Finding Derivatives with Limits

At this point we should have at least a basic understanding of limits and how to find some limits.  However, I have only really discussed limits by themselves and not how they relate to the rest of calculus.  They are very important in calculus because they are used to define the most important calculus topics.

For example, the main topic which will be discussed for quite some time is derivatives.  Derivatives will come up in a lot of different settings, like finding rate of change, instantaneous rate of change, velocity, slope, and a few others.  The main thing to realize is that a derivative is generally used to find out how quickly, or slowly, something is changing.

I will go further into all of these things later, but for now I want to focus on the definition of derivatives and how to find a derivative using the definition.

The definition of a derivative

If we have some function, \(f(x)\), we would write “the derivative of f” as \(f'(x)\).  And we would define the derivative of f by using this limit:

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

This limit can be a bit confusing, so there’s something I would like to point out before we actually begin working with this limit.  The confusing thing here is that we have \(x\) and \(h\) in this limit and it looks as if they are both variables.  However, when we find this limit, we can only treat \(h\) as a variable.  We will need to treat x as a constant and h as the only variable.

The reason for this is that we are finding the limit as \(h\) goes to \(0\).  This tells us that \(h\) is moving in toward \(0\).  It does not tell us that \(x\) is changing at all.  Therefore, when we are working with the limit, we will act as if \(x\) is a number, or a constant.  This means that once we find the limit, our answer may have \(x\) in it still and this is completely fine since \(x\) isn’t the variable in this case.  Now let’s try an example.

Example 1

Consider the function \(f(x) = 4x^2 – 7x + 12\).  We will use the limit definition to find the derivative of this function, but first let’s break it down and consider each part on its own.

Finding f(x+h)

The fist thing we need to find is \(f(x+h)\).  This notation basically just means that we need to look at our function \(f\), and plug in \((x+h)\) wherever we see the input.  In other words, we need to replace all of the \(x\)’s in the function with \((x+h)\)’s.  So,

$$f(x+h) = 4(x+h)^2 – 7(x+h) + 12.$$

Then we will want to expand this out so it’s easier to work with.  Remember \((x+h)^2\) is the same as \((x+h)(x+h)\), which means we need to foil it.

$$f(x+h) = 4(x+h)(x+h) – 7(x+h) + 12$$

$$=4(x^2 + xh + xh + h^2) – 7(x+h) + 12$$

$$=4(x^2 + 2xh+ h^2) – 7(x+h) + 12$$

$$=4x^2 + 8xh+ 4h^2 – 7x – 7h + 12$$

Since there aren’t any like terms we will leave it at that for now.

Putting it all together

Now we can put that into the rest of the equation.  Since we now know \(f(x+h)\) and \(f(x)\), we can plug those into the equation.  I would recommend surrounding each of them with a set of parenthesis so you don’t forget to distribute the negative sign in front of the \(f(x)\).  This is a very common mistake, so be careful not to forget that because it will give you the wrong answer.

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$

$$= \lim_{h \to 0} \frac{(4x^2 + 8xh+ 4h^2 – 7x – 7h + 12) – (4x^2 – 7x + 12)}{h}$$

Solving the limit

When I first see a limit, the first thing I usually consider is whether we can simply plug in \(0\) for \(h\). Essentially, I try to treat this function as if it were continuous at \(h=0\) (remember \(h\) is the variable here).

However, if we do this here we will get \(0\) on the denominator.  Since you cannot divide by 0, this will not work.  So our strategy will be to simplify this fraction to a point where we can plug in \(0\) for \(h\).  The simplest way to do this is to rearrange the numerator so that we can cancel an \(h\) from the numerator and denominator and get rid of our fraction all together.

$$f'(x)= \lim_{h \to 0} \frac{4x^2 + 8xh+ 4h^2 – 7x – 7h + 12 – 4x^2 + 7x – 12}{h}$$

$$= \lim_{h \to 0} \frac{8xh+ 4h^2 – 7h}{h}$$

At this point I would like to point something out. Notice, after simplifying the numerator of the fraction, each term remaining contains an \(h\) in it.  This is important because it allows us to factor the \(h\) out and cancel it with the \(h\) in the denominator, getting rid of the fraction.  This will be an extremely common strategy to use for finding the derivative of a function using the limit definition.

$$f'(x)= \lim_{h \to 0} \frac{h(8x+ 4h – 7)}{h}$$

$$= \lim_{h \to 0} 8x+ 4h – 7$$

Now we have simplified to a point that we can solve this limit by plugging \(0\) in for \(h\).

$$f'(x)= 8x+ 4(0) – 7$$

$$= 8x – 7$$

So we have just shown that if \(f(x)=4x^2-7x+12\), then \(f'(x)=8x-7\).  We will later learn shortcuts like the product rule, quotient rule, and chain rule that will make finding derivatives like this much simpler and faster, but you will need to know how to find these using the limit definition.  The pattern shown in this problem is a common one.  It won’t work for all derivatives, but it’s a good thing to try first.

  • It’s generally a good idea to see if you can reengage the top of the fraction in such a way that every term has \(h\) as a factor.
  • Then you can factor out the \(h\), and cancel it with the \(h\) on the bottom of the fraction.
  • This usually leaves you with a function that you can directly plug \(0\) into \(h\) and simplify from there, leaving you with a function that doesn’t contain any \(h\)’s, but usually contains \(x\)’s.

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I would recommend checking out the other material I have about derivatives.  As I mentioned before, there are several shortcuts and methods that make this whole process a lot easier.  Go check out what I’ve written about on the derivatives page.  If you have a question that isn’t answered there just let me know by emailing me at jakesmathlessons@gmail.com.  I’ll do my best to answer any questions you send me and I may even post a lesson or full solution on it!

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