If `y=f(x)` is a differentiable function x such that invrse function `x=f^(-1)` y exists, then prove that x is a differentiable function of y and `(dx)/(dy)=(1)/((dy)/(dx))`
where `(dy)/(dx) ne0`
Hence, find `(d)/(dx)(tan^(-1)x)`.

(d)/(dx)[sec(tan^(-1)x)]=

d/dx(tan^(-1)x) is :

If y =f(u) is differentiable function of u, and u=g(x) is a differentiable function of x, then prove that y= f [g(x)] is a differentiable function of x and (dy)/(dx)=(dy)/(du)xx(du)/(dx) .

(d)/(dx)(tan^(-1)(birth x))

If y = f(u) is a differentiable functions of u and u = g(x) is a differentiable functions of x such that the composite functions y = f[g(x) ] is a differentiable functions of x then (dy)/(dx) = (dy)/(du) . (du)/(dx) Hence find (dy)/(dx) if y = sin ^(2)x

If x=f(t), y=g(t) are differentiable functions of parameter 't' then prove that y is a differentiable function of 'x' and (dy)/(dx)=((dy)/(dt))/((dx)/(dt)),(dx)/(dt) ne 0 Hence find (dy)/(dx) if x=a cos t, y= a sin t.

If x and y are differentiable functions of t, then (dy)/(dx)=(dy//dt)/(dx//dt)," if "(dx)/(dt)ne0 .

If y = f(x) is a derivable function of x such that the inverse function x = f^(-1)(y) is defined, then show that (dx)/(dy)=(1)/((dy//dx)) , where (dy)/(dx) ne 0 .

Solve the following differential equation: (dy)/(dx)=tan(x+y)

If x = f(t) and y = g(t) are differentiable functions of t so that y is a differentiable functions of x and if dx//dt ne 0 then (dy)/(dx) =(dy)/((dt)/((dx)/(dt)) . Hence find dy/dx if x= 2sint and y=cos2t