`(1+x^(2))(dy)/(dx)+y=tan^(-1)x` को हल कीजिए।

(dy)/(dx)=1+x tan(y-x)

y^(2)-(dy)/(dx)=x^(2)(dy)/(dx) A) y^(-1)+tan^(-1)x=c B) x^(-1)+tan^(-1)y=c C) y+tan^(-1)x=c D) x^(-1)+y^(-1)=tan^(-1)x+c

(dy)/(dx) if a=tan^(-1)(x^(2)+y^(2))

tan ^ (- 1) ((dy) / (dx)) = x + y

(x (dy) / (dx) -y) tan ^ (- 1) ((y) / (x)) = x

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 .

e^(-x) (dy)/(dx) = y(1+ tanx + tan^(2) x)

y=(x^(2)+1)tan^(-l)x find (dy)/(dx)

y = (1 -tan^(2) (x//2))/(1 + tan^(2)(x//2)) , " find " (dy)/(dx)