(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2)),y=0" where "x=1

(1 + x^(2))(dy)/(dx) + 2xy = (1)/(1 + x^(2)), y = 0 when x= 1

(1 + x^(2))(dy)/(dx) + 2xy = (1)/(1 + x^(2)), y = 0 when x= 1

the particular solution of (1 + x^(2))(dy)/(dx) + 2xy = (1)/(1 + x^(2)), y = 0 when x= 1 is

(1+x^(2))(dy)/(dx)+2xy=(1)/(1+x^(2));y=0 if x=1

(1+x^(2)) dy/dx + 2xy = 4x^(2), given y = 0 , when x = 0

(1 + x^(2)) (dy)/(dx) + 2xy = 4x ^(2) , given that y = 0 , when x =0

(dy)/(dx)-(xy)/(1-x^(2))=(1)/(1-x^(2))

If y = f(x) and x = g(y), where g is the inverse of f, i.e., g = f^(-1) and if (dy)/(dx) and (dx)/(dy) both exist and (dx)/(dy) ne 0 , show that (dy)/(dx) = (1)/((dx//dy)) . Hence, (1) find (d)/(dx) (tan^(-1)x) (2) If y=sin^(-1)x, -1lexle1, -(pi)/(2)leyle(pi)/(2) , then show that (dy)/(dx)=(1)/(sqrt(1-x^(2))) where |x| lt 1 .