Tip:
Highlight text to annotate it
X
The stalist, most boring example that one can imagine ,
and that's why it's so wonderful.
Because it illustrates the idea with a
minimum of fuss and calculation
and that example which has probably been given
billions of times
is
I want to look at
f(x)
equals x squared
for x equals
one .
So, a quick picture. Here it is.
and here's the
point one
and that's again P
at (1,1) .
And what I'm interested in
is this tangent line, and I'm going to calculate the slope using the idea of
derivative.
And I think it's always a good idea
to get some investment before you get into a problem
and so I would like to think about
when I get the answer
what are some things that should be true of it ?
Like could the answer be that the slope of this thing is -7?
No, it couldn't.
So, I would expect it to be what?
I would expect... this is my expectations.
that f prime at one
would be positive.
And by the way,
I wouldn't expect it, for instance, to be a hundredth, either, would I?
because it's clearly not gonna
get that close to zero when it cuts across .
Alright, so I've got some investment in this problem.
and uh... let's see what we actually get.
What I need to do... I'll actually write it on the board again. I need to look at
f(x+h) - f(x) divided by h
f(x+h) - f(x) divided by h
and that's going to be (x+h) squared - x squared
all divided by h
Correct?
And that gives me
x squared + 2xh + h squared minus x squared
all over h.
And that gives me
the x squareds go away. That gives me 2xh + h squared
all over h.
And please notice, this is the slope of these secant lines.
So, those were those green secant lines that you may remember from before
and for those uh...
a whole bunch of them
And those secant lines are all for h's not zero. So, this h is not zero.
So, that's why I can divide by it, and in fact, when I do
the h goes away
and I come up with
2x + h
2x + h
That's the slope of the secant line
And now...
I let h approach zero
and when that happens
what happens to the
to this expression here ?
it gets closer and closer to 2x
from which I can conclude
that the derivative f'(x)
is equal to 2x
And in particular, I said I wanted it at 1.
So, f'(1) = 2 .
So, f'(1) = 2 .
So, f'(1) = 2 .
And that's it.
Now, suppose I actually want to get the equation of that tangent line.
The equation of that line
well, it's y-1 = 2(x-1)
because it goes through the point (1,1).
And so I have y...
let me do a little mental algebra here .
y equals
2x
minus 2
plus 1
So, that's the equation of the tangent line,
and let me just see if that agrees with my picture .
If I let...
if i let y be zero to get the x intercept,
for y = 0
I get 0 = 2x - 1 or x is equal to