find dy/dx if x=tan(x+y)
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Jan 8, 2006 #2 d/dx[x=tan(x+y)]\displaystyle d/dx[x = tan(x+y)]d/dx[x=tan(x+y)] 1=sec2(x+y)(1+dy/dx)\displaystyle 1 = sec^2(x+y)(1 + dy/dx)1=sec2(x+y)(1+dy/dx) 1=sec2(x+y)+sec2(x+y)∗(dy/dx)\displaystyle 1 = sec^2(x+y) + sec^2(x+y)*(dy/dx)1=sec2(x+y)+sec2(x+y)∗(dy/dx) can you finish?
d/dx[x=tan(x+y)]\displaystyle d/dx[x = tan(x+y)]d/dx[x=tan(x+y)] 1=sec2(x+y)(1+dy/dx)\displaystyle 1 = sec^2(x+y)(1 + dy/dx)1=sec2(x+y)(1+dy/dx) 1=sec2(x+y)+sec2(x+y)∗(dy/dx)\displaystyle 1 = sec^2(x+y) + sec^2(x+y)*(dy/dx)1=sec2(x+y)+sec2(x+y)∗(dy/dx) can you finish?
