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www.mathcentre.ac.uk/.../8.5%20Product%20Rule.mp4

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    Sometimes we given functions
    that are actually products of
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    functions. That means two
    functions multiplied together.
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    So an example would be Y equals
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    X squared. Times by the cosine
    of 3 X so there we've got a
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    product X squared is one
    function multiplied by the
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    cosine of 3 XA second function.
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    Usually we write this
    as UV.
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    And there is a formula which we
    can use to be able to
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    differentiate this DY by the X
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    is UDVIDX. Plus
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    V times by du by
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    DX. So every time we've
    got a product, we can use
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    this formula.
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    We've identified you with X
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    squared. We've identified
    V with calls 3X, So what
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    we need to do now is
    write down those
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    derivatives. So do you
    buy the X?
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    You, we've identified
    as X squared. So do
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    you buy the X is 2 X.
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    DV by DX, we identified V as
    costs 3X, so we need to write
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    down the derivative of DV by DX,
    and we know that that will be
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    minus three sign 3X.
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    Now we can put these two.
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    Together with these two
    into this formula. So
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    let's do that. Why by
    DX is equal to.
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    You
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    X squared.
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    Times DV by the X so
    it's times by minus three
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    sign 3X.
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    Plus the.
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    That's cause 3X.
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    Times by du
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    by DX2X. Now
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    this. Looks ugly and really we
    need to tidy it up a little bit.
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    We want to write these terms in
    a nice order, but at the same
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    time we want to try and identify
    any common factors that there
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    are. The reason is that what you
    might want to do is solve an
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    equation like this by putting it
    equal to 0 to find solutions for
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    Maxima and minima. So it's
    important to be able to spot the
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    common factors and take them
    out. And if we look in this
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    term, we've gotten X squared,
    and in this term we've gotten X,
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    so we actually got a common
    factor of X, and we can take
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    that out and put it at the front
    of the bracket so X and then a
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    bracket. Let's think what we've
    got left here. We've got minus
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    three and an X, so we have minus
    three X sign 3X.
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    Plus now we took out the X, so
    we've got the two left to
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    multiply by the cost 3X, and so
    we have two cars 3X and we just
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    close the bracket there.
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    We've got that finished and it's
    in a nice factorize form so that
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    if we wanted to do something
    more with it, as I said, find a
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    maximum or minimum we could go
    on and do that.
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    Let's have a look
    at another example.
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    This time, let's have a
    look at one.
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    Where we've got all just
    functions of X know trig
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    functions in just nice functions
    of X except will make this one
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    have a fractional index will
    make it be to the power 1/2. In
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    other words, it's a square root.
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    So why is a U times of
    E? So let's try and get into
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    the habit of identifying these
    functions very specifically.
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    So we've identified the
    two bits that would be
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    multiplied together to
    make the product U&V.
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    Now let's do that derivatives.
    Do you buy the axis
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    three X squared?
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    Member multiplied by the index
    and take one off it so that
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    gives us a three X squared.
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    TV by The X.
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    Now this one is going to be a
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    little bit different. We've got
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    a half there. We've gotta minus
    X in there.
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    Making use of our idea of
    function of a function, we
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    differentiate this inside, so
    that's the derivative of minus.
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    X is minus one.
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    And then we must differentiate
    this so we get a half.
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    Times 4 minus X and we
    take one away from 1/2.
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    That gives us minus 1/2.
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    Let's quote our formula DY by
    the X is equal to.
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    And we know that it's
    UDV by the X plus
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    VU by The X.
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    And we can now plug in
    the bits that we need,
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    so we you is X cubed.
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    Times by this minus 1/2
    of 4 minus X to
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    the minus 1/2.
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    Plus VDU by DX.
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    That's 4 minus
    X to the half.
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    Times by three X
    squared.
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    Now again, this doesn't look a
    nice lump of algebra really. We
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    might have to do something with
    it later on, and if we did have
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    to do something with it later
    on, we need it to look a lot
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    better. We need it to look a lot
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    nicer. So I'm going to go
    into a new sheet and I'm
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    going to rewrite this at the
    top of the page, and then
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    we're going to have a look at
    how we might simplify it.
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    So the why by DX?
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    Equal 2X cubed times
    minus half 4 minus
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    X to the minus
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    1/2. Plus now we have
    V which was 4 minus X to the
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    half times EU by DX, which was
    three X squared and we want to
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    put all this together.
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    It's this term which doesn't
    look awfully good. Let's
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    remember that. To the power
    minus 1/2 means that it's
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    in the denominator, so we
    have minus X cubed over
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    2 * 4 minus X
    to the half plus.
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    Haven't done anything with
    this side yet, so we can
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    leave it as it is what we
    want to do is put everything
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    over this denominator. 2 * 4
    minus X to the half. We treat
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    this as though it's over a
    denominator of one, so I'll
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    common denominator.
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    Will be 2, four minus X
    to the half.
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    This stays just the same no
    problem. We haven't done
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    anything with the denominator,
    so we don't do anything with the
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    top. Plus this we have
    multiplied a one here.
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    By this so we must multiply
    everything here. So two times
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    the three is going to give us
    six X squared, and then we have
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    4 minus X to the power half
    multiplied by 4 minus X to the
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    power half. So that just gives
    us 4 minus X, 'cause if we have
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    the half and half together,
    that's one, and to the power one
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    is just traditionally written is
    just 4 minus X.
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    Equals the denominator can stay
    the same. We've now got it all
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    over this common denominator.
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    And we can look at what we've
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    got on the top. Again, we're
    looking for the idea of a
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    common factor. Can we take out
    a common factor? And yes, we
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    can. There's an X squared
    there, and there's an X
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    squared inside that X cubed X
    cubed is X squared times by X,
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    so we can pull out that X
    squared.
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    Then what are we left with? If I
    take out the X squared out of
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    this bracket? I've got 6 times
    by 4 which is 24 and I've got 6
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    times by minus X which is minus
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    6. Thanks, but I mustn't forget
    if I take an X squared out of
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    here. I've also got another
    minus X left. So altogether
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    that's minus 7X.
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    And again, that's in a nice tidy
    form, so that if we need too
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    next time, we can actually do
    something else with it. We could
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    put this equal to 0 and take the
    top and solve it.
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    We just look at one more
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    example. Let's take Y
    equals 1 minus X cubed
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    all times by E to
    the 2X.
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    First of all, let's identify our
    U and our V, so will put that
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    you is going to be equal to 1
    minus X cubed.
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    The V is going to be equal to
    E to the 2X.
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    And now we differentiate you.
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    So we have DU by the X
    is equal to the
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    derivative of one is 0
    because one is a constant
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    and the derivative of
    minus X cubed is minus
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    three X squared.
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    We want the derivative of V.
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    V is E to the 2X and remember
    that in order to differentiate
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    the exponential function, we
    differentiate the power.
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    And we multiply the derivative
    of that power by E to the 2X
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    in this case, so that DV by the
    X is equal to the derivative of
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    the power. The derivative of 2X,
    which is just two times by E to
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    the 2X. And let's remember
    our formula that if Y is
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    equal to U times by V
    then D why by DX is
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    equal to you?
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    TV by the
    X plus VD
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    you by X.
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    So we've got U. We've got DV
    by DX. We've got V and we've
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    got du by DX here, so let's
    make those substitutions DY
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    by ZX is equal to.
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    U to begin with, that's one
    minus X cubed.
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    Times by and we want the
    derivative of VDV by DX, so
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    that's here 2 times E to the
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    power 2X. Plus
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    and now we want V. That's E to
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    the 2X. And now we want the
    ubaidi X, and that's minus three
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    X squared. That's times by minus
    three X squared. And as we've
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    done before, let's look for any
    common factors that there might
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    be. And here we've got a common
    factor of E to the 2X in each
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    term, so let's take that out E
    to the 2X, which will leave us
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    with. Well, here we've got 2
    times by one minus X cubed, so
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    let's do that multiplication. So
    we've got 2.
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    2 times by 1 -
    2 times by X cubed.
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    And then from this one we've got
    minus three X squared. So we put
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    that minus three X squared and
    its usual. When you've got an
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    expression like this, a
    polynomial in terms of X to
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    write it so that we've got the
    powers of X in some kind of
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    order. And in this case I'll
    write them in ascending order
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    from the smallest powers of X up
    to the largest, so that would be
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    2 - 3 X squared.
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    Minus two X cubed and being
    factorized. That derivative is
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    now in a position where we can
    use it for other things, perhaps
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    to solve the why by DX equals 0.
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www.mathcentre.ac.uk/.../8.5%20Product%20Rule.mp4
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