I need to find dx/dy using logarithmetic differentiation. Thanks for any help!!

1. y = sqrt((x-1) (x-2) (x-3))

2. y= sqrt((x^2-2)/(x^2+2))

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- Nov 3rd 2007, 12:53 PMSMAlvarezlogarithmetic differentiation
I need to find dx/dy using logarithmetic differentiation. Thanks for any help!!

1. y = sqrt((x-1) (x-2) (x-3))

2. y= sqrt((x^2-2)/(x^2+2)) - Nov 3rd 2007, 02:33 PMKrizalid
Shouldn't it be $\displaystyle \frac{dy}{dx}\,?$

$\displaystyle y=\sqrt{(x-1)(x-2)(x-3)}=\exp\left(\frac12\ln\Big[(x-1)(x-2)(x-3)\Big]\right).$

Now take the derivative. - Nov 3rd 2007, 02:39 PMJhevon
Krizalid is correct. but in my experience, logarithmic differentiation is done this way:

$\displaystyle y = \sqrt{(x - 1)(x - 2)(x - 3)} = [(x - 1)(x - 2)(x - 3)]^{\frac 12}$

take the log of both sides

$\displaystyle \Rightarrow \ln y = \ln [(x - 1)(x - 2)(x - 3)]^{\frac 12}$

$\displaystyle \Rightarrow \ln y = \frac 12 \left[ \ln (x - 1) + \ln (x - 2) + \ln (x - 3) \right]$

Now differentiate implicitly.

of course Krizalid's method is equivalent, the difference is, you do not have to differentiate implicitly with his method, explicit differentiation works - Nov 3rd 2007, 04:14 PMSMAlvarez
so is the final answer :

1/2[(1/x-1)+(1/x-2)+(1/x-3)]

?

How would I start the second one? - Nov 3rd 2007, 04:17 PMJhevon