Question

How do you differentiate `tan x + sec y - y = 0`?

Answer 1

`dy/dx = sec^2 x /(1- sec y tan y )`

Explanation:

`d(tan x + sec y - y = 0)`

`implies sec^2 x \ dx + sec y tan y \ dy - dy = 0`

`implies sec^2 x \ dx +dy \ ( sec y tan y -1) = 0`

`implies dy/dx = - sec^2 x /( sec y tan y -1)`

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