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Using the Logarithmic Differentiation find the derivative of $y=\sqrt{x(x-1)/(x-2)}$...so I tried,but the result is not correct..can you show me a hint?

so $\ln y= 0.5\ln[x(x-1)/(x-2)]$

$\ln y=0.5[\ln x+\ln(x-1)-\ln(x-2)]$

$\ln y=0.5\ln x+0.5\ln(x-1)+0.5\ln(x-2)$

$y'=0.5\ln x+0.5\ln(x-1)-0.5\ln(x-2)[\sqrt{x(x-1)/(x-2)}]$

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    Your last line should be $y'=\frac 12 \left(\frac 1x+\frac 1{x-1}-\frac 1{x-2}\right)\sqrt{x(x-1)/(x-2)}$ (you forgot to differentiate the $\ln$ terms and the parenthesis).2012-11-20

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Hint: Remember that the derivative of $\ln y$ is $\frac{y'}{y}$, by the chain rule. Also, you seem to have flipped a sign in your last step; the last term of the third line should be $-0.5 \ln(x-2)$.

Edit: After seeing what you did, you forgot to differentiate the terms with $x$! Suppose we have $\ln y = g(x)$. Then, differentiating both sides, $\frac{y'}{y} = g'(x)$, or $y'(x) = g'(x) y(x)$. You forgot to differentiate $g$, which in this case is $g(x) = \frac12[\ln x + \ln(x-1) - \ln(x-2)]$.

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    wait,i will edit it..2012-11-20
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Your error is that you didn't actually differentiate $\ln x$ and the other logarithms on the right side. $ \frac{d}{dx} (0.5\ln x) = 0.5\cdot\frac1x = \frac{0.5}x. $