
In this tutorial, we're going to
look at differentiating x to the

power n from first principles.

Now n could be a positive
integer, n could be a fraction.

It could be negative, or it
could be 0.

So we're going to start by taking
the case where n is a positive integer.

So we'll be looking at things
like x squared, x to the power 7.

Even x to the power 1.

So we have y equals x
to the power n.

Our definition of our derivative
function is dy by dx equals

the limit as delta x approaches
0, of f of x + delta x,

minus our function of x,
divided by delta x.

So let's just look at this part
first of all. Our f of x + delta x

is going to equal x + delta x,
all to the power n.

And this is a binomial.

So what we're going to start by
doing is actually just expanding

a + b to the power n.

And that is a to the power n,

plus n, times a to the power of n  1,
multiplied by b,

plus and there are lots of other terms in
between both containing powers of a and b

along to our final term, which is
b to the power n.

Well, now let's look at what we've got.

Instead of our a, we've got x, and
instead of the b, we've got delta x.

So, we've got x + delta x
to the power n.

So our a, we've got x to the power n,

plus n x to the power n  1
and our b is delta x plus, and

again, all these terms which
will be in terms of x and delta x

Up to our last term, which is
delta x to the power n.

Now let's substitute this in
into our derivative here.

So our dy by dx. is the limit as
delta x approaches 0,

of our function of x + delta x.

Which is x to the power n,
plus, n x to the n  1, delta x, plus,

and again all these terms in both x
and delta x, plus delta x to the power n.

Minus our function of x,
which is x to the power n.

All divided by delta x.

Oh, here we have x to the power n
takeaway x to the power n.

So our derivative is the limit
as delta x approaches 0.

Of n x to the n  1, delta x
plus all the terms in x and delta x.

Plus delta x to the power n.
All divided by delta x.

Now all these terms, have
delta x in them.

And we're dividing by delta x,
so we can actually cancel the delta x.

So that we have the limit as
delta x approaches 0.

Of n x to the power of n 1,
plus, now all these terms here

have delta x squared, delta x cubed
and higher powers all the way up

to delta x to the power n.
So when we've divided by delta x,

all of these have still got a
delta x in them up to our

final term of delta x to the
power of n 1.

So when we actually take the
limit of delta x approaches 0.

All of these terms are going to
approach 0.

They've all got delta x in them
so they'll all be 0.

So our derivative is just n times
x to the power of n  1.

So at derivative of x to the power n,

is n times x to the power of n  1.

Let's have a look at some examples.

Let's say y equals x squared.

So our dy by dx equals

Well, the power comes down
in front, so it's

2 times x to the power of 2
take away 1 which gives us

2 times x to the power 1,
which is just 2x.

y equals x to the power 7.

dy by dx equals...
the power comes down

7 times x to the power of 7  1,
so we get 7x to the power of 6.

Now.

What happens when we
have y equals just x?

x to the power of 1?

Well, we already know the
derivative of this, but let's

see how it fits with the rule.

Bring down the power 1 times x
to the power of 1 take away 1,

so we have 1 times x to the
power of 0.

Well, x to the power of 0
is 1, so it's 1 times 1, so

we end up with the
derivative of 1, which is

exactly what we expected
because we know the

gradient function of y
equals x is 1.

So that was when x was a positive
integer. What happens when x is 0?

Well, let's have a look: y
equals x to the power 0.

Well, we just saw here that x to
the power 0 is actually 1.

And if we find the derivative of
y equals 1. Well, y equals 1 is a

horizontal line, so the
derivative is 0.

So y equals x to the power 0
when n is 0. The derivative is 0.

Let's have a look at some more
complicated examples.

Let's try y equals 6 x cubed, minus
12 x to the power 4, plus 5.

dy by dx is equal to...

If that's 6 multiplied by 3 as we
bring the power down,

x to the power, take one from the power,

minus 12 times,
bring the power down, 4,

x to power of 4 take away 1 and
our derivative of 5 is 0.

(I'll put the plus 0 there.)

So three 6s... 18 x to the power
3  1 is 2

minus four 12s... 48 x to the power
4  1 is 3.

So there's our derivative.

Let's try another one.

y equals x minus 5x to the power 5
+ 6 x to the power 7 + 25.

So our derivative dy by dx
equals... Now this is x to the power 1.

So, 1 times x to the power 1  1,

minus 5 times, bring the power down,
5 times x to the power of 5  1,

plus 6 times, bring the power down,
7 multiplied by x to power 7  1,

plus the derivative of 25, again 0.

So we have 1 times x to the power
of 0, which is just 1,

minus, five 5s are 25, times x
to the power 5  1 is 4,

plus six 7s, 42 times x to the
power 7  1 which is 6.

Now we've proved this result
when n is a positive integer,

but it actually works also when
n is a fraction or when it's a

negative number. Now we're not
going to do the proof of this

because it requires a more
complicated version of the

binomial expansion, but we're
still going to use the result,

so let's have a look at y equals
x to the power a half.

Our dy by dx is going to be a half,
as we bring down the power,

x to the power of a half take away 1
which equals

a half times x, and a half take away 1,
is minus a half.

Let's have a look now, when n is
a negative number, so

y equals x to the power of minus 1. So dy
by dx equals, let's bring the power down,

minus 1 times x to the power of
minus 1 take away 1, so we have...

Minus x to the power 1  1 is 2.

Now, before we finish, let's
look at two more complicated

examples where we need to do a
little bit of rewriting in

index notation before we can
carry out the differentiation.

So let's have a look at
y equals 1 over x plus 6x

minus 4x to the power of 3
over 2 plus 8.

Now we need to rewrite this in
index notation so that we can

easily differentiate it. So
that's x to the power of minus 1

plus 6x minus 4x to the power 3
over 2 plus 8.

So let's differentiate dy by dx equals...

Bring the power down, minus 1 x to
the power of minus 1 take away 1,

plus derivative of 6x to the power 1,
that's 6 times x to the power 1  1,

minus... now 4 times 3 over 2,

multiplied by x to the power of
3 over 2 take away 1,

plus, the derivative of 8,
which is 0.

So here we have minus x,
1  1 is 2.

Plus 6 x to the power 1  1
is 0 which is 1, so it's just plus 6.

Minus... three 4s are 12 divided by 2
gives us 6,

x to the power of 3/2 minus 1...

So that's 1 and a half minus 1, so we end
up with x to the power a half.

And there's our derivative.

OK, one more example, y equals
4x to the power of one third,

minus 5x plus 6 divided by x cubed.

Now again, we've got a mixture
of notations and to

differentiate it we need to
write it all in index notation

rather than having the division.
So this will be 6x to the power of 3,

So, our dy by dx equals...

4 multiplied by the power,
which is a third,

x to the power of 1/3 take away 1,

minus, this is 5 x to the power 1,
so it's 5 times 1,

multiplied by x to the
power of 1  1,

plus 6 multiplied by minus 3 times
x to the power of 3  1.

So let's tidy it all up.

We get 4 thirds x to the power
1/3 take away 1 is minus 2/3,

so it's x to the power of minus 2/3,

minus, now here x the power of 1  1
is x to the power 0, which is 1,

so it's just minus 5,

Six times 3 is 18 and x to the power
3  1 is 4.

So, there's our derivative.