I am starting a thread on Differentiation. Please do not post here, but use the discussion thread for any questions or comments you may have. This will be about differentiating any equation; we will start with understanding what differentiation represents in mathematics and in physics. This will be non-technical, so for the mathematically inclined, I will not use limits for the moment.

Let's get a simple equation of a parabola:



This translates to the following image:



The derivative of this is :



Mathematically, the derivative gives the slope, that is the inclination, of the line that is tangent at point x. I will show you an example of this with the above, so you can see what I am talking about. The ' after f, that is f' means the first derivative. There are a couple of notations used in physics and mathematics on this one, but I will explain them as I use them. So the second derivative will be noted as f''(x), the third one as f'''(x); however after the 3rd derivative we mostly use superscript numbers like this: f4(x)

Now, one point of the parabola is (1,1), that is the coordinates are x=1, and y=1. Coordinates always are written in the form (x,y). I take this point because it is very easy to calculate.

Now, a line has equation



where m=slope, and c=intercept with the vertical (y) axis in the graph. Remember that the horizontal axis, is called the x-axis, and the vertical axis is called the y-axis.

So, we need to find the tangent to the parabola at the point (1,1). Then we can find m, the slope from:



Putting x=1 as is the x-coordinate of our point:



And so the slope m=2. Now to find the intercept c, by substituting the coordinates of the point (1,1) in the equation of a line:





And so we have the equation of the line:



If we put this into the first graph we get:



As you can see the line touches the parabola, a curve in just one point: (1,1).

In physics, a derivative shows the rate of change of a quantity; for example, if you look in the equations thread in Physics, you will see that the derivative of position, is velocity. Velocity is the rate of change of position; and similarly, the derivative of velocity is acceleration, which is the rate of change of velocity.

In physics, and other sciences, most times we need to see how the rate of change of a quantity behaves; velocity may not behave linearly for example, and the derivative function shows how velocity changes, and what we can expect.

Tomorrow I will start with the simple rules of differentiation: constants and powers.

Now, since we understand a couple of basic things about differentiation, let's start seeing how to do it.

Rule #1: Any constant's derivative is equal to 0.

A constant is a definite quantity like 5, or 45 or maybe 567; so for example:



Its derivative (with respect to x), is:



You can change the constant to any value you wish; its derivative is still 0. And when in an equation you are told that a certain symbol is to be treated like a constant, then its derivative will be 0. For example:



And you are told that c is a constant by definition. Then the function will have derivative 0, since c is considered to be a constant (of unknown quantity).



More examples I will do more at the end of this post.

Rule #2: if you have a function like [math]f(x)=x^n, then its derivative will be [math]f'(x) =nx^{n-1}

Let's take the parabola of my first post:



Notice that n=2; so n-1 = 1, and the resulting derivative will be:



One should always remember that [math]x^1=x; there is no need to put the power of 1 to any variable or constant. Also we should remember that [math]x^0=1 no matter what the variable, quantity is.

So, if we have:



then n=1, so n-1 = 0, then the derivative of the above is [math]f'(x)=1.

Let's do a couple more examples:



So, we have n=19, and n-1 = 19-1 = 18, so its derivative will be [math]f'(x)=19x^18. How about the second derivative though? The rule is that when you have a constant before a power that constant remains unaffected in differentiating. When I will explain more complex functions like [math]x\times sin(x)or [math]x(sin(x^2)), you will see why this constant is not differentiated to 0. For the moment, just take it as part of the definition.

The definition then is amended to:



Its derivative will be:



So, if we want to make the second derivative of the above what happens?



We have a=19, n=18, n-1=17. So plugging in the numbers we have:




Another very important thing to remember is the following: if a function is made up of addition of other functions, then its derivative will be the addition of the derivatives of those functions.

In mathematical terms:



then the derivative will be:

For the following functions, please find the first and second derivatives:

1. [math]f(x)=x^3+2
2. [math]f(x)=5x^7+x^{-5}
3. [math]f(x)=8x^{-3}-9x^7-4
4. [math]f(x)=bx^f+c
   where b,f and c are constants

Let's start solving:

1. We are given the equation [math]f(x)=x^3+2. Remember that the derivative of this is the addition of each term, as each term can be seen as a separate function. So, we start with the first term: [math]x^3. What is its derivative? Remembering the rule, a=1, n=3 and n-1=2. Then the derivative of this is: [math]3x^2. The second term is 2, which is a constant and so its derivative is equal to 0. Adding those two we get the following:
   
   
   [math]f'(x)=3x^2
   
   
   There is no need to add the 0, the derivative of the 2 term.
   
   We also need the second derivative; just follow the same rule and we have: a=3, n=2 and n-1=1. So, the second derivative is:
   
   
   [math]f''(x)=3\times 2\times x
   
   
   [math]f''(x)=6x