Calculus Help

[tfinch2]

Find the derivative new problem:

y = (tan x)^(1/x)

 

[nycxandy]

Textbook.

 

[Goosemaster]

chain rule

 

[Legendary]

Product rule + chain rule

 

[IAteYourMother]

-2Cos[x]*Cot[Sin[x]]*Csc[Sin[x]]^2

 

[th3dumbguy]

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...

 

[Goosemaster]

Originally posted by: IAteYourMother
-2Cos[x]*Cot[Sin[x]]*Csc[Sin[x]]^2

:thumbsup:

 

[th3dumbguy]

dam crap he beat me..

 

[tfinch2]

bump

 

[PurdueRy]

No effort from you = no effort from me

 

[Rip the Jacker]

Originally posted by: PurdueRy
No effort from you = no effort from me

LOL.

 

[tfinch2]

Originally posted by: PurdueRy
No effort from you = no effort from me

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

 

[PurdueRy]

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

 

[SilentZero]

Wow im surprised people are actually doing the work for you. This is not like ATOT'ers at all!