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    y=\frac{tan3x+cos2x}{tanx-sec3x} this is the question, find the gradient of the curve when x=0 the answer is -4

    i used the quotient rule

    i did

    \frac{(3sec^23x-2sin2x)(tanx-sec3x)-(tan3x+cos2x)(sec^2x-3secxtanx)}{tan^2x-sec^23x}

    i then attempted to cancel

    \frac{(3sec3x-2sin2x)(-3x)-(3x+cos2x)(secx-3)}{tanx-sec3x}

    can some show me a full solution
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    this is wrong in a few places The most obvious way is the (tan(x))^2 - (sec(3x))^2, you do realise how that is wrong, right?

    I would advise doing this again systematically and carefully

    let u = tan(3x) +cos(2x)
    v = ... etc
    finding u' and v'

    and then use the quotient rule.
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    (tanx-sec3x)^2 \ne tan^2x-sec^23x
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    (tanx-sec3x)^2 is not tan^2x-sec^23x

    edit: too late
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    Note:

    If you are only differentiating to find the gradient at x=0, there is no requirement to "tidy up" the derivative. Once you have it in the raw form, just substitute straight away.
 
 
 
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Updated: February 13, 2010
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