Welcome to a Proof of the Power Rule
of Differentiation.
Which is the derivative of x raised to the power of n
with respect to x equals n times x raised to the power
of n minus 1.
This proof uses the binomial theorem.
So let's begin by reviewing this.
The binomial theorem is used to expand binomials
when we have a binomial raised to a whole number power.
So if you have the quantity x plus y raised to the nth,
the binomial theorem gives us this expansion.
Where notice how the first term would be n choose zero
times x to the nth, times y to the zero.
And then we have plus n choose 1, times x to the power
n minus 1, times y to the first.
Plus, the next term is n choose two,
times x to the power of n minus two,
times y squared and so on.
Notice as we move to the right,
we are losing factors of x and gaining factors of y
and the coefficient is determined by a combination
where n choose r is equal to n factorial divided by
r factorial times n minus r factorial.
So if we begin with the function f of x equals
x to the nth, by the definition of the derivative
if prime of x is going to be equal to
the limit as h approaches zero of
f of the quantity x plus h,
which in our case would be the quantity
x plus h raised to the nth.
And then we have minus f of x,
which in our case is x to the nth, all divided by h.
And now we use the binomial theorem to expand
the quantity x plus h raised to the nth.
So our next step,
this looks a little intimidating, but all of this...
Is the expansion of the quantity x plus h raised to the nth.
Notice on the far right we still have minus x to the nth.
The next step we want to evaluate these combinations
to determine the first several coefficients.
So here we see that n choose zero is equal to one.
N choose one is equal to n.
N choose two is equal to n times the quantity
n minus one divided by two and so on.
And just for review, let's evaluate these combinations
to verify the coefficients.
So n choose zero,
is equal to n factorial,
divided by r factorial
that'd be zero factorial.
And then times and minus r factorial, which in our case
should be n minus zero factorial or just n factorial.
Zero factorial is defined as being equal to one
and therefore this simplifies to one.
Which is why this first coefficient is one.
Then we have n choose one
which is equal to n factorial divided by one factorial
times n minus one factorial.
If we expand n factorial we can right n factorial
as n times n minus one times n minus two and so on
or just n times n minus one factorial.
Divided by one factorial times n minus one factorial.
Simplifying.
Notice how this simplifies to n,
which is our second coefficient.
And let's look at one more.
N choose two
would be equal to n factorial divided by two factorial
times n minus two factorial.
And now if we begin expanding n factorial
we'd have n times n minus one.
But the remaining terms would be n minus two factorial.
We divide this by two factorial and then we have n minus
two factorial.
So these two simplify to one.
So we have n times n minus one, divided by two factorial.
Which is just two.
So again we have n times n minus one, divided by two
which we see here in our expansion.
Now we want to begin simplifying,
notice here we have x to the nth minus x to the nth.
So these two terms simplify out.
Which gives us the limit as h approaches zero
of this fraction.
Now looking at the numerator notice how all of the terms
contain at least one factor of h.
So for the next step, we'll factor one factor of h
out of the numerator.
So this first line is the same line from the previous slide
and the second line here is where we factored out
one factor of h from the numerator.
And notice once we do that, we can simplify this fraction
h divided by h, simplifies to one.
Which gives us the limit as h approaches zero
of the sum of these terms.
Now if we look at these terms closely,
notice all the terms contain at least one factor of h
except this first term.
We have one factor of h here in the second term.
And remember each term after this is going to have one more
factor of h until this last term
has n minus one factors of h.
So again, all the terms except the first term,
has a factor of h and therefore, as h approaches zero
all the terms except the first term are
going to approach zero and therefore the limit is just
going to be this first term
n times
x raised the the power of n minus one.
And we have our proof of the Power Rule of Differentiation.
The derivative of x raised to the power of n
with respect to x equals
n times x raised to the power of n minus one.
And of all the differentiation rules,
this one is probably used the most in differential calculus.
And now you know where it came from.
I hope you found this helpful.