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Derivatives – Understanding the Basics

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Introduction to Calculus  

Warning: This explanation will not make sense to everybody. You don't need to understand it. However, it may be best to first learn to calculate derivatives, lower down, and then read this.

When I think of calculus, I think infinite precision. That is really the point of calculus. There are many things in the world for which you can obtain an approximate measurement, but calculus will give a precise measurement. Before entering calculus, you have the knowledge to calculate the area of a theoretical square. However, if you want an area that is bounded by a line which is not straight, you have difficulty. You can approximate by dividing it up into a million little squares, and adding up the areas. The more squares, the more precise. Calculus, I mentioned, deals with the infinite. With calculus, you can measure it using an infinite amount of infinitely small squares. 

 
 

You will not be working on this particular aspect yet. The first thing you learn in calculus is how to get an infinitely precise slope. If you are looking at an equation that is curved, at each point there is a different slope. No precalculus equation can solve for it. Derivatives in calculus can. 

Derivatives are rather the easiest thing in the world, once you learn how to do them properly. There a very few things you have to know. But before I try to cover the calculation of derivatives, please allow me to explain more clearly what they are. This is simple - they are the slope of an equation, or the rate of change. It should be evident to you why I equate the two! The slope of an equation is essentially how many Y's change for each change in X. The slope that you have thought of up until now was generally not as precise as we like to be at this level. You previously used (change in y/change in x), which may be perfectly accurate for straight line graphs, but not for curved slopes. We want to get the slope of a curve at a single point (called the instantaneous slope), ideally, at any single point that we choose; the slope when x is 3, or when x is 0, for example. Over any curve that is not a line, the slope of the curve fluctuates, and is different for different values of x. In these cases, you would like to get an equation that will tell you the slope at any point. A derivative does this - it gives you an entirely new equation so that every y of the new equation (written as y', or y-prime) is equal to the slope of the original, and at the corresponding x in the graphs. If you want to know the slope of an equation when x=3, you take the derivative, and plug x=3 into the derivative. You will get the slope of the original equation at the point x=3. If this doesn't make any sense, try glancing it over again. In any case, the following discussion should hopefully clarify things. 

 

Deriving Derivatives from Limits

I will now explain how limits from the last section are used to get the derivative equation. The honest truth is that you are going to forget this section entirely. There are various shortcut methods that can differentiate equations (take derivatives) far more quickly. However, the shortcuts do not explain WHY. You will want to know the method I am about to explain for this reason, and because your teacher is going to test you on it. If neither of these reasons apply, move down the page to "Simple Derivatives". I have merely a very brief subsection here on how derivatives are derived, though. (Love that pun!)

The slope of a line is the change in y over the change in x. The logic is that if for a given increase in x, y increases sharply, the slope will be steep. If, as x moves from 2 to 3, to 4, y is unchanged, the slope is flat. The change in y is 0 for any amount of x. The way to find the slope is to take any two x values, find their corresponding y values, and see how many units of y changed over the amounts of x changed: (y1 - y2)/(x1 - x2). 

So how do you find the slope at a particular point when the slope is changing over x? Let's say this point is when x = 4. If you take the slope (y1 - y2)/(x1 - x2) using two points around 4, you should get a slope reasonably close to the instantaneous slope. The theory is that the closer these two x-values are to 4, the closer your slope will get. The way it is usually done is by making one of the points already at 4, and making the second one get closer and closer. 

This process of getting closer and closer is a limiting process. The limit of the second x value will be 4. Notice that as the x values approach 4, the line created from that point to 4 has a slope closer and closer to the precise slope. (Which is given by a line brushing the curve at x = 4, the tangent line.)

What we now have to do is take a limit of the equation for slope as the gap between x values approaches zero. This is because as the x values approach 4, 
(x1 - x2) becomes closer and closer to (4 - 4) which is zero. We can now see the fundamental difficulty in taking an instantaneous slope: The equation for slope (y1 - y2)/(x1 - x2) will have a 0 in the denominator! This is why we must use limits. 

A rather specific formula has evolved for this limit. Let us forget that I said x equals 4; we need a formula for whatever x equals. So x will remain x. The gap between the two x values is termed h: (x1 - x) = h. This gap should have a limit of 0, so whatever the equation will be, it will have lim h-->0. For each x, there is an f(x). The numerator of the slope equation is (y1 - y), or f(x1) - f(x). For a given x1, the gap will have a certain size. If you look at the graph, you will see that x1 is equal to x - h, or x minus the gap when the second point is x1. So f(x1) is now f(x - h) and f(x1) - f(x) is now f(x - h) - f(x). And this is the numerator of the slope equation.

We now have a complete new version of the slope equation that incorporates the limit:

y' = lim h-->0  [f(x - h) - f(x)]/h

Since h is approaching 0, x - h = x + h. This is a shorthand explanation for why most books use:

y' = lim h-->0  [f(x + h) - f(x)]/h

Here are two examples of how to put this awkward equation to use:

Find the derivative of y = x2:

y' = lim h-->0 [(x + h)2 - x2]/h 
   = lim h-->0 [x2 + 2xh + h2 - x2]/h
   = lim h-->0 (2xh + h2)/h
   = lim h-->0  h(2x + h)/h
   = lim h-->0 2x + h
   = 2x + 0
   = 2x

y' = 2x is an equation that will tell you the slope of y = x2 at any point. When x = 0, the slope is 0. When x = 1, the slope is 2. When x = -3.5, the slope is -7.

Find the derivative of y = 4/(x - 2):

Lim h-->0  y = [4/(x + h - 2) - 4/(x - 2)]/h   
  = [(4x - 8 - 4x - 4h + 8)/(x + h - 2)(x - 2)]/h
  = -4/(x + h - 2)(x - 2)   (I canceled the h's)
  = -4/(x - 2)2    (I got rid of the last h because it was going to zero!)

This is all we will do with limits. I'm not even going to test you on it. The following sections deal with shortcut methods to take derivatives, but you should be aware that any derivative CAN be taken with the method we just used. My example was a bit easy though. 

Simple Derivatives

For any number that includes an xn, to any degree, the form to take the derivative is nxn-1. The derivative of y, on the other hand, is y’ (y-prime). The reason for this is that y is a function of x, and its existence is based on its being a function if x, so it’s derivative is tied to the function it is equal to. If there is an equal sign between two sides of a function, then you must do the same thing to both sides in order for the operation to be legal. So, in the function y=x, the derivative of y is y’, and the derivative of x is 1. This second part I get through use of nxn-1. 1x1-1 = 1x0 = 1. Now I know that in this scenario y’=1. I will soon give several examples of this fascinating phenomenon. Be patient. 

The derivative of any lone number (including e), without a variable attached, is 0. The derivative of 10e is 0, the derivative of 3 is 0, and the derivative of 10e is 0. There is no variable in any of these. These are constants. A constant is defined as any number whose value cannot change during the course of the problem. The value of 5 is 5, the value of e is always 2.71828...

If there is a number in front of any form of x, then the number does not become 0. (This includes a minus sign, which can be considered as a –1) It is treated as a coefficient. When taking the derivative of 3x2, you temporarily remove the 3 for the purposes of the differentiating, solve the derivative of x2, and put it back. The derivative of x2, using nxn-1, is 2x2-1 = 2x. Now we put the 3 back in front, making it 3*2x, which is 6x. After this, you may want to take the derivative of this derivative. The derivative of 6x is 6. This is simple, because it always turns out that any number with an x to the first degree (having an exponent of 1) simply loses the x. The derivative of 6 is 0, as per the rule I explained regarding lone numbers or constants.

The derivative of y’ is y" (y-double-prime). This is the second derivative of function y.

Here are some examples:

y=x32     y’=32x31   y’’=32*31x30

y = 4x3 – 5x2 + 17x – 1      y’ = 12x2 – 10x + 17      y" = 24x – 10

y = √x = x1/2      y’ = (1/2)x(1/2)-1 = x-1/2/2 = 1/(2x1/2)

This last one uses several rules I mentioned in the Introduction. Firstly, a square root is the same as being to the power 1/2. The only way to solve derivatives for square roots is to use that, and convert the equation into a useable form. So I made it x1/2. Then you had to realize that there was a power of –1/2 on the x, after the derivative is taken. This means you can swap it to the bottom of the equation, and lose the negative sign. 

y = ex2      y' = 2ex

y = ex       y' = e

y = 7x6      y' = 7 * 6x5 = 42x5 

y = x/8 = (1/8)x      y' = 1/8  I hope you understood this. x/8 is the same as saying (1/8) times x and then the (1/8) is a coefficient, which remains after the derivative.

Also note that in all equations, the way you take the derivative of a polynomial is by taking the derivative of each segment. They do not interfere with each other, unless they are all part of a bigger fraction, in which case you will probably end up needing the quotient rule. That is in a next section; skipping there before you know this well would not be wise! 

 

When x is the exponent

y = 2x      y’ = 2xln 2   This is an annoying rule. When the x is in the exponent, in order to take the derivative you must multiply the equation by the natural log of the base. In this case, that base is 2. So the derivative of 10x is 10xln10. Simple enough.

y = ex      y’ = exln e = ex Two things two realize here. Firstly, e is NOT a variable, it's just a number, so you treat it like any other. Secondly, as mentioned in the Intro, ln and e cancel out always. Therefore they equal 1, so exln e is equal to ex. This is what is so special about e – the way the derivative (slope) comes out. The derivative of ex is ex.

One more useful thing is that the derivative of ln x is 1/x.

Logarithmic differentiation, something quite a bit more complicated than this, is covered in the section on implicit differentiation. Do not go there unless you understand everything in all the sections before it clearly. 

More basic derivatives that you are required to know are included in my general reference tables.

 

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Copyright 2004 Bruce
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