if tan(x+y)=x, then dy/dx=?

Question

A. tan^2(x+y) B. sec^2(x+y) C. ln(abs(sec(x+y))) D. sin^2(x+y)-1 E. cos^2(x+y)-1

Answer 1

first u should know that dy/dx(tanx)=sec²x ,1/secx =cosx
and the derivative of y = dy/dx
(if y² so the derivative 2ydy/dx)
tan (x+y)=x

sec²(x+y)*(1+dy/dx) = 1 just divided both side by [sec²(x+y)]

(1+dy/dx)= 1/ [sec²(x+y)]

1+ dy/dx = cos²(x+y)

dy/dx =cos²(x+y) -1

Answer 2

Taking derivative
1/cos^2(x+y) *(1+y´)=1
so y´= cos^2(x+y) -1 (E)

Answer 3

answer: E. cos^2(x+y)-1
(1+y')(1+tan^2(x+y))=1

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

hint:
sin^2 (x+y) + cos^2 (x+y)=1

Answer 4

lny = tanx lnx -----> y ' / y = sec^2x lnx + (tanx)/x -----> y ' = (sec^2x lnx + (tanx)/x) y = [sec^2x lnx + (tanx)/x] x^tanx