1. The problem statement, all variables and given/known data Consider the curve satisfying the equation 2(x+1)^(tanx)=(y^2)cosx+y and find dy/dx 2. Relevant equations (tanx)'=sec^2x (lnx)'=1/x 3. The attempt at a solution I've tried taking the natural log of both sides and then taking d/dx of both sides but something seems to go wrong with my algebra each time.. that or I'm getting confused about my natural log / implicit differentiation rules. I'm not actually sure if I'm approaching the problem correctly but here's one of the many attempts I've made: ln2 + tanxln(x+1) = 2lny + lncosx + lny(??) 1/2 + sec^2xln(x+1) + tanx(1/x+1) = 2y'/y + sin(x)/cos(x) + y'/y something should already be wrong so i won't post the rest of the work but i know the answer is supposed to be dy/dx = [2(x+1)^(tanx)[sec^2xln(x+1)+(tanx)/(x+1)]+(y^2)(sinx)]/ all over (2ycosx+1). Need help getting there, thanks!