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hewsmike
 one year ago
Best ResponseYou've already chosen the best response.0I 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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