Question

What is the implicit derivative of `1= e^y-xcosy`?

Answer 1

`y' = (cosy)/ (e^y + x sin y)`

Explanation:

`(1= e^y-xcosy)'`

`\implies 0= e^y \ y' - cosy - x (-sin y) \ y'`

`\implies 0= e^y \ y' - cosy + x sin y \ y'`

`\implies 0= \ y' (e^y + x sin y) - cosy`

`\implies \ y' = (cosy)/ (e^y + x sin y)`