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Math Help - Find the gradient of the normal

  1. #1
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    Apr 2010
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    Find the gradient of the normal

    This question, in a list of past examination questions, has me stuck. It says:


    "For the curve y= \sqrt{\frac{1-x}{1+x } } ,

    Obtain an expression for \frac{dy}{dx }, by first differentiating \frac{1-x}{1+x }.

    Thus, show that the gradient of the normal to the curve at the point (x,y) is given by (1+x)\sqrt{(1-x^2)} ."


    Differentiating \frac{1-x}{1+x } as told, I get \frac{2x}{(1+x)^2}.

    Using this and the chain rule to differentiate y, I get \frac{1}{2}\sqrt{(\frac{1-x}{1+x})}\frac{2x}{(1+x)^2}.

    Some algebraic gymnastics makes this simplify to \frac{x}{(1+x)\sqrt{1-x^2}}, for the gradient of the tangent to the curve...

    But I cannot eliminate the x in the numerator to arrive at the given expression for the gradient of the normal to the curve, even after checking all working carefully. It is produced by the differential to \frac{1-x}{1+x } and that seems correct. Several friends have tried and failed as well. Is there something very obvious that we're missing?
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  2. #2
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    Joined
    Apr 2010
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    Just realized that differentiating (1-x) gives -1, not 1 as I thought. This eliminates the x term in the numerator. The littlest things get you sometimes!
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