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## www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4

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In this tutorial, we're going to
look at differentiating x to the
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power n from first principles.
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Now n could be a positive
integer, n could be a fraction.
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It could be negative, or it
could be 0.
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So we're going to start by taking
the case where n is a positive integer.
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So we'll be looking at things
like x squared, x to the power 7.
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Even x to the power 1.
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So we have y equals x
to the power n.
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Our definition of our derivative
function is dy by dx equals
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the limit as delta x approaches
0, of f of x + delta x,
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minus our function of x,
divided by delta x.
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So let's just look at this part
first of all. Our f of x + delta x
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is going to equal x + delta x,
all to the power n.
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And this is a binomial.
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So what we're going to start by
doing is actually just expanding
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a + b to the power n.
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And that is a to the power n,
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plus n, times a to the power of n - 1,
multiplied by b,
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plus and there are lots of other terms in
between both containing powers of a and b
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along to our final term, which is
b to the power n.
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Well, now let's look at what we've got.
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Instead of our a, we've got x, and
instead of the b, we've got delta x.
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So, we've got x + delta x
to the power n.
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So our a, we've got x to the power n,
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plus n x to the power n - 1
and our b is delta x plus, and
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again, all these terms which
will be in terms of x and delta x
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Up to our last term, which is
delta x to the power n.
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Now let's substitute this in
into our derivative here.
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So our dy by dx. is the limit as
delta x approaches 0,
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of our function of x + delta x.
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Which is x to the power n,
plus, n x to the n - 1, delta x, plus,
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and again all these terms in both x
and delta x, plus delta x to the power n.
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Minus our function of x,
which is x to the power n.
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All divided by delta x.
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Oh, here we have x to the power n
takeaway x to the power n.
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So our derivative is the limit
as delta x approaches 0.
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Of n x to the n - 1, delta x
plus all the terms in x and delta x.
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Plus delta x to the power n.
All divided by delta x.
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Now all these terms, have
delta x in them.
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And we're dividing by delta x,
so we can actually cancel the delta x.
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So that we have the limit as
delta x approaches 0.
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Of n x to the power of n -1,
plus, now all these terms here
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have delta x squared, delta x cubed
and higher powers all the way up
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to delta x to the power n.
So when we've divided by delta x,
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all of these have still got a
delta x in them up to our
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final term of delta x to the
power of n -1.
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So when we actually take the
limit of delta x approaches 0.
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All of these terms are going to
approach 0.
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They've all got delta x in them
so they'll all be 0.
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So our derivative is just n times
x to the power of n - 1.
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So at derivative of x to the power n,
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is n times x to the power of n - 1.
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Let's have a look at some examples.
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Let's say y equals x squared.
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So our dy by dx equals
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Well, the power comes down
in front, so it's
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2 times x to the power of 2
take away 1 which gives us
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2 times x to the power 1,
which is just 2x.
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y equals x to the power 7.
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dy by dx equals...
the power comes down
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7 times x to the power of 7 - 1,
so we get 7x to the power of 6.
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Now.
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What happens when we
have y equals just x?
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x to the power of 1?
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Well, we already know the
derivative of this, but let's
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see how it fits with the rule.
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Bring down the power 1 times x
to the power of 1 take away 1,
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so we have 1 times x to the
power of 0.
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Well, x to the power of 0
is 1, so it's 1 times 1, so
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we end up with the
derivative of 1, which is
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exactly what we expected
because we know the
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gradient function of y
equals x is 1.
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So that was when x was a positive
integer. What happens when x is 0?
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Well, let's have a look: y
equals x to the power 0.
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Well, we just saw here that x to
the power 0 is actually 1.
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And if we find the derivative of
y equals 1. Well, y equals 1 is a
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horizontal line, so the
derivative is 0.
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So y equals x to the power 0
when n is 0. The derivative is 0.
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Let's have a look at some more
complicated examples.
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Let's try y equals 6 x cubed, minus
12 x to the power 4, plus 5.
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dy by dx is equal to...
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If that's 6 multiplied by 3 as we
bring the power down,
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x to the power, take one from the power,
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minus 12 times,
bring the power down, 4,
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x to power of 4 take away 1 and
our derivative of 5 is 0.
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(I'll put the plus 0 there.)
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So three 6s... 18 x to the power
3 - 1 is 2
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minus four 12s... 48 x to the power
4 - 1 is 3.
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So there's our derivative.
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Let's try another one.
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y equals x minus 5x to the power 5
+ 6 x to the power 7 + 25.
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So our derivative dy by dx
equals... Now this is x to the power 1.
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So, 1 times x to the power 1 - 1,
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minus 5 times, bring the power down,
5 times x to the power of 5 - 1,
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plus 6 times, bring the power down,
7 multiplied by x to power 7 - 1,
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plus the derivative of 25, again 0.
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So we have 1 times x to the power
of 0, which is just 1,
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minus, five 5s are 25, times x
to the power 5 - 1 is 4,
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plus six 7s, 42 times x to the
power 7 - 1 which is 6.
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Now we've proved this result
when n is a positive integer,
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but it actually works also when
n is a fraction or when it's a
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negative number. Now we're not
going to do the proof of this
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because it requires a more
complicated version of the
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binomial expansion, but we're
still going to use the result,
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so let's have a look at y equals
x to the power a half.
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Our dy by dx is going to be a half,
as we bring down the power,
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x to the power of a half take away 1
which equals
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a half times x, and a half take away 1,
is minus a half.
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Let's have a look now, when n is
a negative number, so
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y equals x to the power of minus 1. So dy
by dx equals, let's bring the power down,
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minus 1 times x to the power of
minus 1 take away 1, so we have...
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Minus x to the power -1 - 1 is -2.
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Now, before we finish, let's
look at two more complicated
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examples where we need to do a
little bit of rewriting in
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index notation before we can
carry out the differentiation.
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So let's have a look at
y equals 1 over x plus 6x
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minus 4x to the power of 3
over 2 plus 8.
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Now we need to rewrite this in
index notation so that we can
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easily differentiate it. So
that's x to the power of minus 1
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plus 6x minus 4x to the power 3
over 2 plus 8.
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So let's differentiate dy by dx equals...
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Bring the power down, minus 1 x to
the power of minus 1 take away 1,
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plus derivative of 6x to the power 1,
that's 6 times x to the power 1 - 1,
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minus... now 4 times 3 over 2,
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multiplied by x to the power of
3 over 2 take away 1,
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plus, the derivative of 8,
which is 0.
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So here we have minus x,
-1 - 1 is -2.
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Plus 6 x to the power 1 - 1
is 0 which is 1, so it's just plus 6.
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Minus... three 4s are 12 divided by 2
gives us 6,
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x to the power of 3/2 minus 1...
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So that's 1 and a half minus 1, so we end
up with x to the power a half.
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And there's our derivative.
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OK, one more example, y equals
4x to the power of one third,
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minus 5x plus 6 divided by x cubed.
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Now again, we've got a mixture
of notations and to
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differentiate it we need to
write it all in index notation
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rather than having the division.
So this will be 6x to the power of -3,
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So, our dy by dx equals...
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4 multiplied by the power,
which is a third,
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x to the power of 1/3 take away 1,
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minus, this is 5 x to the power 1,
so it's 5 times 1,
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multiplied by x to the
power of 1 - 1,
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plus 6 multiplied by minus 3 times
x to the power of -3 - 1.
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So let's tidy it all up.
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We get 4 thirds x to the power
1/3 take away 1 is minus 2/3,
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so it's x to the power of minus 2/3,
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minus, now here x the power of 1 - 1
is x to the power 0, which is 1,
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so it's just minus 5,
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Six times -3 is -18 and x to the power
-3 - 1 is -4.
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So, there's our derivative.
Title:
www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4
Video Language:
English
Duration:
14:56 EHCliffe edited English subtitles for www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4 EHCliffe edited English subtitles for www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4 EHCliffe edited English subtitles for www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4 mathcentre edited English subtitles for www.mathcentre.ac.uk/.../8.1a%20Diff%201st%20principles.mp4

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