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grobles27

  • 3 years ago

Find the derivative of y= ln ((square root of 4+x^2)/(x))

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  1. hewsmike
    • 3 years ago
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    I assume that is \[y=ln(\frac{\sqrt{4+x^2}}{x})\]expand\[=ln(\sqrt{4+x^2})-ln(x)=\frac{1}{2}ln(4+x^2)-ln(x)\]now derive\[\frac{dy}{dx}=\frac{1}{2}\frac{1}{(4+x^2)}\frac{d}{dx}(4+x^2)-\frac{1}{x}\]\[=\frac{2x}{2(4+x^2)}-\frac{1}{x}=\frac{x}{(4+x^2)}-\frac{1}{x}\]\[=\frac{x^2-(x^2+4)}{x(x^2+4)} =\frac{-4}{x(x^2+4)}\]

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