Calculus: Differentiation Made Easy

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Calculus: Differentiation Made Easy

#1  Postby Darkchilde » Feb 16, 2012 11:28 am

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:

[math]


This translates to the following image:

a8692f6d28486eddebd317d1bc0939a4-2.png
a8692f6d28486eddebd317d1bc0939a4-2.png (18.57 KiB) Viewed 7588 times


The derivative of this is :

[math]


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

[math]


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:

[math]


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

[math]


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:

[math]

[math]

[math]


And so we have the equation of the line:

[math]


If we put this into the first graph we get:

plot2.png
plot2.png (22.09 KiB) Viewed 7588 times


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.
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Re: Calculus: Differentiation Made Easy

#2  Postby Darkchilde » Feb 17, 2012 9:56 am

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:

[math]


Its derivative (with respect to x), is:

[math]


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:

[math]


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).

[math]


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

Rule #2: if you have a function like [math], then its derivative will be [math]

Let's take the parabola of my first post:

[math]


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

[math]


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

So, if we have:

[math]


then n=1, so n-1 = 0, then the derivative of the above is [math].

Let's do a couple more examples:

[math]


So, we have n=19, and n-1 = 19-1 = 18, so its derivative will be [math]. 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]or [math], 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:

[math]


Its derivative will be:

[math]


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

[math]


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

[math]

[math]


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:

[math]


then the derivative will be:

[math]
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Re: Calculus: Differentiation Made Easy

#3  Postby Darkchilde » Feb 17, 2012 10:09 am

EXAMPLES FOR CONSTANTS AND POWERS


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

  1. [math]
  2. [math]
  3. [math]
  4. [math]
    where b,f and c are constants

Let's start solving:

  1. We are given the equation [math]. 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]. What is its derivative? Remembering the rule, a=1, n=3 and n-1=2. Then the derivative of this is: [math]. 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]


    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]

    [math]


Think the other three for today. Post your solutions in spoilers in the discussion thread, and I will give the full solution tomorrow.
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Re: Calculus: Differentiation Made Easy

#4  Postby Darkchilde » Feb 18, 2012 10:56 am

I will not have time to go through all the rest of the examples today; so I will just go through the second equation and solve this one in this installment.

[math]


The above is solved by seeing it as the sum of two terms; each term will be differentiated and their sum will be the first derivative of the above function. The same is true for the second derivative.

So we can say that:

[math] and

[math] then:

[math]


and the derivative will be:

[math]


Let's start with [math]. In this case we have a=5, n=7, and n-1=7-1=6. So, the derivative will be:

[math]

[math]


This one was easy, if you follow the rules about differentiation I gave earlier. How about the second term? It is a negative power. Does that play any role in differentiating? The answer is no. Whether n is positive or negative it does not play any role at all in differentiating powers.

So for [math], we have a=1, n=-5, and n-1=-5-1=-6. Be careful with the signs here, as it is easy to make a mistake, and change the result. So, our result here is:

[math]


Putting the two derivatives together, we get:

[math]


And that's your result, for the first derivative.

The problem asks for the second derivative as well and here we have as terms:

[math]

[math]


So what is the second derivative of the first term? For [math] we have a=35, n=6, n-1=6-1=5. Then this derivative will be:

[math]

[math]


What about the second term? Here we have a=-5, n=-6 and n-1=-6-1=7. Plugging in the numbers we have:

[math]

[math]


And the resulting second derivative will be:

[math]


Tomorrow, I will try and solve the other two examples.
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Re: Calculus: Differentiation Made Easy

#5  Postby Darkchilde » Feb 19, 2012 8:38 am

This post will be about solving the third example:

[math]


Again we use the exact same methodology to differentiate the above example. Let's see it all term by term, and start differentiating:

[math]


So, for the first term we have: a=8, n=-3 and n-1=-3-1=-4; then the derivative for this term is:

[math]

[math]


Proceeding to the second term:

[math]


Here we have: a=-9, n=7 and n-1=7-1=6. Then this derivative will be:

[math]

[math]


The third term is a constant, -4, and so its derivative is equal to 0. So, the first derivative of our function will be equal to:

[math]


We also want to find the second derivative; again we follow the same method, as before.

[math]


We have a=-24, n=-4 and n-1=-4-1=-5; so the second derivative of the above will be:

[math]

[math]


And continuing with the second term of the first derivative:

[math]


Here we have a=-63, n=6 and n-1=6-1=5. Plugging in the numbers we get:

[math]

[math]


Putting the two terms together, we get the second derivative of our function:

[math]


Always remember to check the signs of each function; as I said earlier, a change of sign will give a vastly different result. I will do the last example later this afternoon.
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Re: Calculus: Differentiation Made Easy

#6  Postby Darkchilde » Feb 20, 2012 8:52 am

I know I said, I would go through the last example yesterday, but as usual, rl intervened. So, I am going to go through the example today, and tomorrow I am going to show you how to differentiate the trigonometric functions.

[math]


So, we have b,f and c as constants; which means that those will be treated as numbers, as if they were 2,3,6,-9,15 or whatever number you can think of. So, let's start with the second and easiest term to differentiate: the c term. This is a constant, and as for every constant, its derivative is equal to 0. It does not matter that we have a symbol instead of a number; the problem specifies that b,f and c are constants, and they are going to be treated as constants.

How about the first term? For the first term we have: a=b, n=f and n-1=f-1. So the full derivative of the above function will be:

[math]


You just apply the same rule, whether you have numbers or symbols. And what about the second derivative? For this we have: a=bf, n=f-1, n-1=f-1-1=f-2. Then, the second derivative of the above will be:

[math]


If you have any questions, any comments, please ask them in the discussion thread. Tomorrow we will continue with the rules for differentiating trigonometric functions.
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Re: Calculus: Differentiation Made Easy

#7  Postby Darkchilde » Feb 22, 2012 7:28 am

TRIGONOMETRIC FUNCTIONS


In this installment, I will give you the derivatives of the main trigonometric functions, namely sine, cosine and tangent. Some of you probably do not have a lot of experience with trigonometric functions, but I will not explain about these for the moment. You can read the basics on sine, cosine and tangent here:

Sine at Wolfram
Cosine at Wolfram
Tangent at Wolfram

Just stop at the point where the web pages start involving complex numbers and functions, as this is a bit more complicated and requires knowledge of complex numbers (numbers that involve the square root of negative numbers).

As for the other trigonometric functions, we will derive their derivatives later on. I am keeping those as exercises and examples.

Now, for the following functions:

[math]

[math]

[math]


where a is a constant.

Their derivatives will be:

[math]

[math]

[math]


The trigonometric function secant, or sec(x) is equal to [math].
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Re: Calculus: Differentiation Made Easy

#8  Postby Darkchilde » Feb 22, 2012 8:12 am

E and the Natural Logarithm LN


One of the numbers that comes up a lot in physics, is e and its associated natural logarithm ln. For both of these I will again point you to the Wolfram site to read upon the basics:

e at Wolfram
Natural logarithm ln

e is another irrational number like π; it comes up a lot if you are studying physics or mathematics, and I suggest that you read at least the basics on e. Maybe someone else might take up a discussion on the various mathematical irrational numbers like e and π.

In mathematics e is a lot of times interchanged with exp as a symbol. Let's first see the functions:

[math]

[math]


Then the derivatives will be:

[math]

[math]


If you will notice, in the derivative of the natural logarithm, the constant a does not feature anywhere. If you know anything about logarithms, then you will know that:

[math]


ln(a) is itself a constant, and so its derivative is 0. So we are left with the derivative of the second term, which is ln(x).

Some basic mathematical tricks with the above functions:

[math] for [math]

[math]
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Re: Calculus: Differentiation Made Easy

#9  Postby Darkchilde » Feb 23, 2012 7:37 am

RULES OF DIFFERENTIATION: ADDITION AND MULTIPLICATION


The symbols I will use in this installment are the following:

[math] are functions of x.
[math] is a constant.

Now we have already seen the Sum Rule for differentiation. But let's go over it again:

[math]


Which means that if we have to differentiate a function made of the addition of two functions, then its derivative will be made of the addition of the derivatives of the two functions. And this works for more than two functions.

Next one for today is the Constant Multiple rule:

[math]


Let's do an example of this. Let's just say we have the following function to differentiate:

[math]


We have: [math]. Then its derivative will be:

[math]


Last for today will be how to differentiate two functions that are multiplied with each other, or the Product Rule of Differentiation. The product rule is:

[math]


The derivative of two functions that are being multiplied with each other, is equal to the derivative of the first function multiplied by the second function, plus the first function multiplied with the derivative of the second function. An example will help you understand this more clearly, and don't forget to ask any pertinent questions in the discussion thread.

Our example is the following:

[math]


Here we have: [math]. Following the rules of differentiation then:

[math]

[math]


Tomorrow I will finish up with the Quotient rule and the Composite Rule. Once we have these two in place, I will continue with some examples so that you see differentiation in action.
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Re: Calculus: Differentiation Made Easy

#10  Postby Darkchilde » Feb 24, 2012 9:03 am

QUOTIENT RULE


The quotient rule is about two functions that are divided with each other: so given two functions, f(x) and g(x) that are being divided, how do we differentiate?

[math]


The derivative will be:

[math]


Using words: the derivative of two functions being divided is given by the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator; all this divided by the square of the denominator. Sounds complicated right? However, once you practice it a bit, it will be very easy to do. Let's see a simple example:

[math]


In the above we have:

[math]

[math]


When we differentiate the above two we get:

[math]

[math]


How about the square of the denominator?

[math]

[math]


Then let us apply the rule as above:

[math]


The quotient rule takes some practice. But it is not as difficult as it may seem at first glance. Tomorrow I will finish up with the composite rule; I wanted to do it today, but explaining the quotient rule I saw that it may be better to do the last rule tomorrow, and let you read and understand the quotient rule first.
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Re: Calculus: Differentiation Made Easy

#11  Postby Darkchilde » Feb 27, 2012 8:03 am

COMPOSITE RULE


This is the last differentiation rule you will need to know. What happens when we have nested functions? That means a function like:

[math]


or with an example:

[math]


Well, the rule says that the derivative will be equal to the derivative of the f(x) times the derivative of the g(f(x)). Let's see it in mathematical terms and with our example in action:

[math]


In our example we have:

[math]

[math]


So their derivatives will be:

[math]

[math]


Then combining the two, the derivative will be:

[math]


Despite this one is seemingly difficult, it is very very easy to remember and to differentiate these kinds of functions.

A COUPLE OF LAST THINGS...


If you have a function in more than one variables, let's say f(x,t), and your problem says to differentiate with respect to t (for example), then you treat the x variable as a constant. The same applies for functions of more variables; you differentiate the variable you are told, and you treat the others as constants.

When you have a function made of multiple functions, remember to identify each function separately, and first differentiate those. Then apply one by one the appropriate rules, and start combining the results. Especially at the beginning do not try to do it all at once.

Remember to identify which is the variable and which are constants; sometimes you have a general problem like [math], and you are told that n is a constant. It may be that you may have more than one letters acting as constants.

In this short tutorial, I have mostly used x as the independent variable and y as the dependent variable; in the real world, this may change and you may have different combinations of letters. For example, in physics it is common to t for time, and to differentiate with respect to t. There are a number of other variables used.

I will try to start with a few examples tomorrow, more complex than the ones I have put here.
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