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Thread: Gradient of the curve at a point

  1. #1
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    Gradient of the curve at a point

    Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

    $\displaystyle \frac{dy}{dx} = 2x+x^{-1}$
    $\displaystyle =2(1) + (1)^{-1}$
    $\displaystyle =2+1$
    $\displaystyle = 3$

    I've been told this is incorrect, where did I go wrong?
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  2. #2
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    Quote Originally Posted by jimgreenmore View Post
    Find the gradient of the curve $\displaystyle y = x^2 + \frac{1}{x}$ at the point (1,2)

    $\displaystyle \frac{dy}{dx} = 2x+x^{-1}$
    $\displaystyle =2(1) + (1)^{-1}$
    $\displaystyle =2+1$
    $\displaystyle = 3$

    I've been told this is incorrect, where did I go wrong?
    You derived wrongly the second factor: $\displaystyle (\frac{1}{x})'=-\frac{1}{x^2}$
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  3. #3
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    Quote Originally Posted by felper View Post
    You derived wrongly the second factor: $\displaystyle (\frac{1}{x})'=-\frac{1}{x^2}$
    Okay, so after re-doing it:

    $\displaystyle \frac{dy}{dx} = 2x+-\frac{1}{x^2}$

    $\displaystyle =2(1)+-\frac{1}{(1)^2}$

    $\displaystyle =2+-\frac{1}{1}$

    $\displaystyle =2+-1$

    $\displaystyle =1$

    Is this the answer?
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