if y=(sinx)x+(cosx)tanx,find dy/dx Share with your friends Share 8 Manbar Singh answered this Let sin xx = u and cos xtan x = vthen, y = u + v⇒ dydx = dudx + dvdx ............1Taking u = sin xx⇒ log u = x log sin xdifferentiating both sides with respect to x we get,1u dudx = x ddx log sin x + log sin x ddx x⇒ 1u dudx = x . 1sin x . cos x + log sin x . 1⇒ dudx = u x cot x + log sin x⇒ dudx = sin xx x cot x + log sin xTaking v = cos xtan x⇒ log v = tan x log cos xDifferentiating both sides with respect to x we get, 1v dvdx = tan x ddx log cos x + log cos x ddx tan x⇒ 1v dvdx = tan x . 1cos x - sin x + log cos x . sec2x⇒ dvdx = v sec2 log cos x - tan x sin xcos x⇒ dvdx = v sec2x log cos x - sin2xcos2x⇒ dvdx = cos xtan x sec2x log cos x - tan2xPutting the values of dudx and dvdx in 1 we get,dydx = sin xx x cot x + log sin x + cos xtan x sec2 x log cos x - tan2x 20 View Full Answer