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    Differentiation using product rule

    Hey all,

    I have done this question but don't seem to get the same answer as given, i don't know where I am going wrong? can someone check for me please. Thank you

    Find dy/dx using the product rule when: y=x^2lnxcos3x

    u=x^2
    du/dx=2x

    v=lnxcos3x
    a=lnx, b=cos3x
    dv/dx=a.db/dx+b.da/dx

    da/dx=1/x
    db/dx=-3sin3x
    dv/dx=lnx.-3sin3x+cos3x.1/x
    dv/dx=-3lnxsin3x+cos3x/x

    dy/dx=u.dv/dx+v.du/dx
    dy/dx=x^2.-3lnxsin3x+cos3x/x+lnxcos3x.2x
    dy/dx=-3x^2lnxsin3x+co3x/x+2xlnxcos3x
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    Quote Originally Posted by dadon View Post
    Hey all,

    I have done this question but don't seem to get the same answer as given, i don't know where I am going wrong? can someone check for me please. Thank you

    Find dy/dx using the product rule when: y=x^2lnxcos3x

    u=x^2
    du/dx=2x

    v=lnxcos3x
    a=lnx, b=cos3x
    dv/dx=a.db/dx+b.da/dx

    da/dx=1/x
    db/dx=-3sin3x
    dv/dx=lnx.-3sin3x+cos3x.1/x
    dv/dx=-3lnxsin3x+cos3x/x

    dy/dx=u.dv/dx+v.du/dx
    dy/dx=x^2.-3lnxsin3x+cos3x/x+lnxcos3x.2x
    dy/dx=-3x^2lnxsin3x+co3x/x+2xlnxcos3x
    The error is in the second to the last line: you didn't distribute the x^2.

    How about we write it this way: dy/dx = y' It works out to be a bit neater.

    Then we have
    y = [x^2]*[ln(x)cos(3x)]

    y' = [x^2]' * [ln(x)cos(3x)] + [x^2] * [ln(x)cos(3x)]'

    Now
    [x^2]' = 2x
    and
    [ln(x)cos(3x)]' = [ln(x)]' * [cos(3x)] + [ln(x)] * [cos(3x)]'
    where
    [ln(x)]' = 1/x and [cos(3x)] = -sin(3x) * 3 <-- By the chain rule.

    So
    [ln(x)cos(3x)]' = (1/x)*cos(3x) - 3*ln(x)*sin(3x)

    Thus

    y' = [x^2]' * [ln(x)cos(3x)] + [x^2] * [ln(x)cos(3x)]'

    y' = 2x*ln(x)*cos(3x) + x^2*[(1/x)*cos(3x) - 3*ln(x)*sin(3x)]

    y' = 2x*ln(x)*cos(3x) + x*cos(3x) - 3x^2*ln(x)*sin(3x)

    -Dan
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