Question

How do you implicitly differentiate `-1=xy-tan(x+2y)`?

Answer 1

Differentiate both sides of the equation with respect to `x`:

`0 = y+xy'-(1+2y')/cos^2(x+2y)`

solve for `y'`

`0 = (y+xy')cos^2(x+2y)-(1+2y')`

`y'(2-xcos^2(x+2y)) = ycos^2(x+2y)-1`

`y'= (ycos^2(x+2y)-1)/(2-xcos^2(x+2y))`