Suppose you are downloading a large file to your computer. At $t=0$ you clicked “Save as” in your browser and started the download. Let $f(t)$ represent the size of the downloaded data. The number $f(t)$ is what you would see if, at time $t$, you clicked on the partially-downloaded file and checked how much space it takes on your disk. If you are downloading a $700$[Mb] file, then the download progress bar at time $t$ will correspond to the fraction $\frac{f(t)}{700 \text{[Mb]}}$.

The derivative function $f'(t)$ is a description of how the function $f(t)$ changes over time. In our example $f'(t)$ is the download speed. Indeed, if you are downloading at 100 [kb/s], then the function $f(t)$ must increase by 100[kb] each second. If you maintain this download speed the file size will grow uniformly: $f(0)=0$[kb], $f(1)=100$[kb], $f(2)=200$[kb], $\ldots$, f(100)=10[Mb]. The “estimated time remaining” is calculated by dividing the size of the part remaining to be downloaded by the current download speed: \[ \text{time remaining } = \frac{ 700 - f(t) }{ f'(t) } [\textrm{s}]. \] The bigger the derivative is, the faster the download will finish.

The derivative function $f'(t)$ encodes the information about the slope of the function $f(t)$. If the slope of $f(t)$ is big at some value of $t$, this means that the function changes very quickly at that time. At the other extreme we have the points where $f'(t)=0$ which correspond to locations where the function is flat—it is neither increasing nor decreasing.

Give the formula for the function $f(t)$, we can compute the slope of a function at the point $t=\tau$ by using a “rise over run” calculation. We imagine that we start at the point $(\tau, f(\tau))$ on the graph of the function $f(t)$ and move a short distance to the right along the graph of $f(t)$ to get to the point $(\tau+\Delta t, f(\tau+\Delta t))$. The slope of the line between these two points on the curve is an approximation for the slope of $f(t)$: \[ f^\prime(\tau) = \text{slope}_f(\tau) = \frac{ \text{change in} \ f(t) \ \text{near } t=\tau }{ \text{change in}\ t } = \frac{ f(\tau+\Delta t) - f(\tau) }{ \tau+\Delta t\ \ + \ \ \tau }. \]

Derivatives are used widely in many areas of life so it is good that you know what a derivative is now.

The formal definition of the derivative function is obtained using the limit argument: we imagine that we take an infinitely small $\Delta t$ so that the points $(\tau, f(\tau))$ and $(\tau+\Delta t, f(\tau+\Delta t))$ are very close together. \[ f^\prime(\tau) \ \ = \ \ \text{instantaneous slope of } f \text{ at } t=\tau \ \ = \lim_{\Delta t \to 0} \frac{ f(\tau+\Delta t) - f(\tau) }{ \Delta t }. \]

The derivative function $f'(\tau)$ describes the rate of change of $f(t)$ at $t=\tau$. The graphical way of thinking about derivatives If the slope is big for some value of $t$ this means that the “rise over run” of the function is big there. by the formula: \[ \text{slope}_f(\tau) = \frac{ \text{change in} \ f(t) }{ \text{change in}\ t } = \frac{ f(\tau+\Delta t) - f(t) }{ \Delta t }. \] Suppose you are given the formula of some function $f(t)$ and asked questions about the slope of $f(t)$. If Rise over run works for lines since they have the same slope everywhere, but general functions have a different slope at different values of $t$. The mathematically precise way to say “the slope of $f$ at $t$” is “the derivative of $f$ at $t$”. The slope of the function $f(t)$ at a particular point $t=\tau$ is given So we have three synonyms now: slope, rate of change and derivative. OK, I am about to drop the “d” word on you soon, but first let me give some synonyms for the word “slope”. Instead of “slope of $f$”, we can say “rate of change of $f$”, and to be specific we should say the rate of change “with respect to $t$”. In other words when you do the “rise over run” you look at the change in $f$ divided by the change in $t$: Now imagine that every time you had to talk about the “slope” of a function you had to write out $slope_f(t)$. Wouldn't this get annoying after a while? Let's try to come up with a better notation. Let's just write $f'$ instead of $\text{slope}_f$. How about that? But slope where? We need to explicitly show the value of $t$ for which we are talking about. Remember that only for lines is the slope the same everywhere. Curved functions have different slopes at different places in their graph. Our shorthand will be $f'(t) \equiv \text{slope}_f(t)$. In other words when you do the “rise over run” you look at the change in $f$ divided by the change in $t$: \[ \text{slope}_f(t) = \frac{ \text{change in} \ f(t) }{ \text{change in}\ t } = \frac{ f(t+\Delta t) - f(t) }{ \Delta t }. \]

The entire subject of kinematics can be summarized as the following equation: \[ a(t) \overset{\ \int\!dt }{\longrightarrow} v(t) \overset{\ \int\!dt }{\longrightarrow} x(t). \] No matter what shape the function $a(t)$ has over time, we just add all of it in order to compute the total change in the velocity from $t=0$ until $t=\tau$ $v(\tau)-v(0)=v(\tau)-v_i$. Assuming we know the initial condition $v(0)$ we now have the velocity function of the object $v(\tau)$. We obtain the position function $x(t)$ after a second integration step and by specifying the initial position $x(0)$. Kinematics, therefore. does not consist of three equations to memorize, but a recipe for how to compute $v(t)$ and $x(t)$ starting from the acceleration $a(t)$ function, and the numbers $v_i$ and $x_i$.

The good news is that the equations of motion for the UAM (constant acceleration $a(t)=a$) is the only thing you need to know for the exam. All the kinematics questions will involve uniform acceleration. You don't need to compute the double integrals to find $x(t)$ each time, if you can remember the pre-integrated equations of motion $x(t)=\frac{1}{2}at^2 +v_it+x_i$ and $v(t)=at+v_i$ for the motion with uniform acceleration $a(t)=a$. Should you ever forget the equations of motion, you now know how to derive them from first principles.

For certain functions, it is possible to find an anti-derivative function $F(t)$ which is the “running total” of the area under the curve. The area under $f(t)$ between $a$ and $b$ is computed as the change in $F(\tau)$: \[ A(a,b) = F(b) - F(a). \] The anti-derivative function $F(\tau)$ corresponds to the integral calculation without a definite on the limits of integration. Instead, we have an integral with a variable endpoints. Without loss of generality we can always assume that the integral starts at $t=0$ and so we have: \[ F(\tau) = \int^\tau_0 f(t)\;dt + F(0) = A(0,\tau) + F(0). \] We use the Greek letter $\tau$ (tau) to denote the input of $F$ since $t$ is already used as the integration variable. In particular, we can think of the integral starting from $t=0$ until a number for the upper limit:

The choice of $t=0$ as the starting point is completely arbitrary. We could have used a different starting point and obtained a different anti-derivative: \[ F_2(\tau) = \int^\tau_{42} f(t)\;dt + F_2(42) = A(42,\tau) + F_2(42). \] Both $F$ and $F_2$ are anti-derivatives of the function $f$ and contain all the information about the area under the curve \[ A(a,b) = F(b) - F(a) = F_2(b) - F_2(a). \] The effect of the choice of the starting point is to add or subtract a constant term to the anti-derivative. All anti-derivatives of the function $f(t)$ differ only by an additive constant factor $F=F_2 + C$. We call $C$ the integration constant, and its value is usually specified by external conditions.

To find where the maximum of the function occurs, we use the standard calculus approach. We find the derivative of the function $f'(x)=6.75(3x^2-4x+1)$, which tells us the value of the slope of $f(x)$ for all values. Note that $f'(x)$ can be factored as $f'(x)= 6.75(3x-1)(x-1)$. The maximum of a smooth function is a place where it will have zero slope, so we see the exact location of the maximum must be at $x=\frac{1}{3}$. The value of $f$ at the maximum is $f(\frac{1}{3})=6.75\left(\frac{1}{3}\right)\left(\frac{1}{3}-1\right)^2$ and since $6.75=\frac{27}{4}$ we have $f(\frac{1}{3})=\frac{27}{4}\left(\frac{1}{3}\right)\left(\frac{-2}{3}\right)^2=\frac{27}{4}\cdot \frac{1}{3}\cdot\frac{4}{9}=1$.

Consider the binomial $(a+b)$ raised to different powers: \[ \begin{align} (a+b)^2 & = a^2 + 2ab + b^2, \nl (a+b)^3 & = a^3 + 3a^2b + 3ab^2 + b^3, \nl (a+b)^4 & = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. \end{align} \] There is a regular pattern in the coefficients of the above expansions. The binomial expansion formula gives us a formula for raising $(a+b)$ to the power $n$: \[ (a+b)^n=\sum_{k=0}^n{n\choose k}a^{n-k}b^k \] where ${n\choose k}=\frac{n!}{k!(n-k)!}$.

You might be wondering what is going on with the signs and the dot products in the above integrals. You see, technically we have the definition of work done by gravity is $W = \int \vec{F}_g \cdot d\vec{x}$, where $d\vec{x}$ is the displacement vector. In both of the above scenarios, we were moving the object against the force of gravity so $W$ is negative (the dot product of two vectors opposing each other is negative). If gravity did negative work, then it means that the object worked against gravity, thus $U_g = - W_g$. Potential energy is defined as stored work. It is stored because if you let go of the object they will naturally tend to come back together.

Now you know where the equations for uniform accelerated motion come from. Not only that, but now that you understand calculus, I can give you any kind of acceleration function $a(t)$ and you know that to get the $x(t)$ you would have to do two steps of integration. To fully describe the motion, you also need the values of the initial velocity and the initial position which correspond to the integration constants.

    The polynomial //could// be factored, however, in terms of complex factors
    though $ax^2+bx+c=(x-c_1)(x-c_2)$, where $c_1,c_2 \in \mathbb{C}$.
    But let us not get into complex stuff now.

math buzz can be compared to “smoking some of that funny stuff”, as my Calculus teacher told us when he showed us the

It is therefore worth understanding the concepts of calculus because of the numerous applications which they have.

In order to understand the concept of the integral we will draw an analogy with a scenario you are familiar with. Consider the function $\textrm{ba}(t)$ which represents your bank account balance at time $t$, and the function $\textrm{tr}(t)$ which corresponds to the transactions (deposits and withdraws) on your account. The function $\textrm{tr}(t)$ describes the change in the function $\textrm{ba}(t)$, the same way the function $a(t)$ describes the change in $v(t)$. Knowing the balance of your account at the beginning of the month, you can calculate the balance at the end of the month as follows: \[ \textrm{ba}(30)=\textrm{ba}(0)+\int_0^{30} \textrm{tr}(t)\:dt. \]

Once you learn calculus you will start to use it everywhere, especially if you are in a technical field. Are you interested in physics, chemistry, engineering, biology, economics, programming or business? In each of these fields calculus is used as a tool for solving problems. You need to understand calculus in order to use modern tools. Above all, you need calculus if you want to build and invent new tools.

Now suppose that you are not downloading the file to your home computer, but to a remote server. The hosting company that stores your data charges you 1 cent per gigabyte of storage used each second. At the beginning of the download, you will not pay much storage fees since $f(t)$ is small, so the cost will also be small: \[ \text{cost_per_second}(t) = \frac{0.01}{1000}\times f(t) \ \text{[dollars]}, \] where we assume the file size is given in megabytes[Mb]. As the download progresses, you will be using more and more storage on the server and the cost per second will rise. At the end of the download when $f(t_{fin})=700$[Mb], storing the file at this provider will cost you $0.7$ cents per second.

What will be the total cost incurred during the download? To get the total we have to add up all the contributions from the different times. Because the file size kept changing during the download, you can't just calculate the total as a product of total download time multiplied by something. We really have to step time through each second from $t=0$ all the way to $t=t_{fin}$ and add up the instantaneous cost at each time. This is the concept of an integral: \[ \text{total_cost} = \int_{t=0}^{t=t_{fin}} \!\!\!\!\!\!\!\text{cost_per_second}(t) \ dt = \int_{t=0}^{t=t_{fin}} \ \frac{0.01}{1000} \times f(t) \ dt \ \ \ \text{[dollars]}. \]

Calculus is the study of continuous functions $f(t)$ over the real numbers:

• $\mathbb{R}$: the set of real numbers. This set contains any number you can think of, say $3.4321242..$.
• $f(t)$: A function.

\[ f: \mathbb{R} \to \mathbb{R}, \]

  which means that $f$ takes as input some number 
  (usually we call that number $x$ or $t$) 
  and it produces as an output another number $f(t)$.

Differential calculus is all about derivatives:

• $f^\prime(t)$: the derivative of $f(t)$ is the rate of change of $f$ at $t$.

The derivative is also a function of the form

  \[
     f^\prime: \mathbb{R} \to \mathbb{R},
  \]
  The output of $f^\prime(t)$ represents the //slope// of 
  a line parallel (tangent) to $f$ at the point $(t,f(t))$.

Integral calculus is all about integration:

• $\int_a^b f(t) dt$: the integral of $f(t)$ corresponds to the area

$A(a,b)$ of the region under $f(t)$ from $t=a$ to $t=b$ is given by

  \[
      A(a,b) = \int_a^b f(t)\;dt.
  \]
  The integral sign $\int$  is a mnemonic for //sum//.
  Indeed the integral is a calculation of the total or "sum" of 
  $f(t)$ over that interval $[a,b]$.

You can feel the magnetic force $\vec{F}_b$ if you try to bring to magnets together. Because of the equation $F=ma$, we can write the Newton in terms of the basic units of mass length and time $[N]=\left[ \frac{ \textrm{kg}\:\textrm{m} }{ \textrm{s}^2 } \right]$.

In the next sections we will look a little mode closely at each of these concepts and work out a few examples together. The goal of this chapter is for you to develop the intuitive understanding of slopes and areas under the curve. We will delay the detailed mathematical study until the calculus chapter later in the book.

Computers are powerful too. By using sympy live or Wolfram alpha we can get the computer do calculus for us.

Say you have to compute this integral: \[ \int \log^4(x) \ dx. \]

This is how you would write that down in sympy:

  In [1]: from sympy import *
  In [2]: x = Symbol(’x’)

  In [3]: Integral( log(x)**4, x)
  Out[3]:
  ⌠
  ⎮    4
  ⎮ log (x) dx
  ⌡

But it is not enough to just write it down, you need to do it!

  In [4]: Integral( log(x)**4, x).doit()
  Out[4]:
                                  2             3           4
     24*x - 24*x*log(x) + 12*x*log (x) - 4*x*log (x) + x*log (x)

See this example on Wolfram alpha.

Bear with me for one more minute, because we are getting to the point now. Assuming the slope of the function changes smoothly, then the slope of the function must be equal to zero at $t_{max}$. How else would the slope get from being positive ($f$ increasing) to negative ($f$ decreasing) if you do not pass by being zero in between ($f$ flat)? Thus we have learned a very powerful fact: the slope of the function at a maximum is zero: \[ t_{max} \text{ is a maximum of } f \quad \Rightarrow \quad f'(t_{max})= 0. \] It is important to take a piece of paper and a pen and squiggle some random “smooth” curve. Look at the highest points of the squiggle and check that the slope of the curve changes from positive to negative near each of them. Exactly at the peak, the function is horizontal. Imagine you zoom in many times near the maximum value – it will look flat right?

Using this fact in reverse, we can come up with a whole algorithm for finding the maximum of $f(t)$:

1. Find $f'(t)$, the derivative of $f(t)$.
2. Find all the values of $t$ where the slope is zero:

\[ \text{candidates} = \{t_1, t_2, t_3, \ldots, \} = \{ t_i \in \mathbb{R} \ : \ f'(t_i) = 0 \}. \]

1. Check which of $\{t_1, t_2, t_3, \ldots \}$ is the maximum.

That is it. That's the entire Calculus I course. I am not kidding you. Of course, no textbook publisher will tell you that, because then it wouldn't make sense for you to buy a 400 page calculus book. You see, instead of teaching you about optimization, the textbook publishers are using optimization to find the maximum price they can get from you for teaching you this simple algorithm. To be totally frank with you, I have to say that I want your money too, but at least I am not going to try to bullshit you that this is hard stuff which needs hundreds of pages to be explained.

Dear reader, I have told you that Calculus is really easy, but that doesn't mean that you should get the party bong and the beer funnel out yet. I still have to teach you how to do all these things. Taking derivatives is really easy stuff, but you have to learn how to use the derivative rules and memorize some derivative formulas, and of course we will have to go over the definition of a derivative more rigorously. It will still take us 20 pages or so, and you need to give it at least a few evenings to practice your skills. But don't worry, it is all chill stuff. Just don't fall behind on your assignments.

Assuming you knew the function $f(t)$, is there some mathematical way to find the optimal value of $t$ such that your financial gain is maximal? Yep. That's why they force you to take that class, so let's get right into it. This is the minireference promise: we get to the good stuff right away.

area $A$ of the region under $f(t)$ from $t=a$ to $t=b$ is given by

  \[
      A = \int_a^b f(t) dt = F(b) - F(a).
  \]

Well, this is the deal, you only have to compute the integral once and then simply use the formula anytime a problem involving this integral comes up.

The relationship “$F(x)$ is the integral of $f(x)$” two functions

I want to make an important observation at this point. If you know the slope $f'(t)$ for all values of $t$, then you can reverse engineer the function $f(t)$ up to a constant. Given $f'(t)$, the act of finding $f(t)$ is called “integration” or “taking the integral”, since you are taking all the changes and putting them together to calculate the total change in $f(t)$. Since the derivative contains only information about the change in $f$, taking the integral of $f'(t)$ between two values of $t$ (say points $t_a$ and $t_b$) tells you the total change in $f$ between these two points: $f(t_b)-f(t_a)$.

The are under the function $f(t)=c$ for some for More generally we can have the general formula for \[ F(\tau) = \int_0^\tau b\;dt = c\tau + C. \]

In many problems you will be told the general form of the function in some problem. For example, you are told that there is a linear relationship between the input current $I$ and the resulting voltage $V$ across some electric device. Note

\[ \frac{d}{dx}\left( f(g(h(x))) \right) = \frac{df}{dg}\frac{dg}{dh}\frac{dh}{dx}. \]

SERIES \dokutitleleveltree{Discussion}

The key strategy to prove that series converges or diverges is to compare it to another series which you know to converge. Thus for example if you know that $a_n \leq b_n$ for all $n$, and you know that $\sum b_n$ converges, the it must be that $a_n$ also converges. All the tests are of this form, where you have some

We can use the binomial formula to obtain the power series of $e^x$ by another route. Recall that $e^1 = \lim_{n\to\infty}(1+\frac{1}{n})^n$. From this we get: \[ e^x = \lim_{n\to\infty}(1+\frac{x}{n})^n = 1^n + {n\choose 1}1^{n-1}\left(\frac{x}{n}\right)^1 + {n\choose 2}1\left(\frac{x}{n}\right)^2 + {n\choose 3}1\left(\frac{x}{n}\right)^3 + \]

An arrow is shot from a bow towards a target. Before the arrow hits the target it must travel half the distance, and then it must travel half the remaining distance, and so on and so forth. It seems then, that the arrow will always have some distance (however short) to travel in front of it before it reaches the target. Therefore, concludes Zeno, the arrow will never reach the target.

No. Zeno got that part wrong. Let's say that at time index t=1 (not necessarily equal to 1 sec), the arrow has traveled to mid point of its flight and that at time index 2 at the mid point of the remaining half, and at t=3 it is one eight away of the total distance form the target. The distance as a function of the time index is \[ d(t) = \frac{1}{2^t}. \] In the limit of many time indices the arrow will get closer and closer to the target. In the limit we say, it reaches it: \[ \lim_{t\to \infty} d(t) = \lim_{t\to \infty} \frac{1}{2^t} = 0. \]

relate to 1sec – and derivative / integral example w/ file download

Infinitesimals are the only way to make mathematical statements really precise. For example we say that a function is left continuous at $a$ if we have that \[ \lim_{x\to a} f(x) = f(a), \] which is short hand for saying that for any precision level $\delta > 0$, there exists $\epsilon$ such that \[ | f(x - \epsilon) \]

A derivative can be brought under the intergral sign (see section sec:convf for the required conditions): \[ \frac{d}{dy}\left[\int\limits_{x=g(y)}^{x=h(y)}f(x,y)dx\right]= \int\limits_{x=g(y)}^{x=h(y)}\frac{\partial f(x,y)}{\partial y}dx-f(g(y),y)\frac{dg(y)}{dy}+f(h(y),y)\frac{dh(y)}{dy} \]

For readers who are not enrolled in a Calculus II class, and don't have to sit a final exam, I recommend you skip the integration techniques section as they are kind of tedious and complicated. It is good stuff to know, but in general if you have to take an integral in the real world it will most likely be very complicated so you will be better off to type it into wolfram alpha.

Now imagine you are ship-wrecked on a desert island and the only calculator you have has four buttons on it [+], [-], [$\times$], [$\div$]. No $\sin$. And suppose your life depended on knowing the value of $\sin(40^\circ)$. No seriously. You know that there is an inhabited island exactly at $40^\deg$ with respect to the North. You can shine the laser in the direction of the island and signal for help: \[ . . . \ \ - - - \ \ . . . \] which will either be understood as the morse code for SOS, or alternately the someone might get hit in the eye by your laser and get sufficiently annoyed to get on a boat and kick the ass of whomever is playing with a laser. Either way, if you only knew the direction to shine the laser in you would probably survive. Can you compute $\sin(40^\circ)$ using only the basic calculator?

$\frac{GM}{r^2} \approx 9.81 [\frac{m}{s^2}]$,

Calculus is useful math. Useful for physics, chemistry, biology, business and all kinds of other areas. You need calculus in order to do quantitative analysis of how variables change over time (derivatives) or sum up all kinds of contributions that add up to a total (integration).

The main part of calculus is the study of continuous functions $f(x)$. These are the concepts we will be playing with in this course.

• $\mathbb{R}$: A Real number, basically any number you can think of $3.4321242..$.
• $f(x)$: A function.

The study of calculus is the study

  of the properties of functions
  \[
     f: \mathbb{R} \to \mathbb{R},
  \]
  which means that $f$ takes as input some number (usually we call that number $x$) 
  and it produces as an output another number $f(x)$ (sometimes we also give an alias for the output $y=f(x)$).
* $\lim_{\epsilon \to 0}$: limits are the mathematically rigorous 
  way of speaking about very large and very small numbers.
* $f'(x)$: the derivative of $f(x)$ is the rate of change of $f$ at $x$.
  The derivative is also a function of the form
  \[
     f': \mathbb{R} \to \mathbb{R},
  \]
  The output of $f'(x)$ we sometimes call $m$ and represents the //slope// of 
  a line parallel to $f$ at the point $(x,f(x))$.
  {{ :calculus:280px-integral_as_region_under_curve.svg.png?180|}}
* $\int f(x) dx$: the integral of $f(x)$.
  More precisely we can define the //antiderivative// of $f(x)$ as follows
  \[
     F(x) = \int_0^x f(x) dx \ \ + \ \ F(0).
  \] 
  The area $S$ of the region under $f(x)$ from $x=a$ to $x=b$ is given by
  \[
      \int_a^b f(x) dx = F(b) - F(a).
  \]
  The $\int$ sign is a mnemonic for //sum//.
  Indeed the integral is nothing more than the "sum" of $f(x)$. \\
  The name antiderivative comes from the fact that 
  \[
     F'(x) = f(x),
  \]
  which follows from the //fundamental theorem of calculus//.

A relatively separate topic of sequences and series is also commonly taught in Calculus II. The following new concepts will be used.

• $\mathbb{N}$: A Natural number, which means positive integer, which means

whole number like 1, 2, 3 and 499201.

• $a_n$: sequence of numbers $a_1, a_2, a_3, a_4, \ldots$.

You can also think about each sequence as a funciton

  \[
     a: \mathbb{N} \to \mathbb{R},
  \]
  where the input is $n$ an integer (index into the sequence) and
  the output is $a_n$ which could be any number.
* $\sum$: sum. Means to take the sum of several objects
  put together. The summation sign is the short way to express 
  certain long expressions:
  \[
    a_3 + a_4 + a_5 + a_6 + a_7 = \sum_{3 \leq i \leq 7} a_i.
  \]
  Note that summations could go up to infinity.
* $\sum a_i$: series. The running total of a sequence until $n$.
  \[
     S_n = \sum_{1\leq i \leq n} a_i  = a_1 + a_2 + \ldots + a_{n-1} + a_n.
  \]
* $f(x)=\sum_{i=0}^\infty a_i*x^i$ taylor series is a way to approximate
  any function as an infinitely long polynomial.

The derivatives of Calculus I and integrals of Calculus II can also be generalized to functions of more than one variable $f(x,y)$. We will need the following new concepts to study multivariate calculus.

• $\frac{\partial}{\partial x}$ will take the place of $\frac{d}{dx}$,

in order to indicate that we are taking the slope in the $x$ direction,

  but keeping the function constant in other directions.
* $\frac{\partial}{\partial y}$ is similarly taking the derivative with
  respect to $y$ keeping $x$ constant.
  ($\frac{\partial f}{\partial y}$ is pronounced "del f del y". )
* $\nabla=(\frac{\partial}{\partial x},\frac{\partial}{\partial y})$: 
  the greek symbol nabla, is used to denote derivative 
  in both $x$ and $y$ directions.
  Note that nabla has two components so it is "vector like".
  Some textbooks will make this fact explicit and write $\vec{\nabla}$.
* $\vec{\nabla}f=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$:
  gradient of the function $f(x,y)$. 
  The gradient is the "slope" of a function of two variables.

The last kind of calculus that scientists and engineers learn in university is called Advanced calculus and deals with vector functions $\vec{F}({r})=(F_x({r}),F_y({r}))$. Vector calculus, is very useful for physicists and engineers because it allows you to study precisely phenomena like electromagnetism.