Find the derivative of y = coth^-1(sec(x)). SImplify if possible.

Question

Find the derivative of y = coth^-1 (sec(x)). Simplify if possible.

 

I applied the Chain Rule.

 

-sech(x)tanh(x)

_____________

(1 - sec(x)²)^1/2

 

I don't know where to go from here.

 

#45, p 263

ANSWER:   y' = csc(x)

Answer 1

Hey Mia!

 

I think Hilton's answer might be wrong actually :(

 

 

  (d/dx coth^-1)           (d/dx sec(x) -- we find the derivative of sec due to the Chain Rule)

1/[1 - sec² (x)]   *     sec(x) tan(x)

 

 

sec(x)tan(x)                        

___________            

1 - sec² (x)

 

 

Then we substitute tan and sec's cosine and sine values

 

 

sin(x)

___________________________________

[1 -  1/cos² (x)] cos² (x)

 

 

When you distribute the cos(x)² in the denominator throughout the terms in the parenthesis, you get:

 

 

sin(x)

____________________________

cos²(x) * 1  -  cos²(x) * 1/cos²(x)

 

 

 

sin(x)
____________________________
cos² (x)  - cos² (x)/cos² (x)

 

 

 

sin(x)
______________________
cos² (x) - cos² (x)/cos² (x)

 

 

 

sin(x)
___________
cos² (x) - 1

 

 

 

sin(x)

__________

sin²(x)                         <----------    Remember our properties of sine!   sin² (x) + cos² (x) = 1  so I just subbed in sin²

 

 

 

1/sin(x)    =    csc(x)

 

 

 

Answer 2

Firstly,  coth-1 (sec(x) is not the same as sec x / coth x.

 

Find derivative of y = coth-1 (sec(x))

The derivative of coth-1(x) = 1/(1-x^2)

Let g = sec x        dg/dx = sec x tan x

y = coth-1(g)

dy/dx = 1/(1-g^2). dg/dx = sec x tan x . 1/(1-g^2)

          = (sec x tan x) /(1-(sec x)^2)

          = sec x tan x / -( tan x ) ^2 = - sec x/ tanx

          = - 1/ cos x tan x = - 1/ sin x = - csc x