T tfinch2 Lifer Feb 3, 2004 22,114 1 0 Oct 18, 2005 #1 Find the derivative new problem: y = (tan x)^(1/x)
T th3dumbguy Senior member Feb 4, 2005 242 0 0 Oct 18, 2005 #6 y' = (dy/dx) [cot (sin x)]^2 y' = 2cot(sin x) * (dy/dx) [cot (sin x)] y' = 2cot(sin x) * -csc^2 (sin x) * (dy/dx) sin(x) y' = 2cot(sin x) * -csc^2 (sin x) * cos (x) i think im right...
y' = (dy/dx) [cot (sin x)]^2 y' = 2cot(sin x) * (dy/dx) [cot (sin x)] y' = 2cot(sin x) * -csc^2 (sin x) * (dy/dx) sin(x) y' = 2cot(sin x) * -csc^2 (sin x) * cos (x) i think im right...
G Goosemaster Lifer Apr 10, 2001 48,775 3 81 Oct 18, 2005 #7 Originally posted by: IAteYourMother -2Cos[x]*Cot[Sin[x]]*Csc[Sin[x]]^2 Click to expand... :thumbsup:
Originally posted by: IAteYourMother -2Cos[x]*Cot[Sin[x]]*Csc[Sin[x]]^2 Click to expand... :thumbsup:
R Rip the Jacker Diamond Member Dec 29, 2004 5,415 1 76 Oct 23, 2005 #11 Originally posted by: PurdueRy No effort from you = no effort from me Click to expand... LOL.
T tfinch2 Lifer Feb 3, 2004 22,114 1 0 Oct 23, 2005 #12 Originally posted by: PurdueRy No effort from you = no effort from me Click to expand... I know that you take the natural log of each side... ln y = ln (tan x)^(1/x) 1/y dy/dx = (1/x) ln (tan x) <- not sure if that is right so that's where i get stuck
Originally posted by: PurdueRy No effort from you = no effort from me Click to expand... I know that you take the natural log of each side... ln y = ln (tan x)^(1/x) 1/y dy/dx = (1/x) ln (tan x) <- not sure if that is right so that's where i get stuck
P PurdueRy Lifer Nov 12, 2004 13,837 4 0 Oct 23, 2005 #13 ok because I am in a good mood today.... Where you left off dy/dx = ((1/x)(1/tan(x))(sec^2(x)) + ln(tan x)*(-1/x^2)) * y where y = (tan x)^(1/x) sooo... dy/dx = ((1/x)(1/tan(x))(sec^2(x)) + ln(tan x)*(-1/x^2)) * (tan x)^(1/x) checked for accuracy and it works
ok because I am in a good mood today.... Where you left off dy/dx = ((1/x)(1/tan(x))(sec^2(x)) + ln(tan x)*(-1/x^2)) * y where y = (tan x)^(1/x) sooo... dy/dx = ((1/x)(1/tan(x))(sec^2(x)) + ln(tan x)*(-1/x^2)) * (tan x)^(1/x) checked for accuracy and it works
S SilentZero Diamond Member Apr 8, 2003 5,158 0 76 Oct 23, 2005 #14 Wow im surprised people are actually doing the work for you. This is not like ATOT'ers at all!