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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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I gave your question a more descriptive title, and added the calculus tag. –  Nate Eldredge Nov 20 '12 at 21:04
Check the plusses and minuses in the third row. Also, it'd be helpful if said what you got when you went on, instead of just saying it's not the same as what you want... –  Micah Nov 20 '12 at 21:21
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). –  Raymond Manzoni Nov 20 '12 at 22:05

2 Answers 2

Your error is that you didn't actually differentiate $\ln x$ and the other logarithms on the right side. $$ \frac{d}{dx} (0.5\ln x) = 0.5\cdot\frac1x = \frac{0.5}x. $$

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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.. –  AAS Nov 20 '12 at 21:21

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