I'm having a little trouble with finding the derivative of y=csc-1(sec x). Any help?

Question

csc^-1(sec x)=y

Answer 1

y = csc^-1(secx)

 

 

You will need to use the chain rule along with the inverse trigonometric differentiation.

 

Use the inverse trigonometric differentiation to start off.

 

d/dx[csc^-1(x)] = -1 / [x√(x² - 1)]

 

 

Therefore,

 

dy/dx = (secx tanx) / [- secx√(sec²x - 1]

 

dy/dx = (secx tanx) / [- secx√(tan²x)]

 

dy/dx = (secx tanx) / (- secx tanx)

 

dy/x = - 1