# What is the derivative dy/dx if sin y = tan x*cos y + sec x*cot y

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### 1 Answer

It is given that sin y = tan x*cos y + sec x*cot y. To determine dy/dx use implicit differentiation for the terms on both the sides of the equation and also use the product rule.

`sin y = tan x*cos y + sec x*cot y`

`(sin y)' = (tan x*cos y + sec x*cot y)'`

=> `cos y*(dy/dx) = (tan x*cos y)' + (sec x*cot y)'`

=> `cos y*(dy/dx) = sec^2 x*cos y + tan x*(-sin y)*(dy/dx) `

`+ sec x*tan x*cos y - sec x*cosec^2 y*(dy/dx)`

=> `(dy/dx)(cos x + tan x*sin y + sec x*cosec^2 y) `

`= sec^2x*cos y + sec x*tan x*cos y`

=> `dy/dx = (sec^2x*cos y + sec x*tan x*cos y)/(cos x + tan x*sin y + sec x*cosec^2y)`

**The required value of `dy/dx = (sec^2x*cos y + sec x*tan x*cos y)/(cos x + tan x*sin y + sec x*cosec^2y)` **