}
\int_0^1\sqrt{1+\cos x}dx&\leq\int_0^1\sqrt{2}dx\\
&=\sqrt{2}\\
&\approx 1.4142136\\
&<\frac{3}{2}.
\end{align*}

Using the symmetries of even functions and odd functions, we obtain the following properties.

Theorem. Let $f:[-a,a]\longrightarrow\mathbb{R}$ be a continuous function. Then

1. If $f$ is an even function,
$$\int_{-a}^a f(x)dx=2\int_0^a f(x)dx.$$

2. If $f$ is an odd function,
$$\int_{-a}^a f(x)dx=0.$$

An Application of Definite Integral: Average Value of a Continuous Functions

Let $f: [a,b]\longrightarrow\mathbb{R}$ be a continuous function. Divide the closed interval $[a,b]$ into $n$ equal subintervals. Choose $n$ samples of $f(x)$ on $[a,b]$
$$f(c_1), f(c_2),\cdots,f(c_n)$$
such that $c_k\in[x_{k-1},x_k]$, $k=1,2,\cdots,n$. Then the average value of these $n$ samples is
\begin{align*}
\frac{f(c_1)+f(c_2)+\cdots+f(c_n)}{n}&=\frac{1}{b-a}(f(c_1)+f(c_2)+\cdots+f(c_n))\frac{(b-a)}{n}\\
&=\frac{1}{b-a}\sum_{k=1}^nf(c_k)\frac{(b-a)}{n}.
\end{align*}
Now we increase the number of sample points to infinity:
\begin{align*}
\lim_{n\to\infty}\frac{f(c_1)+f(c_2)+\cdots+f(c_n)}{n}&=\lim_{n\to\infty}\frac{1}{b-a}\sum_{k=1}^nf(c_k)\frac{(b-a)}{n}\\
&=\frac{1}{b-a}\int_a^b f(x)dx.
\end{align*}

Definition. The average or mean value of a continuous function $f(x)$ on $[a,b]$ is given by
\begin{equation}
\label{eq:mv}
\mathrm{av}(f)=\frac{1}{b-a}\int_a^b f(x)dx.
\end{equation}

Example. Find the average value of $f(x)=\sqrt{4-x^2}$ on $[-2,2]$.

Solution. Notice that $y=\sqrt{4-x^2}$ on $[-2,2]$ is the upper semi-circle centered at the origin with radius 2. Hence,
\begin{align*}
\mathrm{av}(f)&=\frac{1}{2-(-2)}\int_{-2}^2\sqrt{4-x^2}dx\\
&=\pi.
\end{align*}