The Definition of a Derivative – Pushing Things To The Limit

Hey guys! In this video we will be taking a look at the definition of a derivative.

So, remember the formula we were given to find the instantaneous rate of change of a function?

V= lim         ?x
?t➝0 _____

This can also be written as lim as x goes to a of f(x) – f(a) / x-a

And to simplify even further, this can be written as:

lim              f(a+h) - f(a)
h→0 ____________

You should, at this point, already be familiar with this limit. This limit will come up all the time. This limit is what is called a derivative.

Here is the official definition:

The derivative of f(x) with respect to x is the function f’(x) and is defined as,
f’(x)= lim f(x+h) - f(x)
h→0 ___________
Where x is equal to a (x=a) from our original instantaneous velocity equation.

Now, let’s find some derivatives using this definition.

Now, typically when they want you to use the definition of a derivative to find a derivative, the question will be worded like this:

Find the derivative of the function using the definition of a derivative.

f(x)= 3×2 – 21x + 48

To help you follow your own work, it can help to start off by just writing out the definition of a derivative.

f’(x)= lim          f(x+h) - f(x)
h→0 ___________

Now, we just need to plug f(x) into our definition wherever we see an f(x), but this (x+h) here (point) lets me know that wherever there is an x in the function f(x), I need to replace it with (x+h). I’ll show you what I mean (start plugging everything in).

f’(x)= lim     (3(x+h)2 - 21(x+h) +48) - (3x2 - 21x + 48)
h→0 ______________________________

So, let’s look back at our definition. We can see here (point) in the first part of our definition that is is telling us to plug in x+h everywhere there is an x in the f(x) function. So, our 3×2 gets changed to 3(x+h)2 and our -21x gets changed to -21(x+h). +48 we can just leave alone, because there is no x variable. Okay, now, it’s telling us to subtract the original f(x) function. So, we just subtract 3×2 – 21x + 48 from our f(x+h) here. And all of this is over h.

Now, we can multiply all of this out, then cancel our like terms.

f’(x)= lim   3x2 + 6xh + 3h2 - 21x - 21h + 48 - 3x2 + 21x - 48
h→0 ______________________________________

So, our 48’s cancel, our 3×2’s cancel, and our 21x’s cancel. That gives us,

f’(x)= lim      3h2 + 6xh -21h
h→0 ___________

We can pull an h out from the top, which would allow us to cancel these h’s.

=lim     h(3h + 6x - 21)
h→0 __________

We are left with,

f’(x)= lim   3h + 6x - 21
Now, we can evaluate the limit by plugging in 0 wherever there is an h.
= 3(0) + 6x -21
So, the derivative is,
f’(x) = 6x - 21

That was simple enough. We just plug all our numbers into the definition, and do some algebra.

Let’s look at another example:

Find the derivative of the function using the definition of a derivative.

B(p) =   p

Let’s write out our definition here.

B’(p)= lim       B(p+h) - B(p)
h→0 ______________

Now, plug in.

B’(p)=lim      p+h            p
h→0 _____ - ___
(p+h)+1 p+1

So, we know that dividing by h is the same thing as multiplying times the reciprocal, but because we are dividing three times, we can just leave it as 1/ h and pull it out here. And when adding or subtracting fractions we need to have the same denominator. So, to get that we need to multiply the top and the bottom here (point) by (p+1) and the top and the bottom here (point) by (p+h+1) to give us the same denominator in each of our fractions.

So, doing that gives us:

B’(p)=lim        1         ( (p+h)(p+1) - (p(p+h+1) )
h→0 __ x _________________
h (p+h+1)(p+1)

Now, we perform all of the operations in our numerator, and try and get some terms to cancel.

B’(p)=lim       1          ( (p2+p +ph +h) - p2-ph-p)
h→0 __ x __________________
h (p+h+1)(p+1)

Combine all of our like terms, which will cause everything but our h to cancel out.

B’(p)=lim        1           ( (p2+p +ph +h - p2-ph-p)
h→0 __ x _____________________
h (p+h+1)(p+1)

Now, we have:

B’(p)=lim      1                  h
h→0 __ x _________
h (p+h+1)(p+1)

Our h’s here cancel out.

Now we have. So, let’s evaluate the limit by plugging in 0 for h.

B’(p)=lim                1                               1
h→0 ___________ = ___________
(p+h+1)(p+1) (p+(0)+1)(p+1)

Therefore, the derivative is,

    B’(p)=      1

It will be super valuable to you to memorize the definition of the derivative as you continue practicing.

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See you next time!



by Mometrix Test Preparation | Last Updated: June 17, 2020