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Equation of a tangent line given x

  1. Jan 29, 2009 #1
    I am having a time trying to figure out this problem given only an x value. The equation is "Find the equation of the tangent line to y= Square root of x^4-2337 .....at x=7. If anybody could help me solve this it would be a great help. Thanks
  2. jcsd
  3. Jan 30, 2009 #2


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    Welcome to PF!

    Hi Drewnlauren06! Welcome to PF! :smile:

    Do you know how to find the slope (the y/x) of the tangent line?​
  4. Feb 2, 2009 #3
    I know to find Y you set x=0 I believe and I know the slope formula is y2-y1/x2-x1. I don't exactly remember how to pull everything together.
  5. Feb 2, 2009 #4


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    yes … y2-y1/x2-x1 is the slope of a line, if you already know two points that it goes through …

    but you only know one point …

    have you done calculus? do you know what the derivative of √(x^4-2337) is? :smile:
  6. Feb 3, 2009 #5
    I'm in Business calculus now. If I was going to take the derivative of √(x^4-2337). I would get rid of the square root by (x^4-2337)^1/2 and the derivative would be 1/2(x^4-2337)^-1/2 or would it be [2x^3-(2337/2)]^-1/2 or neither I don't know. I know how to do basic derivatives with no problem but this one is just giving me problems.
  7. Feb 3, 2009 #6


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    Hi Drewnlauren06! :smile:

    (have a square-root: √ and use the X2 tag, just above the reply box :wink:)
    That's right :smile:
    ah, you need to revise the chain rule

    the derivative of √(f(x)) is f'(x) times 1/2√(f(x)) …

    so use that first formula of yours, but multiply it by the derivative of x4 - 2337 :wink:
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