U2S5V05 Power Rule for rational exponents.txt

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Recall the Power Rule, which says
that the derivative of x to the power r
is equal to r times x to the r minus 1.
What we want eventually is to be able to prove this in the case
that r is any real number.
And this is how we've been using the Power Rule.
But so far, we have only proven that this
is true for r any integer.
What we will prove now is the Power Rule in the case
that r is a rational number.
That is, when r is equal to m over n
where both m and n are any non-zero integer.
And the reason I'm going to go ahead and do this proof
is that I think it's a pretty cute use
of implicit differentiation going
in sort of a different direction than we often see it.
So let's get started.
We're going to start and say that y
is equal to x to the m over n.
Now, if we're giving a proof, that
means I can only use things that I've already proven,
which means I can only use the Power Rule on a variable raised
to an integer power.
So I'm going to rearrange this formula so that it only
involves integer powers.
And the easiest way to do this is
to raise both sides to the n-th power.
So this is going to give me that y to the n
is equal to x to the m over n to the n, which
is, of course, x to the m.
So now I have y as a function of x.
And I have a formula that I can implicitly
differentiate using the Power Rule,
because both n and m are integers.
So differentiating both sides with respect to x,
I get that n times y to the n minus 1 times dy/dx
is equal to m times x to the m minus 1.
Now I'm going to solve for dy/dx,
which is, of course, what we're looking for.
And I get that dy/dx is equal to m over
n times x to the m minus 1 over y to the n minus 1.
So now I'm going to go ahead and go back to the original formula
that I had where I said that y was equal to x to the m over n,
and I'm going to substitute that in for y
here so that I can get a formula involving only x.
So that gives me x to the m over n
raised to the n minus 1 power.
Of course, I can rewrite this as m over n times x to the m
minus 1 minus m over n times the quantity n minus 1.
Doing a little bit of algebra and rearranging this,
I get that m minus 1 minus m minus minus m
over n, that is x to the m over n minus 1.
And this is exactly the formula we were looking for.
So we're done.
We used implicit differentiation to prove the Power Rule
in the case where r is a rational number.
Of course, we still need to prove this in the case
that a is any real number, and we'll actually
be able to do that at the very end of this unit.
Now it's time for you to get some practice
using implicit differentiation yourself.