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)`.

Explanation

Let `delta x` be an increment in x.
`deltay` be corresponding increment in y.
`:. (delta y)/(delta x)xx(delta x)/(delta y)=1`
`:. (delta x)/(delta y)=(1)/(((delta y)/(delta y))),(delta y)/(delta x) ne 0`
`=(1)/(lim_( delta x to 0)((delta y)/(delta x)))`
As `delta x to 0 rArr delta y to0`
`:.lim_(delta y to 0)(deltax)/(delta y)=(1)/(lim _(delta x to0)((delta y)/(delta x)))`
`:. lim _(delta y to 0) (delta x)/(delta y)=(1)/((dx)/(dx)),(dy)/(dx) ne0`
Since the limit on R.H.S. exists, limit on L.H.S also exists adn is equa to `(dx)/(dy)`
`:. (dx)/(dy)=(1)/((dx)/(dx)),(dy)/(dx) ne0`
`:.x` is a differentiable function of y.
Let `y= tan^(-1)x`
`:. x = tan y`
Differentiating w.r.t. 'x'
`(dx)/(dy)= sec^(2) y= 1+ tan^(2) y= 1+x^(2)`
Now, `(dy)/(dx)=(1)/((dx)/(dy))=(1)/(1+x^(2))`
`:. (d)/(dx)(tan^(-1)x)=(1)/(1+x^(2))`