Monday, March 18, 2013
Update
Hey guys, just recapping what we have gone over so far. We have covered basic derivatives, and dealt with how the functions inside them are added, subtracted, and multiplied. From here one, we will discuss division (the quotient rule), the power rule, and jump into limits. Hopefully this blog has been helpful to some. If you enjoy or dislike the blog, I'd love to hear about it either way! Please subscribe and tell all your friends where to learn the very basics of calculus!
Saturday, March 16, 2013
Derivatives: The Product Rule
Up to this point, we have learned a lot about derivatives. We know you to find them, multiple times, and how to apply these rules in a large function using addition and subtraction. Now, lets talk about a different situation. If you are given this function:
f(x) = 5x4 * 2x
What do you do? The logical answer would be the same way we differentiated functions with addition and subtraction. However, its not quite that simple. In order to differentiate this function, we have to use something called the product rule.
The product rule looks something like this:
f(x) = g(x) * h(x)
f '(x) = [g(x) * h'(x)] + [h(x) * g'(x)]
So, in words, what happens here is you have the first argument times the derivative of the second, added to the second times the derivative of the first. It may sound confusing to begin with, but you'll get the hang of it quickly. Some people may prefer to flip-flop the addends in this equation. That is completely fine given the standard rules of addition. As long as you are multiplying one times the derivative of the other, you're fine. Now back to the example. Given what we have just learned, lets put that rule to use.
f(x) = 5x4 * 2x
f '(x) = 5x4(2) + 2x(20x3)
f '(x) = 10x4 + 40x4
As you will see, I have stopped showing the simple steps of the derivative. At this point, I am assuming you're a champ at differentiation so I won't keep showing you the steps. If you have a question about one, however, feel free to ask or refer to the prior pages about simple derivatives. I hope this gives you the basic understanding of finding derivative using the product rule. Just remember, first times the derivative of the second plus the second times the derivative of the first. A lot of math and memorization just takes practice, so do a few problems and you'll get the hang of it. Next post, we will discuss the missing operand thus far: division.
f(x) = 5x4 * 2x
What do you do? The logical answer would be the same way we differentiated functions with addition and subtraction. However, its not quite that simple. In order to differentiate this function, we have to use something called the product rule.
The product rule looks something like this:
f(x) = g(x) * h(x)
f '(x) = [g(x) * h'(x)] + [h(x) * g'(x)]
So, in words, what happens here is you have the first argument times the derivative of the second, added to the second times the derivative of the first. It may sound confusing to begin with, but you'll get the hang of it quickly. Some people may prefer to flip-flop the addends in this equation. That is completely fine given the standard rules of addition. As long as you are multiplying one times the derivative of the other, you're fine. Now back to the example. Given what we have just learned, lets put that rule to use.
f(x) = 5x4 * 2x
f '(x) = 5x4(2) + 2x(20x3)
f '(x) = 10x4 + 40x4
As you will see, I have stopped showing the simple steps of the derivative. At this point, I am assuming you're a champ at differentiation so I won't keep showing you the steps. If you have a question about one, however, feel free to ask or refer to the prior pages about simple derivatives. I hope this gives you the basic understanding of finding derivative using the product rule. Just remember, first times the derivative of the second plus the second times the derivative of the first. A lot of math and memorization just takes practice, so do a few problems and you'll get the hang of it. Next post, we will discuss the missing operand thus far: division.
Friday, March 15, 2013
Second Derivatives and Beyond!
Now that we have been introduced to the basic idea of derivatives, we will discuss multiple differentiation. If you are given a function and asked to find the second derivative, you do just that. Lets say this is your function:
f(x) = 3x4 - 5x2 + 17x
Now, the first thing you will do is differentiate this function. So, following the guidelines we have discussed, that would look something like this:
f(x) = 3x4 - 5x2 + 17x
f '(x) = (4*3)x4-1 - (2*5)x2-1 + (1*17)1-1
f '(x) = 12x3 + 10x + 17
Easy enough, right? Now that you have found your first derivative, we can move on the the second. A second derivative is generally noted by two apostrophe characters. In our example, it will be f ''(x), or "F double-prime of x" in words. To find the second derivative, you simply differentiate your first derivative. So lets continue from our original function:
f '(x) = 12x3 + 10x + 17
f ''(x) = (3*12)x3-1 + (1*10)x1-1 + 0
f ''(x) = 36x2 + 10
And there we have it, our second derivative. This concept is fairly easy to follow. If you wanted to find the third derivative, you would follow the same steps. Take your second derivative, and differentiate. Just to prove the point, we'll continue with the function:
f ''(x) = 36x2 + 10
f '''(x) = (2*36)x2-1 + 0
f '''(x) = 72x
We now have our third derivative. You could go on and on, as far as necessary, repeating the same steps to find further derivatives. In our case, anything beyond the third derivative in a function will not be necessary. So, I hope this post left you understanding how to find multiple derivatives of the same function. Post for any further questioning.
f(x) = 3x4 - 5x2 + 17x
Now, the first thing you will do is differentiate this function. So, following the guidelines we have discussed, that would look something like this:
f(x) = 3x4 - 5x2 + 17x
f '(x) = (4*3)x4-1 - (2*5)x2-1 + (1*17)1-1
f '(x) = 12x3 + 10x + 17
Easy enough, right? Now that you have found your first derivative, we can move on the the second. A second derivative is generally noted by two apostrophe characters. In our example, it will be f ''(x), or "F double-prime of x" in words. To find the second derivative, you simply differentiate your first derivative. So lets continue from our original function:
f '(x) = 12x3 + 10x + 17
f ''(x) = (3*12)x3-1 + (1*10)x1-1 + 0
f ''(x) = 36x2 + 10
And there we have it, our second derivative. This concept is fairly easy to follow. If you wanted to find the third derivative, you would follow the same steps. Take your second derivative, and differentiate. Just to prove the point, we'll continue with the function:
f ''(x) = 36x2 + 10
f '''(x) = (2*36)x2-1 + 0
f '''(x) = 72x
We now have our third derivative. You could go on and on, as far as necessary, repeating the same steps to find further derivatives. In our case, anything beyond the third derivative in a function will not be necessary. So, I hope this post left you understanding how to find multiple derivatives of the same function. Post for any further questioning.
Derivatives: Addition & Subtraction
This post will elaborate more on the basics of derivatives. As you may recall, we discussed in the previous post that the derivative of axn, a being any constant, is (n*a)xn-1. When dealing with a larger function, this rule still applies. Consider this example:
f(x) = 3x4 + 2x + 18
We have learned how to find the derivative of these individually, but what would you do if there are lots of these added together? Quite simple, actually. You take the derivative of each addend individually. So for the solution to the above problem:
f '(x) = (4*3)x4-1 + (1*2)x1-1 + 0
f '(x) = 12x3 + 2
As you can see, these will give you no problem as long as you are familiar with the simple derivatives. The same approach is taken for subtraction. Here is an example:
f(x) = 4x3 - 8x - 8
f '(x) = (3*4)x3-1 - (1*8)x1-1 - 0
f '(x) = 12x2 - 8
Again, very simple. Since the rule applied to both is the exact same, you could very easily differentiate a function that included both addition and subtraction. Just to be clear, I will list another example.
f(x) = 5x2 - x + 4
f '(x) = (2*5)x2-1 - (1*1)x1-1 + 0
f '(x) = 10x - 1
This concludes addition and subtraction using differentiation. We will get into multiplication and division in later posts, but it is slightly more complex. In the next tutorial we will cover how to take multiple derivatives of the same function. Hope this was helpful and stay tuned!
f(x) = 3x4 + 2x + 18
We have learned how to find the derivative of these individually, but what would you do if there are lots of these added together? Quite simple, actually. You take the derivative of each addend individually. So for the solution to the above problem:
f '(x) = (4*3)x4-1 + (1*2)x1-1 + 0
f '(x) = 12x3 + 2
As you can see, these will give you no problem as long as you are familiar with the simple derivatives. The same approach is taken for subtraction. Here is an example:
f(x) = 4x3 - 8x - 8
f '(x) = (3*4)x3-1 - (1*8)x1-1 - 0
f '(x) = 12x2 - 8
Again, very simple. Since the rule applied to both is the exact same, you could very easily differentiate a function that included both addition and subtraction. Just to be clear, I will list another example.
f(x) = 5x2 - x + 4
f '(x) = (2*5)x2-1 - (1*1)x1-1 + 0
f '(x) = 10x - 1
This concludes addition and subtraction using differentiation. We will get into multiplication and division in later posts, but it is slightly more complex. In the next tutorial we will cover how to take multiple derivatives of the same function. Hope this was helpful and stay tuned!
Thursday, March 14, 2013
The Basic Derivative
The first topic we will cover is the derivative, and we will start on the most basic level. Generally speaking, most Calculus 1 classes will begin their material discussing the derivative. Lets first discuss how to find the derivative. If you are given a function, lets say axn, the derivative of this function would be
(n*a)xn-1.
In words, you are multiplying the exponent by the coefficient. Then, you subtract 1 from the exponent. For example, if you are given the following function:
f(x) = 3x2
f '(x) = (3*2)x2-1
f '(x) = 6x1
f '(x) = 6x
This process is called differentiation. Differentiation is very useful in applied mathematics. More or less, the derivative of a function will give you the slope at whichever x-value you plug in. It is something that never goes away as long as you are studying calculus!
There are a couple of easy, special cases to be aware of. Given the function:
f(x) = 2x
, the derivative is quite simple.
f '(x) = (1*2)x1-1
f '(x) = 2x0
f '(x) = 2(1)
f '(x) = 2
As you can see here, the exponent subtracts to zero. Using the order of operations, we see that the x0 is executed first Therefore, anything raised to the power of 0 is 1. This gives us our constant times 1. So just an easy rule to remember, the derivative of ax will always be a, a being any constant. Also, the derivative of a constant, 5 for example, will always be 0. This covers the very basics of the derivative. In the next post, we'll discuss differentiation of a basic equation acting as a function. Feel free to leave comments or ask questions.
(n*a)xn-1.
In words, you are multiplying the exponent by the coefficient. Then, you subtract 1 from the exponent. For example, if you are given the following function:
f(x) = 3x2
f '(x) = (3*2)x2-1
f '(x) = 6x1
f '(x) = 6x
This process is called differentiation. Differentiation is very useful in applied mathematics. More or less, the derivative of a function will give you the slope at whichever x-value you plug in. It is something that never goes away as long as you are studying calculus!
There are a couple of easy, special cases to be aware of. Given the function:
f(x) = 2x
, the derivative is quite simple.
f '(x) = (1*2)x1-1
f '(x) = 2x0
f '(x) = 2(1)
f '(x) = 2
As you can see here, the exponent subtracts to zero. Using the order of operations, we see that the x0 is executed first Therefore, anything raised to the power of 0 is 1. This gives us our constant times 1. So just an easy rule to remember, the derivative of ax will always be a, a being any constant. Also, the derivative of a constant, 5 for example, will always be 0. This covers the very basics of the derivative. In the next post, we'll discuss differentiation of a basic equation acting as a function. Feel free to leave comments or ask questions.
Welcome!
Hello, and welcome to my blog. This blog is aimed to help people in need of learning basic principles of calculus. The topics covered will be applicable building blocks to any level of the calculus study. This will range from limits, derivatives, integrals, and beyond! If there are any requests on a tutorial or you are in need of individual help, feel free to leave comments. Look forward to helping you along your mathematical journey!
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