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Math Help - equation of tangent

  1. #1
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    equation of tangent

    Given that e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+... and (1+3x)^{-\frac{1}{2}}=1-\frac{3}{2}x+\frac{27}{8}x^2+..., deduce the equation of the tangent to the curve y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: equation of tangent

    Quote Originally Posted by Punch View Post
    e^{tan^{-1}x}=1+x+\frac{1}{2}x^2+...
    Actually, e^x=1+x+\frac{1}{2}x^2+...

    deduce the equation of the tangent to the curve y=\frac{e^{tan^{-1}x}}{\sqrt{1+3x}}
    Write it as y=(1+3x)^{-\frac{1}{2}}e^{tan^{-1}x} and use the product rule. (Make sure you use the correct formula for e^{tan^{-1}x}.)
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  3. #3
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    Re: equation of tangent

    Quote Originally Posted by alexmahone View Post
    Actually, e^x=1+x+\frac{1}{2}x^2+...



    Write it as y=(1+3x)^{-\frac{1}{2}}e^{tan^{-1}x} and use the product rule. (Make sure you use the correct formula for e^{tan^{-1}x}.)
    So i will be able to find the expression of dy/dx. Next, ill need to use y-y1=m(x-x1) but what coordinates do i use for y1 and x1?
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: equation of tangent

    Quote Originally Posted by Punch View Post
    So i will be able to find the expression of dy/dx. Next, ill need to use y-y1=m(x-x1) but what coordinates do i use for y1 and x1?
    y(0) = 1. So you may take y1 = 1 and x1 = 0.
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  5. #5
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    Re: equation of tangent

    Quote Originally Posted by alexmahone View Post
    y(0) = 1. So you may take y1 = 1 and x1 = 0.
    But the equation of the tangent at different point are different. But question did not state which point of the curvethe tangent is required. Can i assume it is asking for the point u stated?
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  6. #6
    MHF Contributor alexmahone's Avatar
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    Re: equation of tangent

    Quote Originally Posted by Punch View Post
    But question did not state which point of the curvethe tangent is required.
    Unless the question states this, it cannot be answered.

    Can i assume it is asking for the point u stated?
    No.
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