How to learn calculus for beginners?

Question

As a precalculus student interested in teaching myself calculus, where should I start and how should I go about learning?

This question is different than past questions as I am not solely interested in finding a book.

Answer 1

Look at KhanAcademy, they have great explanation for calculus (start by order). https://www.khanacademy.org/math/calculus

Also see MIT | Big Picture courses on calculus. http://www.youtube.com/playlist?list=PLA61F034AC6BB546D

For textbook I recommend this one: Calculus Made Easy - Silvanus P. Thompson

http://www.amazon.com/Calculus-Made-Easy-Silvanus-Thompson/dp/0312185480

It costs only 17 box and has great reviews for example: This book is fantastic! If only I had it in college. I happened across Calc Made Easy in a bookstore on a business trip. In college, I was a math major and, at first, struggled through the concept of calculus and was intimidated by the terminology. I haven't done calculus in 25 years, but I read this whole book in one afternoon covering both limits, as well as differential and integral calculus. I cannot believe how easy this author makes these concepts.

Every 1st year calculus student should read this book. After grasping the concepts in Calc Made Easy, then you can concentrate on the algebraic manipulations you need to practice in order to solve the integrals and differencials in you're classroom textbooks. Reading this book you will look like a genius. But, more importantly you'll really appreciate the beauty of calculus and probably be motivated to do well in this class. Good luck and enjoy it.

Also you must buy the paper edition and not the kindle edition since this last one doesn't show off the symbols and graphs...

Answer 2

$ \newcommand{\dd}[2]{\frac{\mathrm{d}#1}{\mathrm{d}#2}} \newcommand{\limh}{\lim_{h \to 0}} $ Disclaimer: This is kind of long because I copied a lot (not all) of it from an old TeX file I wrote. This will be a lot to take in for a beginner, but it provides a much more rigorous foundation in calculus than most textbooks offer. It's best that this be read in supplement to a work explicitly intended for the instruction of basic calculus. Enjoy, OP.

Okay, so the first thing anyone learns about calculus is derivatives. You should, by now, be familiar with the equation $$ m=\frac{\Delta y}{\Delta x} $$ Where $m$ is the slope of a line. We can rewrite this: $$ m=\frac{f(x_1)-f(x_0)}{x_1-x_0} $$ In calculus, we want to generalize the slope. We don't just want to have the slope of a straight line, because that is very limited. So what about the "slope" of a curve? Surely we can approximate it at least. What is the slope of $f(x)=x^2$ at $x=1$? well, we could try $$ m=\frac{f(1+1)-f(1)}{1}=3 $$ and that would get us an approximation, but that's kind of weak. Let's make the difference smaller! $$ m=\frac{f(1+.1)-f(1)}{.1}=2.1 $$ Hmm. That should be closer. Can we make the difference arbitrarily small? Since you've done precalc, you should be familiar with limits. So let's say $$\begin{align} m&=\lim_{h\to0}\frac{(1+h)^2-1^2}{h}\\ &=\lim_{h\to0}\frac{2h+h^2}{h}\\ &=\lim_{h\to0}2+h=2 \end{align}$$

So we would say the "slope" at $x=1$ of $f(x)=x^2$ is $2$. In general, we say $$ \dd{f(x)}{x}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} $$ Where $\dd{f(x)}{x}$ is our "slope," known as a derivative (the shorthand for the derivative of $f(x)$ is $f'(x)$). The next section will prove important properties of derivatives, such as the chain rule and product rule, and generally proves the derivative of $x^n$ with respect to $x$.

Note: When we have $\dd yx$, the d's serve much the same purpose as the $\Delta$ signs do in $\frac{\Delta y}{\Delta x}$. $\text dy$ is actually an infinitesimal number denoting the change in $y$, and $\dd yx$ notes that change in proportion to the change in $x$ over an infinitely small interval.

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First we will prove the chain rule, which states $$\dd{f(g(x))}{x}=f'(g(x))\cdot g'(x)$$ Consider functions $f(x)$ and $g(x)$. $$f'(x)=\limh \frac{f(x+h)-f(x)}{h}$$ $$g'(x)=\limh \frac{g(x+h)-g(x)}{h}$$ Let us define u and v as follows: $$u=\frac{f(x+h)-f(x)}{h}-f'(x)$$ $$v=\frac{g(x+h)-g(x)}{h}-g'(x)$$ Note that as $h \to 0$, $u,v \to 0$.

Now, we rearrange the equations: \begin{align*} u&=\frac{f(x+h)-f(x)}{h}-f'(x) \\ u+f'(x)&=\frac{f(x+h)-f(x)}{h} \\ h(u+f'(x))&=f(x+h)-f(x) \\ h(u+f'(x))+f(x)&=f(x+h) \end{align*} Through the same process, we can obtain that $$h(v+g'(x))+g(x)=g(x+h)$$ Now, consider the derivative of $f(g(x))$: \begin{align*} \dd{f(g(x))}{x}&=\limh \frac{f(g(x+h))-f(g(x))}{h} \\ &=\limh \frac{f\left(g(x)+h(v+g'(x))\right)-f(g(x))}{h} \\ &=\limh \frac{f(g(x))+(h(v+g'(x)))(u+f'(g(x)))-f(g(x))}{h} \\ &=\limh \frac{h(v+g'(x))(u+f'(g(x)))}{h} \\ &=\limh (v+g'(x))(u+f'(g(x))) \\ &=g'(x)\cdot f'(g(x)) \end{align*}

Now, let $f(x)=e^x$ for some constant $e$, and allow $f(x)=f'(x)$: \begin{align*} e^x&=\dd{(e^x)}{x} \\ e^x&=\limh \frac{e^{x+h}-e^x}{h} \\ e^x&=\limh \frac{e^xe^h-e^x}{h} \\ e^x&=\limh \frac{e^x(e^h-1)}{h} \\ 1&=\limh\frac{e^h-1}{h} \\ \limh h&=\limh e^h-1 \\ \limh h+1&=\limh e^h \\ \limh (h+1)^{\frac{1}{h}}&=e \end{align*} Substituting $x=\frac{1}{h}$, we obtain: $$e=\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x$$ We define $f^{-1}(x):=\log{x}$ such that $\log{e^x}=e^{\log{x}}=x$.

We will now consider the derivative of $\log{x}$: $$ \dd{(e^x)}{x}=e^x $$ Substituting $u=e^x$, we get the following: \begin{align*} \dd{u}{(\log{u})}&=u \\ \dd{(\log{u})}{u}&=\frac{1}{u} \end{align*}

Now, we consider the function $x^n$ for some $n \in \mathbb{C}$. We know that $x^n=e^{n \log{x}}$. Consider the derivative of $x^n$ with respect to $x$: $$\dd{(x^n)}{x}=\dd{(e^{n \log{x}})}{x} $$ Here, we let $u=n \log{x}$: \begin{align*} \dd{(x^n)}{x}&=\dd{(e^u)}{u}\cdot \dd{(n \log{x})}{x} \\ &=e^u\cdot n\cdot \frac{1}{x} \\ &=e^{n \log{x}}\cdot \frac{n}{x} \\ &=x^n\cdot \frac{n}{x} \\ &=n x^{n-1} \end{align*}

It's also worth noting that $\dd{f(x)g(x)}x=f'(x)g(x)+f(x)g'(x)$, which is called the product rule. The proof is as follows: $$\begin{align} \dd {f(x)g(x)}x&=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)(g(x+h)-g(x))+g(x)(f(x+h)-f(x))}{h}\\ &=\lim_{h\to0}f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h}\\ &=f(x)g'(x)+g(x)f'(x) \end{align}$$