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Math Help - This is a pretty straight foward question (Trig.)

  1. #1
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    Post This is a pretty straight foward question (Trig.)

    Find the gradient of the tangent to the curve y = x sin x^2 at the point where x = π/3
    Last edited by sweetG; July 20th 2008 at 09:47 AM.
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  2. #2
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    Hello,

    Quote Originally Posted by sweetG View Post
    Find the gradient of the tangent to the curve y = x sin x2 at the point where x = π/3
    Let f(x)=x \sin x^2

    The gradient of the tangent at the point where x=\frac{\pi}{3} is f'(\pi/3).

    So first of all, calculate f'(x) (applying product rule & chain rule) and then take x=\frac{\pi}3
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  3. #3
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    Umm I'm really bad at differentiating trigonometric functions but could the answer possibly be: f '(x) = sinx^2 + 2x cos x^2
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    Quote Originally Posted by sweetG View Post
    Umm I'm really bad at differentiating trigonometric functions but could the answer possibly be: f '(x) = sinx^2 + 2x cos x^2
    Nearly !!

    (uv)'=u'v+uv'
    (x sin x)'=(x)' sin x + x (sin x)'

    you forgot this red x
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  5. #5
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    f '(x) = sinx^2 + x(2x cos x^2)

    Would that be the correct answer ?

    And well if it is correct does that mean I just substitute x = π/3 into all the x's ?

    Oh I'm so bad at this >_<
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  6. #6
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    Quote Originally Posted by sweetG View Post
    f '(x) = sinx^2 + x(2x cos x^2)

    Would that be the correct answer ?

    And well if it is correct does that mean I just substitute x = π/3 into all the x's ?

    Oh I'm so bad at this >_<
    yes, and yes !

    You seem not to be bad since you know the basic rules This was just a careless mistake ^^ Am I wrong ?
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  7. #7
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    Yup it was a careless mistake
    but am this is the part that confuses me as well
    coz am i supposed to substitute it into every x value there ?

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

    so when x = π/3

    Am I supposed to put x = π/3 in to all x's coz I dunno if its just me but that looks wrong ?! =S I'm so confused

    Where do I sub in π/3 >_< sorry its pretty late here and I'm just not thinking properly at the moment
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  8. #8
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    I'm just saying that well if I substituted π/3 in every single x I saw then that would be wrong, right ?

    Argh I hate trigonometry >_<
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  9. #9
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    Hi,

    Given function is y=xsin(x^2)

    to find the slope=m=y'=differentiate with respect to x by using product rule formula

    y'=sinx^2+xcosx^2.2x

    y'=sinx^2+2x^2cosx^2

    gradient of the give curve at the point x=pi/3

    y'=sin(pi/3)^2+2(pi/3)^3cos(pi/3)^2

    here u can use calculator to get the answer .
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