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/dxne0`

Hence if `y=sin^-1x, -1<=x<=1 , -pi/2<=y<=pi/2`

then show that `dy/dx=1/sqrt(1-x^2)`, where  `|x|<1`

 

Answer 1

‘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 `deltax->0, deltay->0`

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

`(deltax)/(deltay)=1/((deltay)/(deltax)), (deltay)/(deltax)ne0`

Taking `lim_(deltax->0)` on both sides, we get

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

Since ‘y’ is a differentiable function of ‘x’,

`lim_(deltax->0)((deltay)/(deltax))`

As `deltax->0,deltay->0`

`lim_(deltay->0)((deltax)/(deltay))=1/(lim_(deltax->0)((deltay)/(deltax)))`        ..................(i)

limits on R.H.S. of (i) exist and are finite. 
Hence, limits on L.H.S. of (i) also should exist and be finite.

`lim_(deltay->0)((deltax)/(deltay))=dx/dy `exists and its finite

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

`y=sin^-1x,-1<=x<=1, -pi/2<=y<=pi/2`

`x=siny`

Differentiating w.r.t. y, we get

`dx/dy=cosy`

`dy/dx=1/cosy`

`dy/dx=1/(+-sqrt(1-sin^2y))`

`dy/dx=1/(+-sqrt(1-x^2))`

since `-pi/2<=y<=pi/2`, y lies in I or IV quadrant.

cosy is positive

`dy/dx=1/sqrt(1-x^2), |x|<1`