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December 21, 2014

December 21, 2014

Posted by **Jesse Stone** on Friday, January 27, 2012 at 9:29am.

NOTE:

that logx is a sub x

- Calculus -
**Reiny**, Friday, January 27, 2012 at 11:21amlet y = log

_{x}(x^2 - 5x + 6)

x^y = x^2 - 5x + 6

ln both sides

ln (x^y) = ln(x^2 - 5x + 6)

y lnx = ln(x^2 - 5x + 6)

differentiate implicityly

y(1/x) + (lnx) dy/dx = (2x-5)/(x^2-5x+6)

dy/dx = ((2x-5)/(x^2-5x+6) - y/x)/lnx

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