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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`
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Solution
‘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`
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