# The Product and Quotient Rules

Product Rule: Let $u=f(x)$ and $v=g(x)$ be differentiable functions. Then $$(fg)’(x)=f(x)g’(x)+f’(x)g(x)$$ or $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+\frac{du}{dx}v.$$

Proof. \begin{eqnarray*}(fg)’(x)&=&\lim_{\Delta x\to 0}\frac{fg(x+\Delta x)-fg(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}g(x+\Delta x)+f(x)\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}\\&=&f’(x)g(x)+f(x)g’(x).\end{eqnarray*} Note that $\displaystyle\lim_{\Delta x\to 0}g(x+\Delta x)=g(x)$ because $g(x)$ is continuous.

Example. Using the product rule, differentiate $(x^2+2x-1)(x^3-4x^2)$.

Solution. \begin{eqnarray*}\frac{d}{dx}[(x^2+2x-1)(x^3-4x^2)]&=&\frac{d(x^2+2x-1)}{dx}(x^3-4x^2)+(x^2+2x-1)\\&&\frac{d(x^3-4x^2)}{dx}\\&=&(2x+2)(x^3-4x^2)+(x^2+2x-1)(3x^2-8x)\\&=&(2x^4-6x^3-8x^2)+(3x^4-2x^3-19x^2+8x)\\&=&5x^4-8x^3-27x^2+8x.\end{eqnarray*} Multiplyng first, \begin{eqnarray*}(x^2+2x-1)(x^3-4x^2)&=&x^5-4x^4+2x^4-8x^3-x^3+4x^2\\&=&x^5-2x^4-9x^3+4x^2.\end{eqnarray*} The derivative of this is $5x^4-8x^3-27x^2+8x$ by the power rule and differentiation formulas we discussed here.

Reciprocal Rule (Baby Quotient Rule): Let $v=g(x)$ be a differentiable function with $g(x)\ne 0$. Then $$\left(\frac{1}{g}\right)’(x)=\frac{-g’(x)}{[g(x)]^2}$$ or $$\frac{d}{dx}\left(\frac{1}{v}\right)=-\frac{1}{v^2}\frac{dv}{dx}.$$

Proof. \begin{eqnarray*}\left(\frac{1}{g}\right)’(x)&=&\lim_{\Delta x\to 0}\frac{\frac{1}{g(x+\Delta x)}-\frac{1}{g(x)}}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{\frac{g(x)-g(x+\Delta x)}{g(x+\Delta x)g(x)}}{\Delta x}\\&=&-\lim_{\Delta x\to 0}\frac{\frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x+\Delta x)g(x)}\\&=&-\frac{g’(x)}{[g(x)]^2}.\end{eqnarray*}

Example. Differentiate $\frac{1}{\sqrt{x}+2}$.

Solution. \begin{eqnarray*}\frac{d}{dx}\frac{1}{\sqrt{x}+2}&=&-\frac{d(\sqrt{x}+2)/dx}{(\sqrt{x}+2)^2}\\&=&-\frac{1}{2\sqrt{x}(\sqrt{x}+2)^2}.\end{eqnarray*}

Using the product rule and the reciprocal rule, we can prove

Quotient Rule: Let $u=f(x)$ and $v=g(x)$ be differentiable functions and assume that $g(x)\ne 0$. Then $$\left(\frac{f}{g}\right)’(x)=\frac{f’(x)g(x)-f(x)g’(x)}{[g(x)]^2}$$ or $$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\frac{du}{dx}v-u\frac{dv}{dx}}{v^2}.$$

Proof. \begin{eqnarray*}\left(\frac{f}{g}\right)’(x)&=&\left(f\frac{1}{g}\right)’(x)\\&=&f’(x)\frac{1}{g(x)}+f(x)\left(\frac{1}{g}\right)’(x)\ (\mbox{the product rule is applied})\\&=&\frac{f’(x)}{g(x)}-f(x)\frac{g’(x)}{[g(x)]^2}\ (\mbox{the reciprocal rule is applied})\\&=&\frac{f’(x)g(x)-f(x)g’(x)}{[g(x)]^2}.\end{eqnarray*}

Example. Find the derivative of $h(x)=\frac{2x+1}{x^2-2}$.

Solution. \begin{eqnarray*}h’(x)&=&\frac{(2x+1)’(x^2-2)-(2x+1)(x^2-2)’}{(x^2-2)^2}\\&=&\frac{2(x^2-2)-(2x+1)2x}{(x^2-2)^2}\\&=&\frac{2x^2-4-4x^2-2x}{(x^2-2)^2}\\&=&-\frac{2x^2+2x+4}{(x^2-2)^2}.\end{eqnarray*}

Example. Differentiate $\frac{x^2+2}{x^8}$.

Solution. Since the function is a rational function, you may hastily try to use the quotient rule to differentiate it. There is nothing wrong with that except there may be a simpler way to differentiate the function. In fact the function can be written as $$\frac{x^2+2}{x^8}=\frac{x^2}{x^8}+\frac{2}{x^8}=\frac{1}{x^6}+2x^{-8}=x^{-6}+2x^{-8}.$$ Thus the derivative is $$-6x^{-7}-16x^{-9}=-\frac{6}{x^7}-\frac{16}{x^9}.$$

# Continuity versus Differentiability

There is a close relatioship between continuity and differentiability, namely

Theorem 18. If $f’(x_0)$ exists then $f(x)$ is continuous at $x_0$; i.e. $\displaystyle\lim_{x\to x_0}f(x)=f(x_0)$. However the converse need not be true.

Proof. \begin{eqnarray*}\lim_{x\to x_0}[f(x)-f(x_0)]&=&\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)\\&=&f’(x)\cdot 0\\&=&0.\end{eqnarray*}

Example. [A Counterexample for the Converse] The function $f(x)=|x|$ is continuous at $x=0$ but has no derivative at $x=0$.

Proof. \begin{eqnarray*}\lim_{x\to 0+}\frac{f(x)-f(0)}{x-0}&=&\lim_{x\to 0+}\frac{|x|}{x}\\&=&\lim_{x\to 0+}\frac{x}{x}\\&=&1,\end{eqnarray*} while \begin{eqnarray*}\lim_{x\to 0-}\frac{f(x)-f(0)}{x-0}&=&\lim_{x\to 0-}\frac{|x|}{x}\\&=&\lim_{x\to 0-}\frac{-x}{x}\\&=&-1.\end{eqnarray*} Hence, $f’(0)=\displaystyle\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}$ does not exist.

# Basic Differentiation Formulas

Let us recall the definition of the derivative $$f’(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.$$ Replace $h$ by $\Delta x$. Then $f’(x)$ is rewritten as $$f’(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$ In mathematics, $\Delta$ often means an increment. So $\Delta x$ means an increment of $x$. Note that $\Delta x$ could be positive or negative. Denote by $\Delta y$ the difference $f(x+\Delta x)-f(x)$. $\Delta y$ is called an increment of $y$. Hence, the average rate of change of $y$ with respect to $x$ in the interval $[x,x+\Delta x]$ is the difference quotient $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$  In the view of Gottfried Leibniz, one could think of $\Delta x$ as becoming infinitesimal. The resulting quantity is denoted by $dx$. When $\Delta x$ becomes the infinitesimal $dx$, $\Delta y$ simultaneously becomes the infinitesimal $dy$. The infinitesimals $dx$ and $dy$ are called differentials. Hence the ratio $\frac{\Delta y}{\Delta x}$ becomes $\frac{dy}{dx}$ accordingly, and it is exactly equal to $f’(x)$. The quantity $\frac{dy}{dx}$ can be viewed as the ratio of differentials or as s synonym for $f’(x)$.

Leibniz Notation: If $y=f(x)$, the derivative $f’(x)$ can be written $$\frac{dy}{dx},\frac{df(x)}{dx},\ \mbox{or}\ \frac{d}{dx}f(x).$$ This is just a notation and does not represent a division. Using Leibniz notaion, the value $f’(a)$ of $f’(x)$ at a specific point $x=a$ can be written $$\left.\frac{dy}{dx}\right|_{x=a}\ \mbox{or}\ \left.\frac{df(x)}{dx}\right|_{x=a}.$$

Calculating derivatives using the limit defintion can be really laborious. In actual practice, special rules and formulas are derived for differentiating certain types of functions. The following are such rules and they can be proved striaghtforwardly by the definition of the derivative.

Theorem 15. Let $c$ be a constant, and $f(x)$ and $g(x)$ be two differentiable functions of $x$.

1. $\displaystyle\frac{dc}{dx}=0$
2. $\displaystyle\frac{d(cf(x))}{dx}=c\frac{df(x)}{dx}$
3. $\displaystyle\frac{d[f(x)+g(x)]}{dx}=\frac{df(x)}{dx}+\frac{dg(x)}{dx}$

The converse of the first rule is also true, namely if $f’(x)=0$ for all $x$ in the domain then $f(x)$ is a constant function. This will be proved later.

Lemma 16. [Binomial Theorem] \begin{eqnarray*}(a+b)^n&=&\begin{pmatrix}n\\0\end{pmatrix}a^nb^0+ \begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\begin{pmatrix}n\\2\end{pmatrix}a^{n-2}b^2+\cdots+\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k+\\&&\cdots+\begin{pmatrix}n\\n-1\end{pmatrix}ab^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}a^0b^n\\&=&a^n+\begin{pmatrix}n\\1\end{pmatrix}a^{n-1}b+\begin{pmatrix}n\\2\end{pmatrix}a^{n-2}b^2+\cdots+\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k+\\&&\cdots+\begin{pmatrix}n\\n-1\end{pmatrix}ab^{n-1}+b^n,\end{eqnarray*} where $$\begin{pmatrix}n\\k\end{pmatrix}=\frac{n!}{k!(n-k)!}.$$ $\begin{pmatrix}n\\k\end{pmatrix}$ is also denoted by $n{\mathrm C}k$. The binomial coefficients $\begin{pmatrix}n\\k\end{pmatrix}$ can be also easily obtained by Pascal’s triangle. For details see here and here.

Theorem 17. [Power Rule] $\displaystyle\frac{dx^n}{dx}=nx^{n-1}$

Proof. Let $y=x^n$. Then by the Binomial Theorem \begin{eqnarray*}\frac{dx^n}{dx}&=&\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{(x+\Delta x)^n-x^n}{\Delta x}\\&=&\lim_{\Delta x\to 0}\frac{\left[x^n+nx^{n-1}\Delta x+\frac{n(n-1)}{2!}x^{n-2}(\Delta x)^2+\cdots+(\Delta x)^n\right]-x^n}{\Delta x}\\&=&\lim_{\Delta x\to 0}\left[nx^{n-1}+\frac{n(n-1)}{2!}x^{n-2}\Delta x+\cdots+(\Delta x)^{n-1}\right]\\&=&nx^{n-1}.\end{eqnarray*}

Example. If $y=3x^5$, then by Property 2 of Theorem 15 and Power Rule (Theorem 17), we have \begin{eqnarray*}\frac{dy}{dx}&=&3\frac{dx^5}{dx}\\&=&3(5)x^{5-1}\\&=&15x^4.\end{eqnarray*}

Remark. In Theorem 17 Power Rule is established only for the case when $n$ is a positive integer. The formula is indeed valid for all real $n$’s.

Example. If $y=8x^{-\frac{3}{4}}$, then we have $$\frac{dy}{dx}=8\left(-\frac{3}{4}\right)x^{-\frac{3}{4}-1}=-6x^{-\frac{7}{4}}.$$

Using Theorems 15 and 17, we can now find the derivative of any polynomial function.

Example. If $y=2x^4-x^3-2x+7$, then \begin{eqnarray*}\frac{dy}{dx}&=&\frac{d(2x^4)}{dx}-\frac{d(x^3)}{dx}-2\frac{dx}{dx}+\frac{d(7)}{dx}\\&=&8x^3-3x^2-2.\end{eqnarray*}

Example. The function $f(x)=\displaystyle\frac{3x^3-4}{x^2}$ does not appear to be a power function, but it actually can be written as a power function. $$f(x)=\frac{3x^3-4}{x^2}=\frac{3x^3}{x^2}-\frac{4}{x^2}=3x-4x^{-2}.$$ Hence we have $$f’(x)=\frac{d(3x)}{dx}-\frac{d(4x^{-2})}{dx}=3+8x^{-3}.$$

Example. The function $y=\root 3\of{x^2}-3\root 3\of{x}-5$ can be also written as a power function. \begin{eqnarray*}y&=&\root 3\of{x^2}-3\root 3\of{x}-5\\&=&(x^2)^{\frac{1}{3}}-3x^{\frac{1}{3}}-5\\&=&x^{\frac{2}{3}}-3x^{\frac{1}{3}}-5.\end{eqnarray*} Hence, $$\frac{dy}{dx}=\frac{2}{3}x^{-\frac{1}{3}}-x^{-\frac{2}{3}}.$$

# Homology 3: Cycle Groups and Boundary Groups

Let us use $\langle\cdots\rangle$ for an unoriented simplex and $(\cdots)$ for an oriented simplex.

Examples. 1. $(p_0p_1)=-(p_1p_0)$.

2. \begin{eqnarray*}\sigma_2&=&(p_0p_1p_2)=(p_2p_0p_1)=(p_1p_2p_0)\\-(p_0p_2p_1)&=&-(p_2p_1p_0)=-(p_1p_0p_2).\end{eqnarray*}

Let $K=\{\sigma_\alpha\}$ be an $n$-dimensional simplicial complex of oriented simplexes.

Definition. The $r$-chain group $C_r(K)$ of a simplicial complex $K$ is a free abelian group generated by $r$-simplexes of $K$. If $r>\dim K$, $C_r(K):=0$. An element of $C_r(K)$ is called an $r$-chain.

Let there be $N_r$ $r$-simplexes in $K$. Denote them by $\sigma_{r,i}$ ($1\leq i\leq N$). Then $c\in C_r(K)$ is expressed as $$c=\sum_{i=1}^{N_r}c_i\sigma_{r,i},\ c_i\in\mathbb Z.$$ The integers $c_i$ are called the coefficients of $c$. The addition of two $r$-chains $\sum_ic_i\sigma_{r,i}$ and $c’=\sum_ic_i’\sigma_{r,i}$ is $$c+c’=\sum_i(c_i+c_i’)\sigma_{r,i}.$$ The unit element is $0=\sum_i0\cdot\sigma_{r,i}$. The inverse element of $c$ is $-c=\sum_i(-c_i)\sigma_{r,i}$. Hence we see that $C_r(K)$ is a free abelian group of rank $N_r$ $$C_r(K)\cong\stackrel{N_r}{\overbrace{\mathbb Z\oplus\mathbb Z\oplus\cdots\oplus\mathbb Z}}.$$

Denote the boundary of an $r$-simplex $\sigma_r$ by $\partial_r\sigma_r$. Since a 0-simplex has no boundary, $$\partial_0p_0=0.$$ For a 1-simplex $(p_0p_1)$, $$\partial_1(p_0p_1):=p_1-p_0.$$ Let $\sigma_r=(p_0\cdots p_r)$ ($r>0$) be an oriented $r$-simplex. The boundary $\partial_r\sigma_r$ of $\sigma_r$ is an $(r-1)$-chain defined by $$\partial_r\sigma_r:=\sum_{i=0}^r(-1)^i(p_0p_1\cdots\hat{p}_i\cdots p_r)$$ where the point $p_i$ under $\hat{}$ is omitted. For example, \begin{eqnarray*}\partial_2(p_0p_1p_2)&=&(p_1p_2)-(p_0p_2)+(p_0p_1),\\\partial_3(p_0p_1p_2p_3)&=&(p_1p_2p_3)-(p_0p_2p_3)+(p_0p_1p_3)-(p_0p_1p_2).\end{eqnarray*} The boundary $\sigma_r$ defines a homomorphism called the boundary operator $$\partial_r: C_r(K)\longrightarrow C_{r-1}(K);\ c=\sum_i c_i\sigma_{r,i}\longmapsto\partial_rc=\sum_ic_i\partial_r\sigma_{r,i}.$$

Let $K$ be an $n$-dimensional simplicial complex. Then there exists a sequence of free abelian groups and homomorphisms $$0\stackrel{i}{\hookrightarrow}C_n(K)\stackrel{\partial_n}{\longrightarrow}C_{n-1}(K)\stackrel{\partial_{n-1}}{\longrightarrow}\cdots\stackrel{\partial_2}{\longrightarrow}C_1(K)\stackrel{\partial_1}{\longrightarrow}C_0(K)\stackrel{\partial_0}{\longrightarrow}0.$$ This sequence is called the chain complex associated with $K$ and is denoted by $C(K)$.

Definition. $Z_r(K):=\ker\partial_r\subset C_r(K)$ is called the $r$-cycle group. The elements of $Z_r(K)$ are called $r$-cycles. If $c\in Z_r(K)$, i.e. if $c$ is an $r$-cycle, $\partial_rc=0$. If $r=0$, $\partial_rc=0$ for all $c\in C_0(K)$, so $C_0(K)=Z_0(K)$.

Definition. Let us consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)$ and let $c\in C_r(K)$. If there exists $d\in C_{r+1}(K)$ such that $c=\partial_{r+1}d$, then $c$ is called an $r$-boundary. The set of $r$-boundaries $B_r(K)$ ($=\partial_{r+1}C_{r+1}(K)={\rm Im}\partial_{r+1}$) is a subgroup of $C_r(K)$ called the $r$-boundary group. If $K$ is an $n$-dimensional simplicial complex, $B_n(K)=0$.

Consider $C_{r+1}(K)\stackrel{\partial_{r+1}}{\longrightarrow}C_r(K)\stackrel{\partial_r}{\longrightarrow}C_{r-1}(K)$. Then the following lemma holds.

Lemma. The composite map $\partial_r\partial_{r+1}:C_{r+1}(K)\longrightarrow C_{r-1}(K)$ is a zero map.

Proof. Since $\partial_r$ is a linear operator on $C_r(K)$, it suffices to prove the identity $\partial_r\partial_{r+1}=0$ for the generators of $C_{r+1}(K)$. If $r=0$, $\partial_0\partial_1=0$ since $\partial_0$ is a zero operator. Let us assume that $r>0$. Take $\sigma=(p_0\cdots p_rp_{r+1})\in C_{r+1}(K)$. \begin{eqnarray*}\partial_r\partial_{r+1}\sigma&=&\partial_r\sum_{i=0}^{r+1}(-1)^i(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\partial_r(p_0\cdots \hat{p}_i\cdots p_{r+1})\\&=&\sum_{i=0}^{r+1}(-1)^i\{\sum_{j=0}^{i-1}(-1)^j(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{j=i+1}^{r+1}(-1)^{j-1}(p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\}\\&=&\sum_{i=0}^{r+1}\sum_{j=0}^{i-1}(-1)^{i+j}(p_0\cdots \hat{p}_j\cdots\hat{p}_i\cdots p_{r+1})+\\&&\sum_{i=0}^{r+1}\sum_{j=i+1}^{r+1}(-1)^{i+j-1}p_0\cdots \hat{p}_i\cdots\hat{p}_j\cdots p_{r+1})\\&=&0.
\end{eqnarray*}

Theorem. $B_r(K)\subset Z_r(K)$ or equivalently $\mathrm{Im}\partial_{r+1}\subset\ker\partial_r$.

Proof. Let $c\in B_r(K)$ Then $c=\partial_{r+1}d$ for some $d\in C_{r+1}(K)$. By Lemma, $\partial_rc=\partial_r\partial_{r+1}d=0$. Hence, $c\in\ker\partial_r=Z_r(K)$.

# A Physical Meaning of Derivative: Velocity and Acceleration

Let us assume that a particle is moving along a straight line and that the function $s=f(t)$ describes the position of moving particle at the time $t$. In physics, such a function $s=f(t)$ is called a motion.

Suppose the particle passes the points $P$ and $Q$ at the times $t$ and $t+\Delta t$, respectively. If $s$ and $s+\Delta s$ are the respective distances from some fixed point $O$, then the average velocity of the particle during the time interval $\Delta t$ is $$\frac{\Delta s}{\Delta t}=\frac{f(t+\Delta t)-f(t)}{\Delta t}=\frac{\mbox{Distance Traveled}}{\mbox{Time Elapsed}}.$$ The instantaneous velocity $v$ of the particle at the time $t$ is then given by the derivative of motion $s=f(t)$ $$v=\frac{ds}{dt}=\lim_{\Delta t\to 0}\frac{\Delta s}{\Delta t}.$$ In physics, the intantaneous velocity is also denoted by $\dot{s}$ or $\dot{f}(t)$. This dot notation was introduced by Sir Issac Newton.

Similaryl, if $\Delta v$ is the change in the velocity of the particle as it moves from $P$ to $Q$ during the time interval $\Delta t$, then $$a=\frac{dv}{dt}=\lim_{\Delta t\to 0}\frac{\Delta v}{\Delta t}$$ is the acceleration of the particle at the time $t$. Using dot notation, the acceleration is also denoted by $\dot{v}$, $\ddot{s}$, $\ddot{f}(t)$, or $\frac{d^2 s}{dt^2}$. The last notation $\frac{d^2 s}{dt^2}$ is due to Gottfried Leibniz.

If a body is thrown vertically upward with a certain initial velocity $v_0$, its distance $s$ from the starting point is given by the formula $$s(t)=v_0t-\frac{1}{2}gt^2,$$ where $g$ is the gravitational constant $g=9.8\mbox{m}/\mbox{sec}^2=32\mbox{ft}/\mbox{sec}^2.$

Example. From the top of a building 96 feet high, a ball is thrown directly upward with a velocity of 80 feet per second. Find (a) the time required to reach the highest point, (b) the maximum height attained, and (c) the velocity of the ball when it reaches the ground.

Solution. $v_0=80$ ft/sec and $g=32\mbox{ft}/\mbox{sec}^2$, so $$s=80t-16t^2$$ and $$v=\frac{ds}{dt}=80-32t.$$

(a) At the heighest point, $v=0$ that is $0=80-32t$. So, $t=\frac{5}{2}$.

(b) $s\left(\frac{5}{2}\right)=80\left(\frac{5}{2}\right)-16\left(\frac{5}{2}\right)^2=100$ft. Hence the height of the ball above the ground is 196 feet.

(c) Since the ball will reach the ground when $s=-96$, it follows that $-96=80t-16t^2$ or $16(t-6)(t+1)=0$. Hence $t=6$ and the velocity is $v(6)=80-32\cdot 6=-112$ft/sec when the ball strikes the ground. The negative sign merely indicates that the velocity of the ball is directed downward.