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Math Help - tangent to a curve (Derivatives)...

  1. #1
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    tangent to a curve (Derivatives)...

    so i have here

    Find an equation for th tanget to the curve y=x+2/x at the point (1,3)

    i dont understand why/how the slope of the curve is 1-2/x^2



    so what would be the slope of the equ. y=8/x^2+4.....
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  2. #2
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    Quote Originally Posted by cyberdx16 View Post
    so i have here

    Find an equation for th tanget to the curve y=x+2/x at the point (1,3)

    i dont understand why/how the slope of the curve is 1-2/x^2
    The derivative of y=x+2/x=x+2x^(-1)
    Is, (using the power rule)
    y'=1-2x^(-2)=1-2/x^2
    Thus, at x=1 the slope is,
    1-2/(1)^2=1-2=-1
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