Question

How do you implicitly differentiate `2=e^(x-y)cosy`?

Answer 1

`dy/dx=(cosy)/(siny+cosy)`

Explanation:

`"differentiate "e^(x-y)cosy" using the "color(blue)"product rule"`

`0=e^(x-y)(-siny)dy/dx+cosye^(x-y)(1-dy/dx)`

`color(white)(0)=-e^(x-y)sinydy/dx+cosye^(x-y)-cosye^(x-y)dy/dx`

`rArrdy/dx(-e^(x-y)siny-cosye^(x-y))=-cosye^(x-y)`

`rArrdy/dx=(-cosye^(x-y))/(-e^(x-y)(siny+cosy))=(cosy)/(siny+cosy)`