rteng Messages 26 Reaction score 0 Thread starter Oct 28, 2007 #1 using logarithmic differentiation I can get it to a point where I have... ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)] I do not know what to do at that point
using logarithmic differentiation I can get it to a point where I have... ln(y)=20[ln(x+sqrt(x))-ln(x^2-2^x)] I do not know what to do at that point
rocomath Messages 1,752 Reaction score 1 Oct 28, 2007 #2 just take the derivative as you normally would also you will have [tex]\frac{y'}{y}= ...[/tex] bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.
just take the derivative as you normally would also you will have [tex]\frac{y'}{y}= ...[/tex] bring Y to the other side and plug in what Y is. the rest is straight forward differentiation.
rteng Messages 26 Reaction score 0 Oct 28, 2007 #3 yeah but then you have a term that is: [1/(x^2-2^x)]*(2x-d/dy2^x) thats where I get stuck
rocomath Messages 1,752 Reaction score 1 Oct 28, 2007 #4 i'm not really following you mean on the [tex]x^{2}-2^{x}[/tex] part? how would you take it's derivative?
i'm not really following you mean on the [tex]x^{2}-2^{x}[/tex] part? how would you take it's derivative?
rteng Messages 26 Reaction score 0 Oct 28, 2007 #5 [tex]y=((x+sqrt(x))/x^{2}-2^{x})^{20}[/tex] [tex]lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x}))[/tex] [tex]y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)[/tex]
[tex]y=((x+sqrt(x))/x^{2}-2^{x})^{20}[/tex] [tex]lny=20(ln(x+sqrt(x))-ln(x^{2}-2^{x}))[/tex] [tex]y'/y=(20/(x+sqrt(x))*(1+1/2(x)^{-1/2}))-(20/x^{2}-2^{x})*(2x-derivative of 2^x)[/tex]