Define a function h : [0, 1] ? [-1, 1] as h(x) = sin p(2x^2-1/2)\nUsing Inverse function Theorem, find d/dx h inverse at x = 0 , x =-1/root 2.\nHint: Show that h(x) is a bijective function.

As x belongs to [0,1], (2x^2 - 1)/2 goes from [-1,1].
Take f(x) = f(y) for some x,y in domain,
sin pi(2x^2-1/2) =sin p(2y^2-1/2)
It implies that 2x^2 - 1 / 2 = 2y^2 - 1 / 2
THerefore, x = y
Also, for all b in range, there is some a in the domain.
Hence, h(x) is a bijective function.
d/dx h^-1(x) = 1/ (d/dx h(x) ) = pi(2x)/ cos pi(2x^2-1/2)
At x=0,
Answer = 0
At x=-1/v2
Answer = -v2 pi