Definition of the Derivative (Difference Quotient) Example 2

Uploaded by TheIntegralCALC on 26.05.2011

Definition of the Derivative (Difference Quotient) Example 2
Hi, everyone.
Welcome back to
Today were going to be using the definition of the derivative to find the...
derivative of the function f of x equals x squared plus five.
So the formula we're given here is often called the difference quotient.
It’s the definition of derivative.
It's also the formula that allows us to find instantaneous rate of change.
You’ll hear all three ways but its always the same formula
and we're always going to be using the same process.
So what we need to know about this formula is that there are three components to it.
Well I guess you could say four.
We've got this component here, the limit as h approaches zero.
And we’ll come back to that in a second.
We've got f of x plus h, we've got f of x, and we've got h here in our denominator.
So let's go through it piece by piece.
First of all, we're going to be taking the limit of this whole function here as h approaches zero.
So we're going to come up with this function and then we're going to plug in zero for h once we've simplified.
So f prime of x will be our derivative function.
Inorder to find it, we'll need to take the limit as h approaches zero of this function here.
The first component of this function is f of x plus h.
And all this means...
this whole piece right here
is that we're goin to be plugging in x plus h everywhere where we have the variable x in our original function.
So in our original function f of x, we've got only one variable right there, x.
so we're going to be plugging in x plus h for that variable.
So instead of x squared plus five, we'll get the quantity x plus h squared plus five.
That takes care of this whole piece right here.
Then as you can see from your formula, we'll subtract.
And then we will plug in f of x
which you can see f of x is just our original function exactly as it is without doing anything to it.
So x squared plus five, we'll put in for f of x and that part's now done.
Then we divide this whole thing by h.
That's the last component of the difference quotient formula.
So we divide it by h and now we've applied the difference quotient.
All we need to do is simplify this function as much as we can and then plug in zero for h to get our final answer for the derivative.
So the derivative f prime of x will be the limit again as h approaches zero.
and now we're just going to be simplifying.
So let's go ahead and square this x plus h here.
When we do, we'll get x squared plus two xh plus h squared.
And then we've got, of course, plus five.
We'll get minus x squared, minus five when we distribute that negative sign.
And then this whole thing is divided by h.
So now, if we simplify our numerator,
you can see we've got a positive x squared and a negative x squared which will cancel.
We've a positive five and a negative five which will also cancel.
And we'll be left with the limit as h approaches zero of two xh plus h squared divided by h.
And as you can see now, we can cancel one h term or one h variable from each of our terms here
because we've got h and all three of terms
so we'll get the limit as h approaches zero of...
this h will go away, this h will go away and this exponent will go away because instead of h squared, we'll just have one h.
So it will be two x plus h with no denominator because this whole thing's going away.
So that's our function, simplified as much as we can.
So now that we've simplified as far as we can go, we just plug in zero for h because again h approaches zero.
So when we do that, we'll get two x plus zero...
which obviously simplifies to two x.
And that's our derivative.
So we can go ahead and say f prime of x equals two x, our derivative.
So that's our final answer.
I hope that video helped you, guys.
And I will see you in the next one.