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, Remember back, remember back to the
power rule. Well, what do the power rules

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say? It's that the derivative of x to the
n, and some real number not zero, the

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derivative of x to the n is n times x to
the n minus one. The power rule isn't just

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something we just made up. It's a
consequence of the definition of

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derivative. But how do we know it's
actually true? Well remember, we've

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already worked this out when n is a
positive whole number. Then, the

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derivative of x to the n is the limit of
this difference quotient. This is the

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limit that calculates the derivative. If n
is a positive whole number, I can expand

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out x plus h to the n, and I get x to the
n plus n x to the n minus 1 times h plus

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things with lots of h's minus x to the n
all over h. Now, the x to the n and the

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minus x to the n cancel, the h here
cancels this h here, and I'm left with a

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bunch of h's divided by h, there's still a
lot of h's in this. And the limit of this

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constant, as far as h is concerned plus a
thing where h is in it, well, this goes to

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zero. And I'm left with n times x to the n
minus 1, which is the derivative of x to

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the n. And this is a completely valid
argument and as long as n is a positive,

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whole number. But there's plenty of
numbers which aren't positive, whole

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numbers. What if n equaled negative 1?
Let's figure out the derivative of x to

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the minus first power. Actually, the
derivative of 1 over x. This is a problem

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that we can attack directly using the
definition of derivative. Here, I've

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written the limit of the function of x
plus h minus the function over h. Now, to

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calculate this limit, I'll first put this
part in the numerator over a common

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denominator. So, this is the limit as h
goes to zero, whole thing's over h. But

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the numerator is now x minus x plus h,
over common denominator for the things in

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the numerator, x plus h times x. Now,
what's x minus x plus h? Well, in that

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case, this x and this x cancel. And what
I'm left with is just negative h up there.

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So, this is the limit as h goes to zero of
negative h over x plus h times x all over

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h. Great.
Now, the h down here and the h up here

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cancel. What am I left with? I'm left with
the limit as h goes to zero of negative 1

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over x plus h times x. Now, how can I deal
with this? Well, as h goes to zero, the

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numerator's just 1 but the denominator is
approaching x squared. So, this limit is

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minus 1 over x squared. What we've
calculated here is the derivative of 1

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over x, the derivative of 1 over x is
negative 1 over x squared. Now, I can use

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this fact. The fact that the derivative 1
over x is negative 1 over x squared to

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compute using the change rule the
derivative of 1 over x to the n. This is a

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composition of two functions, the
composition of the 1 over function and the

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x to the n function. The derivative of 1
over is negative 1 over the thing squared.

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So, it's the derivative of the outside
function at the inside times the

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derivative of the inside function. The
derivative of x to the n, if n says

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positive, whole number, I already know
this, it's n times x to the n minus 1. And

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x to the n squared is x to the 2 n. So,
I've got negative 1 over x to the 2n times

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n times x to the n minus 1. Now, a minor
miracle happens. The x to the 2n and the x

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to the n minus 1, they're interacting, so
that I'm left with the x to the n plus 1

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in the denominator, minus 1 times integer
minus n in the numerator. Now, this movie

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doesn't look so great. But remember that
at 1 over x to the n is just another name

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for x to the negative nth power. And this,
if I rewrote this as x to a power, I could

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rewrite this as negative n times x to the
negative n minus 1 power. And look.

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What we've shown is the derivative of x to
the negative n is negative n x to the

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negative n minus 1. This is verifying the
power rule holds even when n is a negative

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number. Pretty good.
We've done it now for all whole numbers.

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But what about rational numbers? So,
here's a question. How are the derivative

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of x to the 21/17 power is 21/17 times x
to that power minus 1, 4/17.

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Implicit di fferentiation to the rescue.
Well, here's maybe a simpler case. y is

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the derivative of x to the 1/17. 1/17
times x to the negative 16/17. Well, let's

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set y equal x to the 1/17, and that means
y to the 17th power is x. And I can apply

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explicit differentiation to y to the 17
equals x. So the differentiation precisely

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gets 17y to the 16 dy dx equals a
derivative of x, which is 1. Divide both

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sides by 17 times y to the 16th power and
I get that dy dx is 1/17 times 1 over y to

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the 16th power. But, y is x to the 1/17.
So, dy over dx is 1/17 times 1 over y, is

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now x to the 1/17 seventeenth x to the
16/17. But, it's 1 over that, so I could

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write this as 1/17 times x to the negative
16/17. So now the chain rule finishes off

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the problem. if I want to differentiate x
to the 21/17 power, well that's the same

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as differentiating x to the 1/17 power to
the 21st power. It's chain rule. So that's

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the same as 21 times the inside, x to the
1/17 to the 20th. That's the derivative of

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the outside function versus the 21st power
function at the inside, times the

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derivative of the inside function. Now,
good news. We calculated the derivative of

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the outside function. This is 21 times x
to the 1/17 to the 20th power times the

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derivative of x to the 1/17, which is
1/17x to the negative 16/17. well, this is

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21 times x to the 20/17 times 1/17 x to
the negative 16/17. And 20 minus 16 is 4.

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It's 21 times x to the 4/17 over 17, it's
21/17 x to the 4/17. That's exactly what

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the power rule tells you when n is 21/17.
We started off just knowing the power rule

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was true for positive whole number
exponents. And now, after doing a little

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bit of work, we know that the power rule
holds for any rational exponent. What

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about the function f of x equals x to the
square root of 2 power? Whoa. What does

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that even mean? It's a serious objection.
What do I mean by a number raised to the

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square root of 2 power? Well, what can I
do? I can take x and I can raise it to the

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1.4 power, by which I mean, I take x
multiplied by itself 14 times and then

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take the tenth root of that. I can take x
to the 1.41 power, by which I mean I take

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x multiplied by itself 141 times, and then
take the hundredth root of that. And I can

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keep on going, right?
If I want to take x to the 1.414 power,

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I'd multiply x by itself 1,414 times, and
then take the thousandth root of that. If

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I were to take x to the 1.4142 power,
right?

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I take x and multiply by itself 14,142
times and then take the 10,000th root of

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that number. And I can keep doing this,
and I'm getting closer and closer to the

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square root of 2. And that's really what
this function means. It really means to

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take a limit of these functions I actually
understand, functions where I'm taking x

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to a rational exponent. We can handle this
with a logarithm. So, let's set y equals x

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to the square root of 2 power. I want to
calculate dy dx. So, the trick here is

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log. So, I'm going to take a log of both
sides. log of y is log of x to the square

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root of 2 power. But log of something to a
power is that power times log of the base.

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So, I've got log y is the square root of 2
times log x. Now, I differentiate both

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sides and I find out that the derivative
of log y is 1/y dy dx, and the derivative

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of the other side is the square root of 2
times 1/x. Multiply both sides by y, and

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I've got dy dx is the square root of 2
y/x. But I know what y is. y is the x to

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the square root of 2 power. So, this is
the square root of 2 times x to the square

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root of 2 power divided by x. In other
words, it's the square root of 2 times x

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to the square root of 2 minus 1. We're
using logarithms to fill in the gaps in

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the quotient rule. We're not just learning
a bunch of derivative rules, we're

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actually learning why these rules work.
Take a look. The square root of 2 here

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plays no essential role in this argument.
I could go back through this entire thing

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and replace the square root of 2
everywhere I see it by the number n. And

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what I'd see is that this logarithm
argument is justifying the power rule. The

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derivativ e of x to the n is n times x to
the 'n, n minus 1. And so, we're really

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building the foundations of calculus. We
are not just learning how to apply the

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rules to some calculations, we're learning
to justify that these rules are the

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correct rules.