derivatives

[Tueseve728]

did i do these right?

x sin^-1 y=1+x^2
these are my steps:
y=sin((x^2 +1)/x)
cos ((x^2 +1)/x) x d/dx((x^2 +1)/x)-(x^2 -1) d/dx(x)) (whole thing underlined by x^2)
cos((x^2 +1)/x)(x(d/dx(1)+d/dx(x^2))-(x^2 +1) d/dx(x)) (whole thing underlined by x^2)
cos((x^2 +1)(-x^2+(2x+d/dx(1))x-1) (whole thing underlined by x^2)
(x^2 -1) cos ((x^2 + 1)/x) (whole thing underlined by x^2)
(x^2 - 1) cos(x+(1/x)) (whole thing underlined by x^2)

y=x^x
ln y=x lnx
=ln x +x(1/x)=ln x +1
=(ln x +1)y
=(ln x +1)^x

y=ln(2^(tan x))
=(1/2^(tan x))(ln 2)(sec x^2)

 

[stapel]

In the future, please post follow-ups and replies within the original threads. Thank you.

2) xsin^-1(y) = 1 + x²

Yes, you correctly copied and simplified the solution that was provided for you.

3) y = x^x

How did you get that ln(y) = ln(x) + 1? Please clarify your steps. It might help if you started from the nearly-complete solution provided by the tutor.

1) y = ln(2^tan(x))

Please review the complete solution provided by the tutor; it appears you have dropped a factor.

Eliz.

 

[Tueseve728]

i dont understand how to do #3 can u give me a hint? my profess says take the ln of both sides, and dont use the power rule.

 

[stapel]

i dont understand how to do #3 can u give me a hint? my profess says take the ln of both sides, and dont use the power rule.

Please review the worked solution provided in the other thread. The tutor followed your professor's suggestion.

Thank you.

Eliz.