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So I've been requested to do
the proof of the derivative of
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the square root of x, so I
thought I would do a quick
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video on the proof of the
derivative of the
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square root of x.
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So we know from the definition
of a derivative that the
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derivative of the function
square root of x, that is equal
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to-- let me switch colors, just
for a variety-- that's equal to
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the limit as delta
x approaches 0.
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And you know, some people
say h approaches 0,
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or d approaches 0.
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I just use delta x.
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So the change in x over 0.
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And then we say f of x
plus delta x, so in this
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case this is f of x.
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So it's the square root of x
plus delta x minus f of x,
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which in this case it's
square root of x.
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All of that over the change
in x, over delta x.
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Right now when I look at that,
there's not much simplification
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I can do to make this come out
with something meaningful.
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I'm going to multiply the
numerator and the denominator
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by the conjugate of the
numerator is what
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I mean by that.
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Let me rewrite it.
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Limit is delta x approaching
0-- I'm just rewriting
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what I have here.
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So I said the square root
of x plus delta x minus
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square root of x.
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All of that over delta x.
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And I'm going to multiply
that-- after switching colors--
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times square root of x plus
delta x plus the square root of
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x, over the square root of x
plus delta x plus the
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square root of x.
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This is just 1, so I could of
course multiply that times-- if
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we assume that x and delta x
aren't both 0, this is a
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defined number and
this will be 1.
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And we can do that.
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This is 1/1, we're just
multiplying it times this
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equation, and we get limit
as delta x approaches 0.
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This is a minus b
times a plus b.
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Let me do little aside here.
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Let me say a plus b times
a minus b is equal to a
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squared minus b squared.
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So this is a plus b
times a minus b.
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So it's going to be
equal to a squared.
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So what's this quantity squared
or this quantity squared,
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either one, these are my a's.
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Well it's just going
to be x plus delta x.
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So we get x plus delta x.
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And then what's b squared?
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So minus square root of
x is b in this analogy.
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So square root of x
squared is just x.
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And all of that over delta
x times square root of x
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plus delta x plus the
square root of x.
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Let's see what
simplification we can do.
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Well we have an x and
then a minus x, so those
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cancel out. x minus x.
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And then we're left in the
numerator and the denominator,
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all we have is a delta x here
and a delta x here, so let's
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divide the numerator and the
denominator by delta x.
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So this goes to 1,
this goes to 1.
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And so this equals the limit--
I'll write smaller, because I'm
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running out of space-- limit as
delta x approaches 0 of 1 over.
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And of course we can only do
this assuming that delta--
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well, we're dividing by delta
x to begin with, so we know
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it's not 0, it's just
approaching zero.
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So we get square root
of x plus delta x plus
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the square root of x.
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And now we can just
directly take the limit
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as it approaches 0.
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We can just set delta
x as equal to 0.
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That's what it's approaching.
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So then that equals one
over the square root of x.
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Right, delta x is 0, so
we can ignore that.
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We could take the limit
all the way to 0.
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And then this is of course just
a square root of x here plus
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the square root of x,
and that equals 1 over
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2 square root of x.
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And that equals 1/2x
to the negative 1/2.
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So we just proved that x to the
1/2 power, the derivative of it
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is 1/2x to the negative 1/2,
and so it is consistent with
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the general property that the
derivative of-- oh I don't
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know-- the derivative of x to
the n is equal to nx to the n
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minus 1, even in this case
where the n was 1/2.
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Well hopefully
that's satisfying.
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I didn't prove it for all
fractions but this is a start.
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This is a common one you
see, square root of x, and
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it's hopefully not too
complicated for proof.
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I will see you in
future videos.