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What would be the first derivative of f(x)=x-6/x+2
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\[f(x)\frac{ x-6 }{ x+2 }\]
Apply the quotient rule: \[\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\] Letting f(x) = x-6 and g(x) = x+2 you have: f'(x) = 1, g'(x) = 1 \[\implies \frac{d}{dx}\frac{x-6}{x+2} = \frac{1(x+2)-(x-6)(1)}{(x+2)^2}=\frac{8}{(x+2)^2}\]
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