Definition of the Derivative (Difference Quotient) Example 3

Definition of the Derivative (Difference Quotient) Example 3
Hi, everyone.
Welcome back to integralcalc.com.
Today we're going to be doing another problem with the difference quotient.
We're going to be using the definition of the derivative to take the derivative of this function here, f of x.
So we're going to be taking the derivative f prime of x
and we're going to be using the formula for the difference quotient which is here.
And the formula tells us that we have three components to our derivative problem.
The first here is f of x plus h...
where we will plug in x plus h everywhere where we see the variable in our original function.
so we plug in x plus h for this variable and also for this variable.
Then the second component of our formula is the original function itself, f of x.
Notice that our original function is equal to f of x, right?
so we'll plug in our original function here.
Okay. And then we have our third component which is h.
And we'll just be plugging in h to the denominator of the derivative here.
So to take the derivative, we'll say f prime of x is equal to the limit as h approaches zero
and we'll come back to that a second.
And now we've got our three components.
The first is f of x plus h and we plug in x plus h everywhere where we see the variable in our original function.
so we get x plus h divided by x plus h plus three.
and we subtract from that, our original function f of x.
So x divided by x plus three then we're going to divide that entire thing by h, right?
because that's the third component here of our fraction.
So once we've got the derivative set up,
our goal will be to simplify this fraction as much as possible and then evaluate at h equals zero.
So we'll be plugging in zero for h and coming up with our final answer for the derivative.
So inorder to simplify, we're going to need to focus clearly on the numerator
which is by far the most complicated part here of our problem.
So inorder to do that, we need to find a common denominator between both of these fractions.
So we're actually going to multiply this first fraction here by x plus three over x plus three right?
See this? Then we subtract right?
because that minus sign was in there before so we subtract
We multiply the second fraction by the denominator of the other one, x plus h plus three divided by x plus h plus three.
We can do this because it's just like multiplying by one, right?
This fraction here will reduce by one so we could multiply by that to find the common denominators.
So the derivative is going to be equal to the limit as h approaches zero
And now we can go ahead and multiply everything together.
So let's go ahead and multiply x plus h by x plus three.
So when we do that, we can get x squared plus three x plus x h plus three h.
Okay, then we got our minus sign here so we subtract but we need to drop parenthesis here
because that minus sign is going to apply to everything that's about to come.
So now we're going to multiply x by x plus h plus three.
So we'll get x squared plus xh plus three x, alright?
And then once we've done that, we can combine this fraction into one denominator.
So our denominator is now x plus three times x plus h plus three.
That's our common denominator.
And then remember that this whole thing is still all divided by h.
Okay, so now some additional simplification steps.
Let's go ahead and simplify both our numerator and denominator
So we've got in the numerator here, right?
We'll have x squared minus an x squared because we've got this minus sign here.
So those are going to cancel.
We've got a three x minus a three x so those are going to cancel.
We've got an xh minus an xh so those will cancel.
and you can see we'll just be left with three h divided by x plus three times x plus h plus three.
And this whole thing is divided by h.
Now dividing by h is actually the same thing...
we can do this... we can...
instead of dividing by h we can multiply by one over h.
Okay. So we've got something.
We've got this fraction divided by h.
Well we can flip that h upside down to be one over h.
Put the h in the denominator of this new fraction instead of the numerator of the fraction.
It's in the denominator here...
I know that's a little confusing.
But basically just flip it upside down.
It's the inverse but one over h.
And now we can simplify because you can see we've got this h here in the numerator and in the denominator that will cancel.
So what we're left with is equals the limit as h approaches zero of all that's up in the numerator is three and in the denominator, we've got x plus three times x plus h plus three.
And now that we've simplified our fraction as much as possible,
we can go ahead and evaluate at the value h equals zero.
So we plug in zero for h and when we do that, obviously you can see this is the only place where we've got an h in our fraction.
When we plug in zero for that, that's obviously going to disappear
And you'll see that we're left with x plus three times x plus three in the denominator.
So our final answer actually ends up being the derivative f prime of x.
We don't need this notation anymore because we've plugged in zero.
So once we've done that, this whole part here goes away.
And we just say that the derivative is equal to three divided by x plus three squared, right?
This x plus three times x plus three became x plus three squared.
So that's it.
That's our final answer.
I hope that video helped you, guys.
And I will see you in the next one.
Bye!