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Math Help - Pretty easy trig. question

  1. #1
    Junior Member
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    Post Pretty easy trig. question

    Hi I've already posted this question up although its actually getting pretty urgent that I finish it because its due very soon so could someone please help me. Its pretty easy but for some reason I just can't think straight today

    Find the gradient of the tangent to the curve y = x sin x^2 at the point where x = π/3


    I've alreaf '(x) = sinx^2 + x(2x cos x^2)dy done the first bit and found the gradient which is:

    f '(x) = sinx^2 + x(2x cos x^2)

    but for some strange reason I'm real confused about where to substitute x = π/3 because I see a lot of x's there but I'm not too sure on which ones I should sub this in for. I know that this should be very easy but I just can't think straight today >_<
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  2. #2
    MHF Contributor kalagota's Avatar
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    ..sub it to all x's that you see..
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  3. #3
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    f '(x) = sinx^2 + x(2x cos x^2)

    So would it be:

    When x =π/3

    f '(x) = sin(π^2/9) + π/3(2π/3 cos π^2/9)

    Is that correct ?

    Or have I done something wrong ?
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  4. #4
    MHF Contributor kalagota's Avatar
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    yes..
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