, 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.