Question

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

Answer 1

`dy/dx = frac{(cos^2y)(e^(x-y))-cosy}{(cos^2y)(e^(x-y))+xsiny}`

Explanation:

`2=e^(x-y)-frac{x}{cosy}`

`0=(1-dy/dx)e^(x-y)-frac{cosy+xsiny(dy/dx)}{cos^2y}`

`0=e^(x-y)-e^(x-y)(dy/dx)-secy-(xsiny)/cos^2y (dy/dx)`

`0=dy/dx(-e^(x-y)-(xsiny)/cos^2y)+e^(x-y)-secy`

`dy/dx(e^(x-y)+(xsiny)/cos^2y)=e^(x-y)-secy`

`dy/dx = frac{e^(x-y)-secy}{(e^(x-y)+(xsiny)/cos^2y)}`

`dy/dx = frac{(cos^2y)(e^(x-y))-cosy}{(cos^2y)(e^(x-y))+xsiny}`

This fraction could be manipulated more to make terms cancel (eg multiplying by the conjugate of the denominator, etc.), but I have isolated `dy/dx`.