Rough Calculus  Differentiation 
Written by Bryce Ringwood  
First of all  my apologies for having to do this to you. Unfortunately radio circuits have voltages and currents that change all the time. This being the case we need to know how they change, and why. We need to have some language which allows us to describe what we are doing and let us communicate to others exactly what we are doing. I am going to attempt to make this as easy as possible, but it does require some thought on your part. You need to understand the sections on formulae and equations and how to use your calculator. In particular you need to understand graphs. If you don't like the way I portrayed those sections  feel free to browse other websites, in fact I encourage you to do that, because sometimes a single explanation isn't enough to make the proverbial "penny drop". Calculus has a bad rap, there's no doubt about it. Right from the word go, it was criticised as proving things by dividing by zero, and when it was first invented by Liebniz and Newton, this was probably fair. In the intervening years, people have come up with the idea of 'limits"  and it has become an acceptable way to solve problems and describe things. By the way, you will find some new operators don't worry if they all look greek to you  that's because they mostly are greek. They are just symbols and they are explained. Before you beginYou will need to understand algebra, because there are formulae and equations that will be manipulated. You should have gone through the section on "e"  since it is important in calculus, and you should know the basics of complex numbers. I will be using simple graphs and altough I have to introduce complicated looking symbols  don't let them upset you. Your plan of attack might be to understand each section at a time. Understand the words first, then go through the formulae in more detail. Use a calculator to understand the arguments by replacing the algebraic quantities with numbers. Let's begin, like Newton with an apple falling from a tree The Story of the AppleImagine we have a brand new cellphone with a camera that takes pictures every tenth of a second. The split second an apple begins to fall from a tree, it takes pictures, so that every tenth of a second, we can measure how far the apple has fallen. Here is what we record in ten frames:
The first column is the total time taken.The second column is the total distance the apple falls in that time .The third column in the table is the difference in distance travelled divided by the time. In other words the velocity of the apple at the end of each time frame. A change in velocity with time is called an acceleration. The accelleration caused by gravity is a constant 981 cm per second per second, and this is why the last column is a constant. If the distance is $s$ then you can verify the formula $s=0.5at^2$ where $a=981$ using your calculator We can plot this on a graph:
To arrive at the velocity of the apple, we divide a small amount of distance $\Delta s$ by a small amount of time $\Delta t$. Read $\Delta$ as "a small amount of". (Sometimes written $\delta$). Note that if we have a small quantity, such as $.0001$, then $.0001^2 = .0000001$. This is the underlying reason we can ignore what are called second order small quantities.
What happens when the "$\Delta$"s become really small? as in really really small? DerivativesThe answer is  they become "$d$"s,or differentials. Let's take our formula and calculate: {{\Delta s}\over{\Delta t}}= {0.5a}{{t^2 ({{ t+{\Delta t}})^2}}\over{\Delta t}}
={at + 0.5a{\Delta t}}
When $\Delta t$ becomes infinitely small, then we can ignore the second term, giving us a formula for velocity: {ds\over dt}=at
This gives us a simple rule for this type of equation: "To differentiate a simple polynomial, divide by the degree of the expression and decrease the power by one". The $\displaystyle {d\over dt}$ is an operator  meaning "derivative of something with respect to t". Do it again and we get the fact that $\displaystyle{{d{_2}s\over dt^2} = a}$ that is, the acceleration of the apple, which is constant. The little 2s simply signify we have differentiated for a second time. This is not a formal definition of the process of differentiation  its merely a demonstration using a simple polynomial. $f(t)=0.5at^2$. You might like to try it for other polynomials and see that it works. Trigonometric functionsIn electronics, the symbol $\omega$ is used to represent $2\pi f$. It is a constant for a given frequency $f$, like 981 or "a". Now let's see what happens when we apply the above thinking to a trig function: $f(t)=sin(\omega t)$:
{\Delta f(t)\over {\omega \Delta t} }={1\over {\omega \Delta t}}{(sin(\omega t)  sin(\omega (t  \Delta t))}
= {1\over{ \omega \Delta t}}{(sin(\omega t)  sin(\omega t )cos(\Delta{\omega t}) + cos(\omega t)sin(\Delta{\omega t})}=\color \orange {cos(\omega t)}
When the $\Delta$ s approach the infinitely small, then: \color \orange{ {d(sin(\omega t) \over dt} = {\omega cos(\omega t)}}
Note that this works because for small angles $sin(\theta)=\theta$  think about it, and $cos(\theta)=0$. Check it on your calculator. Maybe you can work out the derivative for $cos(\theta)$ with respect to theta. ExponentialsNow let's tackle $e^x$. You will recall that $e=\left (1  {1\over n}\right )^n$ as n approches infinity. Fairly obviously, $e^x=\left (1  {1\over n}\right )^{nx}$ as n approches infinity. We can work out the value of the kth term of the series from the binomial theorem as: $\displaystyle{{{nx} \choose k}{\bigg( {1\over {k!}}\bigg) }{{{ \bigg( {1\over n}} \bigg) }^k}} $.Recall that this can be expressed as: $\displaystyle{ {\left( {1\over{k!}}\right) } {\left( {{nx(nx1)(nx2)...(nxk+1)}\over {n^k}}\right) } }$When n becomes very large, this value becomes simply $\displaystyle{ {{x^k}\over{k!}} }$(Prove it with your calculator). This means we can work out e from the expression: $\displaystyle \color \orange { {e^x}= {1\over {0!}} + {x\over {1!}} + {{x^2}\over{2!}} + ... {{x^k}\over{k!}} .... }$(try this also on your calculator) NOW here's the good bit! If you differentiate each term in the series with respect to $x$, You end up with the same series for e all over again! Try it and see. (Hint: $\displaystyle{d\over dx}{{{x^k}\over {k!}}} = {kx^{k1}\over k!} = {{x^{k1}\over{(k1)!}}}$) So \color \orange {{{de^x}\over{ dx}}={e^x}}
This leads to a wonderful equation called Euler's equation. (See next article). The Product RuleBefore we proceed any further, I need to mention the product rule for differentiation. Suppose we have two functions of x, let's call them $u$ and $v$, then: $\displaystyle \color \orange {{d(uv)\over dx} = {u{dv\over dx} + v{du\over dx}} }$, since $\displaystyle{ {\Delta(uv)} = (u+\Delta u)(v + \Delta v) uv = {u\Delta v + v\Delta u + \Delta u\Delta v}}$ and, as per usual, we ignore the second order small quantity $\displaystyle {\Delta u\Delta v}$ You might also want to look up the chain rule and the division rule. Summary
A little problemCan you show that if $y=log{_e}(x)$, then $\displaystyle \color \orange {{dy\over dx}={1\over x}}$ ? One way of proing this might be to use the fact that $e^y = x$ ......
