y=tan^(-1)[(ax-b)/(bx+alpha)](dy)/(dx)=0

y= tan^(-1) ((ax-b)/(bx + a)) " then " (dy)/(dx)|_(x= -1) = ………

If y=tan^(-1)[(ax-b)/(bx+a)] then prove that (dy)/(dx)=(1)/(1+x^(2))

(d)/(dx)[tan^(-1)((ax-b)/(bx+a))]=

If y=tan ^(-1) ((a+x) /(1-ax) ),then (dy)/(dx) =

If the derivative of tan^(-1)(a+bx) takes the value (dy)/(dx)=1 at x=0 , then prove that b=1+a^(2)

If tan^(-1)(x^(2)+y^(2))=alpha then (dy)/(dx) is equal to

Solve the differential equation: (i) (1+y^(2))+(x-e^( tan ^(-1)y))(dy)/(dx)=0 (ii) x(dy)/(dx)+cos^(2)y=tan y(dy)/(dx)

If tan^-1(x^2+y^2)=alpha,then(dy)/(dx) is equal to

If tan^-1(x^2+y^2)=alpha,then(dy)/(dx) is equal to

y=log(x/(1+bx)) then (dy)/(dx)