# logarithmetic differentiation

• November 3rd 2007, 12:53 PM
SMAlvarez
logarithmetic 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))
• November 3rd 2007, 02:33 PM
Krizalid
Shouldn't it be $\frac{dy}{dx}\,?$

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

Now take the derivative.
• November 3rd 2007, 02:39 PM
Jhevon
Quote:

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

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

Krizalid is correct. but in my experience, logarithmic differentiation is done this way:

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

take the log of both sides

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

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

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
• November 3rd 2007, 04:14 PM
SMAlvarez
so is the final answer :

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

?

How would I start the second one?
• November 3rd 2007, 04:17 PM
Jhevon
Quote:

Originally Posted by SMAlvarez
so is the final answer :

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

?

How would I start the second one?

no. recall that we have ln y on the left hand side. you must differentiate implicitly and simplify accordingly

the second one is done in exactly the same way, start by logging both sides and splitting up the logs