Question

If y = f(x) is a differentiable function of x such that inverse 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≠0`

 

Answer 1

Let `delta`y be the increment in y corresponding to an increment `delta`x in x.

as `deltax->0,deltay->0`

Now y is a differentiable function of x.

`therefore lim_(deltax->0)(deltay)/(deltax)=dy/dx`

Now `(deltay)/(deltax)xx(deltax)/(deltay)=1`

`therefore (deltax)/(deltay)=1/((deltay)/(deltax))`

Taking limits on both sides as `deltax->0, we get`

`lim_(deltax->0)(deltax)/(deltay)=lim_(deltax->0)[1/((deltay)/(deltax))]=1/(lim_(dx->0)(deltay)/(deltax))`

`lim_(deltax->0)(deltax)/(deltay)=1/(lim_(dx->0)(deltay)/(deltax))`   ....[as `deltax->0,deltay->0`]

Since limit in R.H.S. exists 

limit in L.H.S. also exists and we have,

`lim_(deltay->0)(deltax)/(deltay)=dx/dy`

`dx/dy=1/(dy/dx)`, where `dy/dxne0`

Let `y=tan^-1x`

`x=tany=>cosy=1/sqrt(1+tan^2y)=1/sqrt(1+x^2)`

`therefore sec^y.dy/dx=1=>dx/dy=sec^2y`

`dy/dx=1/(dx/dy)=1/sec^2y=cos^2y=>dy/dx=cos^y`

`(d(tan^-1x))/dx=cos^2y=(cosy)^2=(1/sqrt(1+x^2))^2`

`therefore d/dx(tan^-1x)=1/(1+x^2)`

Answer 2

'y’ is a differentiable function of ‘x’.

Let there be a small change δx in the value of ‘x’.

Correspondingly, there should be a small change δy in the value of ‘y’.

As δx → 0, δy → 0