Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 19

Note:- 1) If y = cosx ; then dy/dx = -sinx
2) If y = e^x ; then dy/dx = e^x
3) If y = u*v ; where both u & v are functions of 'x' , then
dy/dx = u*(dv/dx) + v*(du/dx)
4) If y = sinx ; then dy/dx = cosx
Now, the given function is :-
(e^y)*cosx = 1 + sin(xy)
Differentiating both sides w.r.t 'x' we get
-(e^y)*sinx + {(e^y)*cosx}*(dy/dx) = cos(xy)*[y + x*(dy/dx)]
[{(e^y)*cosx} - x*cos(xy)]*(dy/dx) = [y*cos(xy) + (e^y)*sinx]
or, dy/dx = [y*cos(xy) + (e^y)*sinx]/[{(e^y)*cosx} - x*cos(xy)]