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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

correct rules.