Using Chain rule to find derivatives....

[darkknight1]

y = (csc(x) + cot(x) )^-1
Find dy/dx

 

[Monoxdifly]

Suppose u = csc(x) + cot x

dy/dx = dy/du × du/dx

 

[HOI]

y = (csc(x) + cot(x) )^-1
Find dy/dx

darkknight, you titled this "using chain rule" and Monoxdifly just told you what that is! Are you able to do this now?

Letting u= csc(x)+ cot(x), y= u^-1. What is dy/du? What is du/dx?
chain rule: dy/dx= (dy/du)(du/dx).

If the difficulty is du/dx, it might help you to write [math]u= csc(x)+ cot(x)= \frac{1}{sin(x)}+ \frac{cos(x)}{sin(x)}= \frac{1+ cos(x)}{sin(x)}[/math] and use the "quotient rule".