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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

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