find the 2nd derivative of y= loge (x/2)
simDS Member Joined Dec 20, 2007 Messages 34 Gender Female HSC 2009 Jul 25, 2009 #1 find the 2nd derivative of y= loge (x/2)
kaz1 et tu Joined Mar 6, 2007 Messages 6,960 Location Vespucci Beach Gender Undisclosed HSC 2009 Uni Grad 2018 Jul 25, 2009 #2 1st derivative= 1/x 2nd= -1/x2
simDS Member Joined Dec 20, 2007 Messages 34 Gender Female HSC 2009 Jul 25, 2009 #3 kaz1 said: 1st derivative= 1/x 2nd= -1/x2 Click to expand... what happens to the loge 2 how do u derive thaT?
kaz1 said: 1st derivative= 1/x 2nd= -1/x2 Click to expand... what happens to the loge 2 how do u derive thaT?
kaz1 et tu Joined Mar 6, 2007 Messages 6,960 Location Vespucci Beach Gender Undisclosed HSC 2009 Uni Grad 2018 Jul 25, 2009 #4 Derivative of a loge is f'(x)/f(x) so the first derivative is (1/2)/(x/2) which simplifies down to 1/x.
Derivative of a loge is f'(x)/f(x) so the first derivative is (1/2)/(x/2) which simplifies down to 1/x.
oly1991 Member Joined Nov 20, 2008 Messages 411 Location Sydney Gender Male HSC 2009 Jul 25, 2009 #5 simDS said: what happens to the loge 2 how do u derive thaT? Click to expand... the derivative of (x/2) is 1/2. so u put that on top of x/2 and it ends up being 1/x.
simDS said: what happens to the loge 2 how do u derive thaT? Click to expand... the derivative of (x/2) is 1/2. so u put that on top of x/2 and it ends up being 1/x.
T Timothy.Siu Prophet 9 Joined Aug 6, 2008 Messages 3,449 Location Sydney Gender Male HSC 2009 Jul 25, 2009 #6 or just do y=ln (x/2)=ln x - ln2 y'=1/x etc.