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U1S4V03 Computation of 2nd power derivative.txt
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U1S4V03 Computation of 2nd power derivative.txt
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#
# File: content-mit-18-01-1x-captions/U1S4V03 Computation of 2nd power derivative.txt
#
# Captions for MITx 18.01.1x module [CCoBesP7Fmk]
#
# This file has 99 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
We're looking at this function f of
x equals x squared, and let me draw its graph for you.
We've got our nice parabola over here,
and we're interested in f prime of 3.
And we know that that's given by this formula right here.
So to calculate this, we can just use our formula for f.
So we've got the limit as b approaches 3,
and then on the top, f of b is just b squared,
and then we're subtracting f of 3, which is 9,
and then all this is over b minus 3.
And when we look at the limit here,
well, our top is going to go to 0,
and our denominator is also going to go to 0.
So we need to do more work to evaluate this limit,
but the good thing is that we can factor this.
So we're going to get the limit as b approaches 3 of, and then
on the top when we factor, we get b minus 3 times b plus 3
all over our b minus 3 denominator,
and we get this cancellation.
So we're going to be left with the limit
as b approaches 3 of b plus 3, and here this
is just continuous.
So when we evaluate the limit, we can just plug in 3 for b,
and we're going to get 6.
So that's our answer, f prime of 3 is 6.
Great.
Now let's remember what that means.
So that's the derivative at the point x equals 3.
So we've got this point on the graph at x equals 3,
and our calculation is telling us that the slope of this one
tangent line is 6.
If we put this information on the graph of f prime,
then it's telling us that this point right here,
3 comma 6 is on the graph of f prime.
So our limit calculation got us the slope
of one tangent line and one point
on the graph of the derivative.
But we don't just want one point,
we want the entire graph.
After all, there's not just a tangent line here
at x equals 3.
There are tangent lines all over the place,
and we'd like to know all of their slopes.
Do we have to do a different limit calculation
for every single one?
That would be really annoying.
The good news is that no, we don't have
to do a separate calculation.
We can do them all at the same time.
So let me erase some of this stuff,
and show you how that works.
If we want f prime, not just at 3, but at an arbitrary point x,
then on the right, we can just erase all of these 3s,
and replace them with x's.
And now we have an expression that's
the derivative of f at an arbitrary x,
and we can just compute this limit.
So when we do that, we're going to apply f,
and we're going to get the limit as b approaches x of b squared
minus x squared over b minus x.
And now as b approaches x, once again the top is going
to go to 0, and the bottom is going to go to 0,
but once again, we can factor.
So we've got the limit as b approaches x of,
then the top is b minus x times b plus x all over b minus x,
and we've got our cancellation.
So we're left with limit as b approaches x of b plus x.
And here, this is once again a polynomial,
so we can just plug in, because it's continuous,
so we're going to plug in x for b,
and we're gonna get x plus x, which is 2x.
So that's our answer.
That's f prime at an arbitrary value of x.
So that's everything.
We know the entire derivative function f prime.
We can graph it.
The graph of 2x is going to look like this,
and we can compute the value of f's derivative
at any point we want.
So maybe we want f prime of minus 2.
We don't need a limit calculation anymore.
Our formula tells us that f prime of any x is twice the x,
so f prime of minus 2 is 2 times minus 2, which is minus 4.
And that's it.
So minus 2 comma minus 4 is a point on the graph of f prime,
and up here on the graph of f at the point
where x equals minus 2, we get a tangent line
whose slope is minus 4.
And that certainly seems reasonable from the picture.
So it all works out.
It's great.
We can do this at any point we want,
and it's incredibly quick.
So we have some questions for you,
and then we're just going to keep
trying to find formulas for the derivatives of as
many functions as we can.
So let's keep going.